Array range queries over range queries
Last Updated :
03 May, 2023
Given an array of size n and a give set of commands of size m. The commands are enumerated from 1 to m. These commands can be of the following two types of commands:
- Type 1 [l r (1 <= l <= r <= n)] : Increase all elements of the array by one, whose indices belongs to the range [l, r]. In these queries of the index is inclusive in the range.
- Type 2 [l r (1 <= l <= r <= m)] : Execute all the commands whose indices are in the range [l, r]. In these queries of the index is inclusive in the range. It's guaranteed that r is strictly less than the enumeration/number of the current command.
Note : The array indexing is from 1 as per the problem statement.
Example 1
Input : 5 5
1 1 2
1 4 5
2 1 2
2 1 3
2 3 4
Output : 7 7 0 7 7
Explanation of Example 1 :
Our array initially is of size 5 whose each element has been initialized to 0.
So now the question states that we have 5 queries for the above example.
- Query 1 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 2 so after the execution of the first our array turns down to be 1 1 0 0 0 .
- Query 2 is of type 1 : As stated above we will simply increment the array indices by 1
the given indices are 4 and 5 so after the execution of the first our array turns down to be 1 1 0 1 1 . - Query 3 is of type 2 : As stated in the definition of this type of query we will execute the queries stated in the range i.e. we will operate the queries instead of the array. The range given is 1 and 2 so we will execute queries 1 and 2 again i.e. we will use repetitive approach for the type 2 queries so we will execute query 1 again and our array will be 2 2 0 1 1. Now when we execute the query we will execute query 2 and our resultant array will be 2 2 0 2 2 .
- Query 4 is of type 2 : As stated in the definition of this type of query we will execute the queries stated in the range i.e. we will operate the queries instead of the array. The range given is 1 and 3 so we will execute queries 1, 2 and 3 again i.e. using repetitive approach queries 1, 2 and 3 will be executed. After the execution of the query 1 again the array will be 3 3 0 2 2 . After the execution of the query 2 again the array will be 3 3 0 3 3 . Now due to query 3 inclusive in the range we will execute query 3 the resultant array will be 4 4 0 4 4 . As explained above.
- Query 5 is of type 2 : The last query will execute the 3rd and 4th query which has been explained above. After the execution of the 3rd query our array will be 5 5 0 5 5 . And after the execution of the 4th query i.e. execution of query 1, 2 and 3 our array will be 7 7 0 7 7 The above is the desired result
Example 2
Input : 1 2
1 1 1
1 1 1
Output : 2
Explanation of the example 2:
Our array initially is of size 1 whose each element has been initialized to 0.
So now the question states that we have 2 queries for the above example.
- Query 1 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 1 so after the execution of the first our array turns down to be 1 .
- Query 2 is of type 1 : As stated above we will simply increment the array indices by 1 the given indices are 1 and 1 so after the execution of the first our array turns down to be 2 . This gives us the desired result
Method 1: This method is the brute force method where by simple recursion is applied on the type 2 queries and for type 1 queries simple increment in the array index is performed.
Implementation:
C++
// CPP program to perform range queries over range
// queries.
#include <bits/stdc++.h>
using namespace std;
// Function to execute type 1 query
void type1(int arr[], int start, int limit)
{
// incrementing the array by 1 for type
// 1 queries
for (int i = start; i <= limit; i++)
arr[i]++;
}
// Function to execute type 2 query
void type2(int arr[], int query[][3], int start, int limit)
{
for (int i = start; i <= limit; i++) {
// If the query is of type 1 function
// call to type 1 query
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
// If the query is of type 2 recursive call
// to type 2 query
else if (query[i][0] == 2)
type2(arr, query, query[i][1], query[i][2]);
}
}
// Driver code
int main()
{
// Input size of array and number of queries
int n = 5, m = 5;
int arr[n + 1];
for (int i = 1; i <= n; i++)
arr[i] = 0;
// Build query matrix
int temp[15] = { 1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 };
int query[5][3];
int j = 0;
for (int i = 1; i <= m; i++) {
query[i][0] = temp[j++];
query[i][1] = temp[j++];
query[i][2] = temp[j++];
}
// Perform queries
for (int i = 1; i <= m; i++)
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
else if (query[i][0] == 2)
type2(arr, query, query[i][1], query[i][2]);
// printing the result
for (int i = 1; i <= n; i++)
cout << arr[i] << " ";
return 0;
}
// Java program to perform range queries
// over range queries.
import java.util.*;
class GFG
{
// Function to execute type 1 query
static void type1(int[] arr, int start, int limit)
{
// incrementing the array by 1 for type
// 1 queries
for (int i = start; i <= limit; i++)
arr[i]++;
}
// Function to execute type 2 query
static void type2(int[] arr, int[][] query,
int start, int limit)
{
for (int i = start; i <= limit; i++)
{
// If the query is of type 1 function
// call to type 1 query
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
// If the query is of type 2 recursive call
// to type 2 query
else if (query[i][0] == 2)
type2(arr, query, query[i][1],
query[i][2]);
}
}
// Driver Code
public static void main(String[] args)
{
// Input size of array and number of queries
int n = 5, m = 5;
int[] arr = new int[n + 1];
// Build query matrix
int[] temp = { 1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 };
int[][] query = new int[6][4];
int j = 0;
for (int i = 1; i <= m; i++)
{
query[i][0] = temp[j++];
query[i][1] = temp[j++];
query[i][2] = temp[j++];
}
// Perform queries
for (int i = 1; i <= m; i++)
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
else if (query[i][0] == 2)
type2(arr, query, query[i][1],
query[i][2]);
// printing the result
for (int i = 1; i <= n; i++)
System.out.print(arr[i] + " ");
System.out.println();
}
}
// This code is contributed by
// sanjeev2552
# Python3 program to perform range
# queries over range queries.
# Function to execute type 1 query
def type1(arr, start, limit):
# incrementing the array by 1
# for type 1 queries
for i in range(start, limit + 1):
arr[i] += 1
# Function to execute type 2 query
def type2(arr, query, start, limit):
for i in range(start, limit + 1):
# If the query is of type 1
# function call to type 1 query
if (query[i][0] == 1):
type1(arr, query[i][1], query[i][2])
# If the query is of type 2
# recursive call to type 2 query
else if (query[i][0] == 2):
type2(arr, query, query[i][1],
query[i][2])
# Driver code
# Input size of array and
# number of queries
n = 5
m = 5
arr = [0 for i in range(n + 1)]
# Build query matrix
temp = [1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 ]
query = [[0 for i in range(3)]
for j in range(6)]
j = 0
for i in range(1, m + 1):
query[i][0] = temp[j]
j += 1
query[i][1] = temp[j]
j += 1
query[i][2] = temp[j]
j += 1
# Perform queries
for i in range(1, m + 1):
if (query[i][0] == 1):
type1(arr, query[i][1],
query[i][2])
else if (query[i][0] == 2):
type2(arr, query, query[i][1],
query[i][2])
# printing the result
for i in range(1, n + 1):
print(arr[i], end = " ")
# This code is contributed
# by mohit kumar
// C# program to perform range queries
// over range queries.
using System;
class GFG
{
// Function to execute type 1 query
static void type1(int[] arr, int start, int limit)
{
// incrementing the array by 1 for type
// 1 queries
for (int i = start; i <= limit; i++)
arr[i]++;
}
// Function to execute type 2 query
static void type2(int[] arr, int[,] query,
int start, int limit)
{
for (int i = start; i <= limit; i++)
{
// If the query is of type 1 function
// call to type 1 query
if (query[i, 0] == 1)
type1(arr, query[i,1], query[i,2]);
// If the query is of type 2 recursive call
// to type 2 query
else if (query[i, 0] == 2)
type2(arr, query, query[i, 1],
query[i, 2]);
}
}
// Driver Code
public static void Main()
{
// Input size of array and number of queries
int n = 5, m = 5;
int[] arr = new int[n + 1];
// Build query matrix
int[] temp = { 1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 };
int[,] query = new int[6,4];
int j = 0;
for (int i = 1; i <= m; i++)
{
query[i, 0] = temp[j++];
query[i, 1] = temp[j++];
query[i, 2] = temp[j++];
}
// Perform queries
for (int i = 1; i <= m; i++)
if (query[i, 0] == 1)
type1(arr, query[i, 1], query[i, 2]);
else if (query[i, 0] == 2)
type2(arr, query, query[i, 1],
query[i, 2]);
// printing the result
for (int i = 1; i <= n; i++)
Console.Write(arr[i] + " ");
Console.WriteLine();
}
}
// This code is contributed by AbhiThakur
<script>
// Javascript program to perform range queries
// over range queries.
// Function to execute type 1 query
function type1(arr,start,limit)
{
// incrementing the array by 1 for type
// 1 queries
for (let i = start; i <= limit; i++)
arr[i]++;
}
// Function to execute type 2 query
function type2(arr,query,start,limit)
{
for (let i = start; i <= limit; i++)
{
// If the query is of type 1 function
// call to type 1 query
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
// If the query is of type 2 recursive call
// to type 2 query
else if (query[i][0] == 2)
type2(arr, query, query[i][1],
query[i][2]);
}
}
// Driver Code
// Input size of array and number of queries
let n = 5, m = 5;
let arr = new Array(n + 1);
for(let i=0;i<arr.length;i++)
{
arr[i]=0;
}
// Build query matrix
let temp = [ 1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 ];
let query = new Array(6);
for(let i=0;i<6;i++)
{
query[i]=new Array(4);
for(let j=0;j<4;j++)
{
query[i][j]=0;
}
}
let j = 0;
for (let i = 1; i <= m; i++)
{
query[i][0] = temp[j++];
query[i][1] = temp[j++];
query[i][2] = temp[j++];
}
// Perform queries
for (let i = 1; i <= m; i++)
if (query[i][0] == 1)
type1(arr, query[i][1], query[i][2]);
else if (query[i][0] == 2)
type2(arr, query, query[i][1],
query[i][2]);
// printing the result
for (let i = 1; i <= n; i++)
document.write(arr[i] + " ");
document.write("<br>");
// This code is contributed by patel2127
</script>
The Time complexity of the above code is O(2 ^ m)
The space complexity of this program is O(m), where m is the number of queries.
Method 2: In this method we use an extra array for creating the record array to find the number of time a particular query is being executed and after creating the record array we simply execute the queries of type 1 and the contains of the record array is simply added to the main array the and this would give us the resultant array.
Implementation:
C++
// CPP program to perform range queries over range
// queries.
#include <bits/stdc++.h>
using namespace std;
// Function to create the record array
void record_sum(int record[], int l, int r,
int n, int adder)
{
for (int i = l; i <= r; i++)
record[i] += adder;
}
// Driver Code
int main()
{
int n = 5, m = 5;
int arr[n];
// Build query matrix
memset(arr, 0, sizeof arr);
int query[5][3] = { { 1, 1, 2 }, { 1, 4, 5 },
{ 2, 1, 2 }, { 2, 1, 3 },
{ 2, 3, 4 } };
int record[m];
memset(record, 0, sizeof record);
for (int i = m - 1; i >= 0; i--) {
// If query is of type 2 then function
// call to record_sum
if (query[i][0] == 2)
record_sum(record, query[i][1] - 1,
query[i][2] - 1, m, record[i] + 1);
// If query is of type 1 then simply add
// 1 to the record array
else
record_sum(record, i, i, m, 1);
}
// for type 1 queries adding the contains of
// record array to the main array record array
for (int i = 0; i < m; i++) {
if (query[i][0] == 1)
record_sum(arr, query[i][1] - 1,
query[i][2] - 1, n, record[i]);
}
// printing the array
for (int i = 0; i < n; i++)
cout << arr[i] << ' ';
return 0;
}
// Java program to perform range queries
// over range queries.
import java.util.Arrays;
class GFG
{
// Function to create the record array
static void record_sum(int record[], int l,
int r, int n, int adder)
{
for (int i = l; i <= r; i++)
{
record[i] += adder;
}
}
// Driver Code
public static void main(String[] args)
{
int n = 5, m = 5;
int arr[] = new int[n];
// Build query matrix
Arrays.fill(arr, 0);
int query[][] = {{1, 1, 2}, {1, 4, 5},
{2, 1, 2}, {2, 1, 3},
{2, 3, 4}};
int record[] = new int[m];
Arrays.fill(record, 0);
for (int i = m - 1; i >= 0; i--)
{
// If query is of type 2 then function
// call to record_sum
if (query[i][0] == 2)
{
record_sum(record, query[i][1] - 1,
query[i][2] - 1, m,
record[i] + 1);
}
// If query is of type 1 then
// simply add 1 to the record array
else
{
record_sum(record, i, i, m, 1);
}
}
// for type 1 queries adding the contains of
// record array to the main array record array
for (int i = 0; i < m; i++)
{
if (query[i][0] == 1)
{
record_sum(arr, query[i][1] - 1,
query[i][2] - 1,
n, record[i]);
}
}
// printing the array
for (int i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
}
}
// This code is contributed
// by Princi Singh
# Python3 program to perform range queries over range
# queries.
# Function to create the record array
def record_sum(record, l, r, n, adder):
for i in range(l, r + 1):
record[i] += adder
# Driver Code
n = 5
m = 5
arr = [0]*n
# Build query matrix
query = [[1, 1, 2 ],[ 1, 4, 5 ],[2, 1, 2 ],
[ 2, 1, 3 ],[ 2, 3, 4]]
record = [0]*m
for i in range(m - 1, -1, -1):
# If query is of type 2 then function
# call to record_sum
if (query[i][0] == 2):
record_sum(record, query[i][1] - 1,
query[i][2] - 1, m, record[i] + 1)
# If query is of type 1 then simply add
# 1 to the record array
else:
record_sum(record, i, i, m, 1)
# for type 1 queries adding the contains of
# record array to the main array record array
for i in range(m):
if (query[i][0] == 1):
record_sum(arr, query[i][1] - 1,
query[i][2] - 1, n, record[i])
# printing the array
for i in range(n):
print(arr[i], end=' ')
# This code is contributed by shubhamsingh10
// C# program to perform range queries
// over range queries.
using System;
class GFG
{
// Function to create the record array
static void record_sum(int []record, int l,
int r, int n, int adder)
{
for (int i = l; i <= r; i++)
{
record[i] += adder;
}
}
// Driver Code
public static void Main(String[] args)
{
int n = 5, m = 5;
int []arr = new int[n];
// Build query matrix
int [,]query = {{1, 1, 2}, {1, 4, 5},
{2, 1, 2}, {2, 1, 3},
{2, 3, 4}};
int []record = new int[m];
for (int i = m - 1; i >= 0; i--)
{
// If query is of type 2 then function
// call to record_sum
if (query[i,0] == 2)
{
record_sum(record, query[i,1] - 1,
query[i,2] - 1, m,
record[i] + 1);
}
// If query is of type 1 then
// simply add 1 to the record array
else
{
record_sum(record, i, i, m, 1);
}
}
// for type 1 queries adding the contains of
// record array to the main array record array
for (int i = 0; i < m; i++)
{
if (query[i, 0] == 1)
{
record_sum(arr, query[i, 1] - 1,
query[i, 2] - 1,
n, record[i]);
}
}
// printing the array
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program to perform range queries
// over range queries.
// Function to create the record array
function record_sum(record,l,r,n,adder)
{
for (let i = l; i <= r; i++)
{
record[i] += adder;
}
}
// Driver Code
let n = 5, m = 5;
let arr = new Array(n);
// Build query matrix
for(let i=0;i<arr.length;i++)
{
arr[i]=0;
}
let query = [[1, 1, 2], [1, 4, 5],
[2, 1, 2], [2, 1, 3],
[2, 3, 4]];
let record = new Array(m);
for(let i=0;i<record.length;i++)
{
record[i]=0;
}
for (let i = m - 1; i >= 0; i--)
{
// If query is of type 2 then function
// call to record_sum
if (query[i][0] == 2)
{
record_sum(record, query[i][1] - 1,
query[i][2] - 1, m,
record[i] + 1);
}
// If query is of type 1 then
// simply add 1 to the record array
else
{
record_sum(record, i, i, m, 1);
}
}
// for type 1 queries adding the contains of
// record array to the main array record array
for (let i = 0; i < m; i++)
{
if (query[i][0] == 1)
{
record_sum(arr, query[i][1] - 1,
query[i][2] - 1,
n, record[i]);
}
}
// printing the array
for (let i = 0; i < n; i++)
{
document.write(arr[i] + " ");
}
// This code is contributed by unknown2108
</script>
The Time complexity of the above code is O(n^2)
Auxiliary Space: O(m)
Method 3: This method has been made more efficient by applying square root decomposition to the record array.
Implementation:
C++
// CPP program to perform range queries over range
// queries.
#include <bits/stdc++.h>
#define max 10000
using namespace std;
// For prefix sum array
void update(int arr[], int l)
{
arr[l] += arr[l - 1];
}
// This function is used to apply square root
// decomposition in the record array
void record_func(int block_size, int block[],
int record[], int l, int r, int value)
{
// traversing first block in range
while (l < r && l % block_size != 0 && l != 0) {
record[l] += value;
l++;
}
// traversing completely overlapped blocks in range
while (l + block_size <= r + 1) {
block[l / block_size] += value;
l += block_size;
}
// traversing last block in range
while (l <= r) {
record[l] += value;
l++;
}
}
// Function to print the resultant array
void print(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver code
int main()
{
int n = 5, m = 5;
int arr[n], record[m];
int block_size = sqrt(m);
int block[max];
int command[5][3] = { { 1, 1, 2 }, { 1, 4, 5 },
{ 2, 1, 2 }, { 2, 1, 3 },
{ 2, 3, 4 } };
memset(arr, 0, sizeof arr);
memset(record, 0, sizeof record);
memset(block, 0, sizeof block);
for (int i = m - 1; i >= 0; i--) {
// If query is of type 2 then function
// call to record_func
if (command[i][0] == 2) {
int x = i / (block_size);
record_func(block_size, block, record,
command[i][1] - 1, command[i][2] - 1,
(block[x] + record[i] + 1));
}
// If query is of type 1 then simply add
// 1 to the record array
else
record[i]++;
}
// Merging the value of the block in the record array
for (int i = 0; i < m; i++) {
int check = (i / block_size);
record[i] += block[check];
}
for (int i = 0; i < m; i++) {
// If query is of type 1 then the array
// elements are over-written by the record
// array
if (command[i][0] == 1) {
arr[command[i][1] - 1] += record[i];
if ((command[i][2] - 1) < n - 1)
arr[(command[i][2])] -= record[i];
}
}
// The prefix sum of the array
for (int i = 1; i < n; i++)
update(arr, i);
// Printing the resultant array
print(arr, n);
return 0;
}
// Java program to perform range queries over range
// queries.
public class GFG {
static final int max = 10000;
// For prefix sum array
static void update(int arr[], int l) {
arr[l] += arr[l - 1];
}
// This function is used to apply square root
// decomposition in the record array
static void record_func(int block_size, int block[],
int record[], int l, int r, int value) {
// traversing first block in range
while (l < r && l % block_size != 0 && l != 0) {
record[l] += value;
l++;
}
// traversing completely overlapped blocks in range
while (l + block_size <= r + 1) {
block[l / block_size] += value;
l += block_size;
}
// traversing last block in range
while (l <= r) {
record[l] += value;
l++;
}
}
// Function to print the resultant array
static void print(int arr[], int n) {
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
// Driver code
public static void main(String[] args) {
int n = 5, m = 5;
int arr[] = new int[n], record[] = new int[m];
int block_size = (int) Math.sqrt(m);
int block[] = new int[max];
int command[][] = {{1, 1, 2}, {1, 4, 5},
{2, 1, 2}, {2, 1, 3},
{2, 3, 4}};
for (int i = m - 1; i >= 0; i--) {
// If query is of type 2 then function
// call to record_func
if (command[i][0] == 2) {
int x = i / (block_size);
record_func(block_size, block, record,
command[i][1] - 1, command[i][2] - 1,
(block[x] + record[i] + 1));
} // If query is of type 1 then simply add
// 1 to the record array
else {
record[i]++;
}
}
// Merging the value of the block in the record array
for (int i = 0; i < m; i++) {
int check = (i / block_size);
record[i] += block[check];
}
for (int i = 0; i < m; i++) {
// If query is of type 1 then the array
// elements are over-written by the record
// array
if (command[i][0] == 1) {
arr[command[i][1] - 1] += record[i];
if ((command[i][2] - 1) < n - 1) {
arr[(command[i][2])] -= record[i];
}
}
}
// The prefix sum of the array
for (int i = 1; i < n; i++) {
update(arr, i);
}
// Printing the resultant array
print(arr, n);
}
}
// This code is contributed by 29AjayKumar
# Python3 program to perform range
# queries over range queries.
import math
max = 10000
# For prefix sum array
def update(arr, l):
arr[l] += arr[l - 1]
# This function is used to apply square root
# decomposition in the record array
def record_func(block_size, block,
record, l, r, value):
# Traversing first block in range
while (l < r and
l % block_size != 0 and
l != 0):
record[l] += value
l += 1
# Traversing completely overlapped
# blocks in range
while (l + block_size <= r + 1):
block[l // block_size] += value
l += block_size
# Traversing last block in range
while (l <= r):
record[l] += value
l += 1
# Function to print the resultant array
def print_array(arr, n):
for i in range(n):
print(arr[i], end = " ")
# Driver code
if __name__ == "__main__":
n = 5
m = 5
arr = [0] * n
record = [0] * m
block_size = (int)(math.sqrt(m))
block = [0] * max
command = [ [ 1, 1, 2 ],
[ 1, 4, 5 ],
[ 2, 1, 2 ],
[ 2, 1, 3 ],
[ 2, 3, 4 ] ]
for i in range(m - 1, -1, -1):
# If query is of type 2 then function
# call to record_func
if (command[i][0] == 2):
x = i // (block_size)
record_func(block_size, block,
record, command[i][1] - 1,
command[i][2] - 1,
(block[x] + record[i] + 1))
# If query is of type 1 then simply add
# 1 to the record array
else:
record[i] += 1
# Merging the value of the block
# in the record array
for i in range(m):
check = (i // block_size)
record[i] += block[check]
for i in range(m):
# If query is of type 1 then the array
# elements are over-written by the record
# array
if (command[i][0] == 1):
arr[command[i][1] - 1] += record[i]
if ((command[i][2] - 1) < n - 1):
arr[(command[i][2])] -= record[i]
# The prefix sum of the array
for i in range(1, n):
update(arr, i)
# Printing the resultant array
print_array(arr, n)
# This code is contributed by chitranayal
// C# program to perform range queries over range
// queries.
using System;
public class GFG {
static readonly int max = 10000;
// For prefix sum array
static void update(int []arr, int l) {
arr[l] += arr[l - 1];
}
// This function is used to apply square root
// decomposition in the record array
static void record_func(int block_size, int []block,
int []record, int l, int r, int value) {
// traversing first block in range
while (l < r && l % block_size != 0 && l != 0) {
record[l] += value;
l++;
}
// traversing completely overlapped blocks in range
while (l + block_size <= r + 1) {
block[l / block_size] += value;
l += block_size;
}
// traversing last block in range
while (l <= r) {
record[l] += value;
l++;
}
}
// Function to print the resultant array
static void print(int []arr, int n) {
for (int i = 0; i < n; i++) {
Console.Write(arr[i] + " ");
}
}
// Driver code
public static void Main() {
int n = 5, m = 5;
int []arr = new int[n]; int []record = new int[m];
int block_size = (int) Math.Sqrt(m);
int []block = new int[max];
int [,]command= {{1, 1, 2}, {1, 4, 5},
{2, 1, 2}, {2, 1, 3},
{2, 3, 4}};
for (int i = m - 1; i >= 0; i--) {
// If query is of type 2 then function
// call to record_func
if (command[i,0] == 2) {
int x = i / (block_size);
record_func(block_size, block, record,
command[i,1] - 1, command[i,2] - 1,
(block[x] + record[i] + 1));
} // If query is of type 1 then simply add
// 1 to the record array
else {
record[i]++;
}
}
// Merging the value of the block in the record array
for (int i = 0; i < m; i++) {
int check = (i / block_size);
record[i] += block[check];
}
for (int i = 0; i < m; i++) {
// If query is of type 1 then the array
// elements are over-written by the record
// array
if (command[i,0] == 1) {
arr[command[i,1] - 1] += record[i];
if ((command[i,2] - 1) < n - 1) {
arr[(command[i,2])] -= record[i];
}
}
}
// The prefix sum of the array
for (int i = 1; i < n; i++) {
update(arr, i);
}
// Printing the resultant array
print(arr, n);
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program to perform range queries over range
// queries.
let max = 10000;
// For prefix sum array
function update(arr,l)
{
arr[l] += arr[l - 1];
}
// This function is used to apply square root
// decomposition in the record array
function record_func(block_size,block,record,l,r,value)
{
// traversing first block in range
while (l < r && l % block_size != 0 && l != 0) {
record[l] += value;
l++;
}
// traversing completely overlapped blocks in range
while (l + block_size <= r + 1) {
let x = Math.floor(l / block_size);
block[x] += value;
l += block_size;
}
// traversing last block in range
while (l <= r) {
record[l] += value;
l++;
}
}
// Function to print the resultant array
function print(arr,n)
{
for (let i = 0; i < n; i++) {
document.write(arr[i] + " ");
}
}
// Driver code
let n = 5, m = 5;
let arr = new Array(n);
for(let i = 0; i < n; i++)
arr[i] = 0;
let record = new Array(m);
for(let i = 0; i < m; i++)
record[i] = 0;
let block_size = Math.floor( Math.sqrt(m));
let block = new Array(max);
for(let i = 0; i < max; i++)
block[i] = 0;
let command = [[1, 1, 2], [1, 4, 5],
[2, 1, 2], [2, 1, 3],
[2, 3, 4]];
for (let i = m - 1; i >= 0; i--) {
// If query is of type 2 then function
// call to record_func
if (command[i][0] == 2)
{
let x = Math.floor(i / (block_size));
record_func(block_size, block, record,
command[i][1] - 1, command[i][2] - 1,
(block[x] + record[i] + 1));
}
// If query is of type 1 then simply add
// 1 to the record array
else {
record[i]++;
}
}
// Merging the value of the block in the record array
for (let i = 0; i < m; i++) {
let check = Math.floor(i / block_size);
record[i] += block[check];
}
for (let i = 0; i < m; i++)
{
// If query is of type 1 then the array
// elements are over-written by the record
// array
if (command[i][0] == 1) {
arr[command[i][1] - 1] += record[i];
if ((command[i][2] - 1) < n - 1) {
arr[(command[i][2])] -= record[i];
}
}
}
// The prefix sum of the array
for (let i = 1; i < n; i++) {
update(arr, i);
}
// Printing the resultant array
print(arr, n);
// This code is contributed by ab2127
</script>
Auxiliary Space: O(m+MAX)
Method 4: This method has been made more efficient by applying Binary Indexed Tree or Fenwick Tree by creating two binary indexed tree for query 1 and query 2 respectively.
Implementation:
C++
// C++ program to perform range queries over range
// queries.
#include <bits/stdc++.h>
using namespace std;
// Updates a node in Binary Index Tree (BITree) at given index
// in BITree. The given value 'val' is added to BITree[i] and
// all of its ancestors in tree.
void updateBIT(int BITree[], int n, int index, int val)
{
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse all ancestors and add 'val'
while (index <= n) {
// Add 'val' to current node of BI Tree
BITree[index] = (val + BITree[index]);
// Update index to that of parent in update View
index = (index + (index & (-index)));
}
return;
}
// Constructs and returns a Binary Indexed Tree for given
// array of size n.
int* constructBITree(int n)
{
// Create and initialize BITree[] as 0
int* BITree = new int[n + 1];
for (int i = 1; i <= n; i++)
BITree[i] = 0;
return BITree;
}
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[]
int getSum(int BITree[], int index)
{
int sum = 0;
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse ancestors of BITree[index]
while (index > 0) {
// Add element of BITree to sum
sum = (sum + BITree[index]);
// Move index to parent node in getSum View
index -= index & (-index);
}
return sum;
}
// Function to update the BITree
void update(int BITree[], int l, int r, int n, int val)
{
updateBIT(BITree, n, l, val);
updateBIT(BITree, n, r + 1, -val);
return;
}
// Driver code
int main()
{
int n = 5, m = 5;
int temp[15] = { 1, 1, 2, 1, 4, 5, 2, 1, 2,
2, 1, 3, 2, 3, 4 };
int q[5][3];
int j = 0;
for (int i = 1; i <= m; i++) {
q[i][0] = temp[j++];
q[i][1] = temp[j++];
q[i][2] = temp[j++];
}
// BITree for query of type 2
int* BITree = constructBITree(m);
// BITree for query of type 1
int* BITree2 = constructBITree(n);
// Input the queries in a 2D matrix
for (int i = 1; i <= m; i++)
cin >> q[i][0] >> q[i][1] >> q[i][2];
// If query is of type 2 then function call
// to update with BITree
for (int i = m; i >= 1; i--)
if (q[i][0] == 2)
update(BITree, q[i][1] - 1, q[i][2] - 1, m, 1);
for (int i = m; i >= 1; i--) {
if (q[i][0] == 2) {
long int val = getSum(BITree, i - 1);
update(BITree, q[i][1] - 1, q[i][2] - 1, m, val);
}
}
// If query is of type 1 then function call
// to update with BITree2
for (int i = m; i >= 1; i--) {
if (q[i][0] == 1) {
long int val = getSum(BITree, i - 1);
update(BITree2, q[i][1] - 1, q[i][2] - 1,
n, (val + 1));
}
}
for (int i = 1; i <= n; i++)
cout << (getSum(BITree2, i - 1)) << " ";
return 0;
}
// Java program to perform range queries over range
// queries.
import java.io.*;
import java.util.*;
class GFG
{
// Updates a node in Binary Index Tree (BITree) at given index
// in BITree. The given value 'val' is added to BITree[i] and
// all of its ancestors in tree.
static void updateBIT(int BITree[], int n, int index, int val)
{
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse all ancestors and add 'val'
while (index <= n)
{
// Add 'val' to current node of BI Tree
BITree[index] = (val + BITree[index]);
// Update index to that of parent in update View
index = (index + (index & (-index)));
}
return;
}
// Constructs and returns a Binary Indexed Tree for given
// array of size n.
static int[] constructBITree(int n)
{
// Create and initialize BITree[] as 0
int[] BITree = new int[n + 1];
for (int i = 1; i <= n; i++)
BITree[i] = 0;
return BITree;
}
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[]
static int getSum(int BITree[], int index)
{
int sum = 0;
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse ancestors of BITree[index]
while (index > 0)
{
// Add element of BITree to sum
sum = (sum + BITree[index]);
// Move index to parent node in getSum View
index -= index & (-index);
}
return sum;
}
// Function to update the BITree
static void update(int BITree[], int l, int r, int n, int val)
{
updateBIT(BITree, n, l, val);
updateBIT(BITree, n, r + 1, -val);
return;
}
// Driver code
public static void main (String[] args)
{
int n = 5, m = 5;
int temp[] = { 1, 1, 2, 1, 4, 5, 2, 1, 2,
2, 1, 3, 2, 3, 4 };
int[][] q = new int[6][3];
int j = 0;
for (int i = 1; i <= m; i++) {
q[i][0] = temp[j++];
q[i][1] = temp[j++];
q[i][2] = temp[j++];
}
// BITree for query of type 2
int[] BITree = constructBITree(m);
// BITree for query of type 1
int[] BITree2 = constructBITree(n);
// Input the queries in a 2D matrix
/*Scanner sc=new Scanner(System.in);
for (int i = 1; i <= m; i++)
{
q[i][0]=sc.nextInt();
q[i][1]=sc.nextInt();
q[i][2]=sc.nextInt();
}*/
// If query is of type 2 then function call
// to update with BITree
for (int i = m; i >= 1; i--)
if (q[i][0] == 2)
update(BITree, q[i][1] - 1, q[i][2] - 1, m, 1);
for (int i = m; i >= 1; i--) {
if (q[i][0] == 2) {
int val = getSum(BITree, i - 1);
update(BITree, q[i][1] - 1, q[i][2] - 1, m, val);
}
}
// If query is of type 1 then function call
// to update with BITree2
for (int i = m; i >= 1; i--) {
if (q[i][0] == 1) {
int val = getSum(BITree, i - 1);
update(BITree2, q[i][1] - 1, q[i][2] - 1,
n, (val + 1));
}
}
for (int i = 1; i <= n; i++)
System.out.print(getSum(BITree2, i - 1)+" ");
}
}
// This code is contributed by avanitrachhadiya2155
class GFG :
# Updates a node in Binary Index Tree (BITree) at given index
# in BITree. The given value 'val' is added to BITree[i] and
# all of its ancestors in tree.
@staticmethod
def updateBIT( BITree, n, index, val) :
# index in BITree[] is 1 more than the index in arr[]
index = index + 1
# Traverse all ancestors and add 'val'
while (index <= n) :
# Add 'val' to current node of BI Tree
BITree[index] = (val + BITree[index])
# Update index to that of parent in update View
index = (index + (index & (-index)))
return
# Constructs and returns a Binary Indexed Tree for given
# array of size n.
@staticmethod
def constructBITree( n) :
# Create and initialize BITree[] as 0
BITree = [0] * (n + 1)
i = 1
while (i <= n) :
BITree[i] = 0
i += 1
return BITree
# Returns sum of arr[0..index]. This function assumes
# that the array is preprocessed and partial sums of
# array elements are stored in BITree[]
@staticmethod
def getSum( BITree, index) :
sum = 0
# index in BITree[] is 1 more than the index in arr[]
index = index + 1
# Traverse ancestors of BITree[index]
while (index > 0) :
# Add element of BITree to sum
sum = (sum + BITree[index])
# Move index to parent node in getSum View
index -= index & (-index)
return sum
# Function to update the BITree
@staticmethod
def update( BITree, l, r, n, val) :
GFG.updateBIT(BITree, n, l, val)
GFG.updateBIT(BITree, n, r + 1, -val)
return
# Driver code
@staticmethod
def main( args) :
n = 5
m = 5
temp = [1, 1, 2, 1, 4, 5, 2, 1, 2, 2, 1, 3, 2, 3, 4]
q = [[0] * (3) for _ in range(6)]
j = 0
i = 1
while (i <= m) :
q[i][0] = temp[j]
j += 1
q[i][1] = temp[j]
j += 1
q[i][2] = temp[j]
j += 1
i += 1
# BITree for query of type 2
BITree = GFG.constructBITree(m)
# BITree for query of type 1
BITree2 = GFG.constructBITree(n)
# Input the queries in a 2D matrix
# Scanner sc=new Scanner(System.in);
# for (int i = 1; i <= m; i++)
# {
# q[i][0]=sc.nextInt();
# q[i][1]=sc.nextInt();
# q[i][2]=sc.nextInt();
# }
# If query is of type 2 then function call
# to update with BITree
i = m
while (i >= 1) :
if (q[i][0] == 2) :
GFG.update(BITree, q[i][1] - 1, q[i][2] - 1, m, 1)
i -= 1
i = m
while (i >= 1) :
if (q[i][0] == 2) :
val = GFG.getSum(BITree, i - 1)
GFG.update(BITree, q[i][1] - 1, q[i][2] - 1, m, val)
i -= 1
# If query is of type 1 then function call
# to update with BITree2
i = m
while (i >= 1) :
if (q[i][0] == 1) :
val = GFG.getSum(BITree, i - 1)
GFG.update(BITree2, q[i][1] - 1, q[i][2] - 1, n, (val + 1))
i -= 1
i = 1
while (i <= n) :
print(str(GFG.getSum(BITree2, i - 1)) + " ", end ="")
i += 1
if __name__=="__main__":
GFG.main([])
# This code is contributed by aadityaburujwale.
// C# program to perform range queries over range
// queries.
using System;
class GFG{
// Updates a node in Binary Index Tree (BITree)
// at given index in BITree. The given value
// 'val' is added to BITree[i] and
// all of its ancestors in tree.
static void updateBIT(int[] BITree, int n,
int index, int val)
{
// index in BITree[] is 1 more than
// the index in arr[]
index = index + 1;
// Traverse all ancestors and add 'val'
while (index <= n)
{
// Add 'val' to current node of BI Tree
BITree[index] = (val + BITree[index]);
// Update index to that of parent in update View
index = (index + (index & (-index)));
}
return;
}
// Constructs and returns a Binary Indexed
// Tree for given array of size n.
static int[] constructBITree(int n)
{
// Create and initialize BITree[] as 0
int[] BITree = new int[n + 1];
for(int i = 1; i <= n; i++)
BITree[i] = 0;
return BITree;
}
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[]
static int getSum(int[] BITree, int index)
{
int sum = 0;
// index in BITree[] is 1 more than
// the index in arr[]
index = index + 1;
// Traverse ancestors of BITree[index]
while (index > 0)
{
// Add element of BITree to sum
sum = (sum + BITree[index]);
// Move index to parent node in getSum View
index -= index & (-index);
}
return sum;
}
// Function to update the BITree
static void update(int[] BITree, int l, int r,
int n, int val)
{
updateBIT(BITree, n, l, val);
updateBIT(BITree, n, r + 1, -val);
return;
}
// Driver code
static public void Main()
{
int n = 5, m = 5;
int[] temp = { 1, 1, 2, 1, 4, 5, 2,
1, 2, 2, 1, 3, 2, 3, 4 };
int[,] q = new int[6, 3];
int j = 0;
for(int i = 1; i <= m; i++)
{
q[i, 0] = temp[j++];
q[i, 1] = temp[j++];
q[i, 2] = temp[j++];
}
// BITree for query of type 2
int[] BITree = constructBITree(m);
// BITree for query of type 1
int[] BITree2 = constructBITree(n);
// If query is of type 2 then function call
// to update with BITree
for(int i = m; i >= 1; i--)
if (q[i, 0] == 2)
update(BITree, q[i, 1] - 1,
q[i, 2] - 1, m, 1);
for(int i = m; i >= 1; i--)
{
if (q[i,0] == 2)
{
int val = getSum(BITree, i - 1);
update(BITree, q[i, 1] - 1,
q[i, 2] - 1, m, val);
}
}
// If query is of type 1 then function call
// to update with BITree2
for(int i = m; i >= 1; i--)
{
if (q[i,0] == 1)
{
int val = getSum(BITree, i - 1);
update(BITree2, q[i, 1] - 1, q[i, 2] - 1,
n, (val + 1));
}
}
for(int i = 1; i <= n; i++)
Console.Write(getSum(BITree2, i - 1) + " ");
}
}
// This code is contributed by rag2127
<script>
// Javascript program to perform range queries over range
// queries.
// Updates a node in Binary Index Tree (BITree) at given index
// in BITree. The given value 'val' is added to BITree[i] and
// all of its ancestors in tree.
function updateBIT(BITree, n, index, val)
{
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse all ancestors and add 'val'
while (index <= n)
{
// Add 'val' to current node of BI Tree
BITree[index] = (val + BITree[index]);
// Update index to that of parent in update View
index = (index + (index & (-index)));
}
return;
}
// Constructs and returns a Binary Indexed Tree for given
// array of size n.
function constructBITree(n)
{
// Create and initialize BITree[] as 0
let BITree = new Array(n + 1);
for (let i = 1; i <= n; i++)
BITree[i] = 0;
return BITree;
}
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[]
function getSum(BITree, index)
{
let sum = 0;
// index in BITree[] is 1 more than the index in arr[]
index = index + 1;
// Traverse ancestors of BITree[index]
while (index > 0)
{
// Add element of BITree to sum
sum = (sum + BITree[index]);
// Move index to parent node in getSum View
index -= index & (-index);
}
return sum;
}
// Function to update the BITree
function update(BITree, l, r, n, val)
{
updateBIT(BITree, n, l, val);
updateBIT(BITree, n, r + 1, -val);
return;
}
// Driver code
let n = 5, m = 5;
let temp = [ 1, 1, 2, 1, 4, 5, 2, 1, 2, 2, 1, 3, 2, 3, 4 ];
let q = new Array(6).fill(0).map(() => new Array(3))
let j = 0;
for (let i = 1; i <= m; i++) {
q[i][0] = temp[j++];
q[i][1] = temp[j++];
q[i][2] = temp[j++];
}
// BITree for query of type 2
let BITree = constructBITree(m);
// BITree for query of type 1
let BITree2 = constructBITree(n);
// Input the queries in a 2D matrix
/*Scanner sc=new Scanner(System.in);
for (int i = 1; i <= m; i++)
{
q[i][0]=sc.nextInt();
q[i][1]=sc.nextInt();
q[i][2]=sc.nextInt();
}*/
// If query is of type 2 then function call
// to update with BITree
for (let i = m; i >= 1; i--)
if (q[i][0] == 2)
update(BITree, q[i][1] - 1, q[i][2] - 1, m, 1);
for (let i = m; i >= 1; i--) {
if (q[i][0] == 2) {
let val = getSum(BITree, i - 1);
update(BITree, q[i][1] - 1, q[i][2] - 1, m, val);
}
}
// If query is of type 1 then function call
// to update with BITree2
for (let i = m; i >= 1; i--) {
if (q[i][0] == 1) {
let val = getSum(BITree, i - 1);
update(BITree2, q[i][1] - 1, q[i][2] - 1,
n, (val + 1));
}
}
for (let i = 1; i <= n; i++)
document.write(getSum(BITree2, i - 1)+" ");
// This code is contributed by gfgking.
</script>
The Time complexity of Method 3 and Method 4 is O(log n).
Auxiliary Space: O(m+n)
Explore
Basics & Prerequisites
Data Structures
Algorithms
Advanced
Interview Preparation
Practice Problem