Queries to find minimum sum of array elements from either end of an array
Last Updated :
01 Sep, 2021
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Given an array arr[] consisting of N distinct integers and an array queries[] consisting of Q queries, the task is for each query is to find queries[i] in the array and calculate the minimum of sum of array elements from the start and end of the array up to queries[i].
Examples:
Input: arr[] = {2, 3, 6, 7, 4, 5, 30}, Q = 2, queries[] = {6, 5}
Output: 11 27
Explanation:
Query 1: Sum from start = 2 + 3 + 6 = 11. Sum from end = 30 + 5 + 4 + 7 + 6 = 52. Therefore, 11 is the required answer.
Query 2: Sum from start = 27. Sum from end = 35. Therefore, 27 is the required answer.Input: arr[] = {1, 2, -3, 4}, Q = 2, queries[] = {4, 2}
Output: 4 2
Naive Approach: The simplest approach is to traverse the array upto queries[i] for each query and calculate sum from the end as well as from the start of the array. Finally, print the minimum of the sums obtained.
Below is the implementation of the above approach:
// C++ implementation
// of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the minimum
// sum from either end of the arrays
// for the given queries
void calculateQuery(int arr[], int N,
int query[], int M)
{
// Traverse the query[] array
for (int i = 0; i < M; i++) {
int X = query[i];
// Stores sum from start
// and end of the array
int sum_start = 0, sum_end = 0;
// Calculate distance from start
for (int j = 0; j < N; j++) {
sum_start += arr[j];
if (arr[j] == X)
break;
}
// Calculate distance from end
for (int j = N - 1; j >= 0; j--) {
sum_end += arr[j];
if (arr[j] == X)
break;
}
cout << min(sum_end, sum_start) << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
int queries[] = { 6, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(queries)
/ sizeof(queries[0]);
calculateQuery(arr, N, queries, M);
return 0;
}
// Java implementation of the
// above approach
import java.util.*;
import java.lang.*;
public class GFG
{
// Function to calculate the minimum
// sum from either end of the arrays
// for the given queries
static void calculateQuery(int arr[], int N,
int query[], int M)
{
// Traverse the query[] array
for (int i = 0; i < M; i++)
{
int X = query[i];
// Stores sum from start
// and end of the array
int sum_start = 0, sum_end = 0;
// Calculate distance from start
for (int j = 0; j < N; j++)
{
sum_start += arr[j];
if (arr[j] == X)
break;
}
// Calculate distance from end
for (int j = N - 1; j >= 0; j--)
{
sum_end += arr[j];
if (arr[j] == X)
break;
}
System.out.print(Math.min(sum_end, sum_start) + " ");
}
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
int queries[] = { 6, 5 };
int N = arr.length;
int M = queries.length;
calculateQuery(arr, N, queries, M);
}
}
// This code is contributed by jana_sayantan.
# Python 3 implementation
# of the above approach
# Function to calculate the minimum
# sum from either end of the arrays
# for the given queries
def calculateQuery(arr, N, query, M):
# Traverse the query[] array
for i in range(M):
X = query[i]
# Stores sum from start
# and end of the array
sum_start = 0
sum_end = 0
# Calculate distance from start
for j in range(N):
sum_start += arr[j]
if (arr[j] == X):
break
# Calculate distance from end
for j in range(N - 1, -1, -1):
sum_end += arr[j]
if (arr[j] == X):
break
print(min(sum_end, sum_start), end=" ")
# Driver Code
if __name__ == "__main__":
arr = [2, 3, 6, 7, 4, 5, 30]
queries = [6, 5]
N = len(arr)
M = len(queries)
calculateQuery(arr, N, queries, M)
# This code is contributed by chitranayal.
// C# implementation
// of the above approach
using System;
class GFG {
// Function to calculate the minimum
// sum from either end of the arrays
// for the given queries
static void calculateQuery(int[] arr, int N,
int[] query, int M)
{
// Traverse the query[] array
for (int i = 0; i < M; i++) {
int X = query[i];
// Stores sum from start
// and end of the array
int sum_start = 0, sum_end = 0;
// Calculate distance from start
for (int j = 0; j < N; j++) {
sum_start += arr[j];
if (arr[j] == X)
break;
}
// Calculate distance from end
for (int j = N - 1; j >= 0; j--) {
sum_end += arr[j];
if (arr[j] == X)
break;
}
Console.Write(Math.Min(sum_end, sum_start) + " ");
}
}
// Driver code
static void Main()
{
int[] arr = { 2, 3, 6, 7, 4, 5, 30 };
int[] queries = { 6, 5 };
int N = arr.Length;
int M = queries.Length;
calculateQuery(arr, N, queries, M);
}
}
// This code is contributed by divyesh072019.
<script>
// Javascript implementation of the above approach
// Function to calculate the minimum
// sum from either end of the arrays
// for the given queries
function calculateQuery(arr, N, query, M)
{
// Traverse the query[] array
for(let i = 0; i < M; i++)
{
let X = query[i];
// Stores sum from start
// and end of the array
let sum_start = 0, sum_end = 0;
// Calculate distance from start
for(let j = 0; j < N; j++)
{
sum_start += arr[j];
if (arr[j] == X)
break;
}
// Calculate distance from end
for(let j = N - 1; j >= 0; j--)
{
sum_end += arr[j];
if (arr[j] == X)
break;
}
document.write(Math.min(
sum_end, sum_start) + " ");
}
}
// Driver code
let arr = [ 2, 3, 6, 7, 4, 5, 30 ];
let queries = [ 6, 5 ];
let N = arr.length;
let M = queries.length;
calculateQuery(arr, N, queries, M);
// This code is contributed by suresh07
</script>
Output:
11 27
Time Complexity: O(Q * N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using extra space and the concept of prefix sum and suffix sum and to answer each query in constant computational complexity. The idea is to preprocess prefix and suffix sums for each index. Follow the steps below to solve the problem:
- Initialize two variables, say prefix and suffix.
- Initialize an Unordered Map, say mp, to map array elements as keys to pairs in which the first value gives the prefix sum and the second value gives the suffix sum.
- Traverse the array and keep adding arr[i] to prefix and store it in the map with arr[i] as key and prefix as value.
- Traverse the array in reverse and keep adding arr[i] to suffix and store in the map with arr[i] as key and suffix as value.
- Now traverse the array queries[] and for each query queries[i], print the value of the minimum of mp[queries[i]].first and mp[queries[i]].second as the result.
Below is the implementation of the above approach:
// C++ implementation
// of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum sum
// for the given queries
void calculateQuery(int arr[], int N,
int query[], int M)
{
// Stores prefix and suffix sums
int prefix = 0, suffix = 0;
// Stores pairs of prefix and suffix sums
unordered_map<int, pair<int, int> > mp;
// Traverse the array
for (int i = 0; i < N; i++) {
// Add element to prefix
prefix += arr[i];
// Store prefix for each element
mp[arr[i]].first = prefix;
}
// Traverse the array in reverse
for (int i = N - 1; i >= 0; i--) {
// Add element to suffix
suffix += arr[i];
// Storing suffix for each element
mp[arr[i]].second = suffix;
}
// Traverse the array queries[]
for (int i = 0; i < M; i++) {
int X = query[i];
// Minimum of suffix
// and prefix sums
cout << min(mp[X].first,
mp[X].second)
<< " ";
}
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
int queries[] = { 6, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(queries) / sizeof(queries[0]);
calculateQuery(arr, N, queries, M);
return 0;
}
// Java implementation
// of the above approach
import java.util.*;
class GFG
{
static class pair<E,P>
{
E first;
P second;
public pair(E first, P second)
{
this.first = first;
this.second = second;
}
}
// Function to find the minimum sum
// for the given queries
@SuppressWarnings({ "unchecked", "rawtypes" })
static void calculateQuery(int arr[], int N,
int query[], int M)
{
// Stores prefix and suffix sums
int prefix = 0, suffix = 0;
// Stores pairs of prefix and suffix sums
HashMap<Integer, pair > mp = new HashMap<>();
// Traverse the array
for (int i = 0; i < N; i++) {
// Add element to prefix
prefix += arr[i];
// Store prefix for each element
mp.put(arr[i], new pair(prefix,0));
}
// Traverse the array in reverse
for (int i = N - 1; i >= 0; i--) {
// Add element to suffix
suffix += arr[i];
// Storing suffix for each element
mp.put(arr[i], new pair(mp.get(arr[i]).first,suffix));
}
// Traverse the array queries[]
for (int i = 0; i < M; i++) {
int X = query[i];
// Minimum of suffix
// and prefix sums
System.out.print(Math.min((int)mp.get(X).first,
(int)mp.get(X).second)
+ " ");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
int queries[] = { 6, 5 };
int N = arr.length;
int M = queries.length;
calculateQuery(arr, N, queries, M);
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation
# of the above approach
# Function to find the minimum sum
# for the given queries
def calculateQuery(arr, N, query, M):
# Stores prefix and suffix sums
prefix = 0
suffix = 0
# Stores pairs of prefix and suffix sums
mp = {}
# Traverse the array
for i in range(N):
# Add element to prefix
prefix += arr[i]
# Store prefix for each element
mp[arr[i]] = [prefix, 0]
# Traverse the array in reverse
for i in range(N - 1, -1, -1):
# Add element to suffix
suffix += arr[i]
# Storing suffix for each element
mp[arr[i]] = [mp[arr[i]][0], suffix]
# Traverse the array queries[]
for i in range(M):
X = query[i]
# Minimum of suffix
# and prefix sums
print(min(mp[X][0], mp[X][1]), end = " ")
# Driver Code
arr = [ 2, 3, 6, 7, 4, 5, 30 ]
queries = [ 6, 5 ]
N = len(arr)
M = len(queries)
calculateQuery(arr, N, queries, M)
# This code is contributed by avanitrachhadiya2155
// C# implementation
// of the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find the minimum sum
// for the given queries
static void calculateQuery(int[] arr, int N,
int[] query, int M)
{
// Stores prefix and suffix sums
int prefix = 0, suffix = 0;
// Stores pairs of prefix and suffix sums
Dictionary<int, Tuple<int,int>> mp =
new Dictionary<int, Tuple<int,int>>();
// Traverse the array
for (int i = 0; i < N; i++)
{
// Add element to prefix
prefix += arr[i];
// Store prefix for each element
mp[arr[i]] = new Tuple<int,int>(prefix, 0);
}
// Traverse the array in reverse
for (int i = N - 1; i >= 0; i--)
{
// Add element to suffix
suffix += arr[i];
// Storing suffix for each element
mp[arr[i]] = new Tuple<int,int>(mp[arr[i]].Item1, suffix);
}
// Traverse the array queries[]
for (int i = 0; i < M; i++)
{
int X = query[i];
// Minimum of suffix
// and prefix sums
Console.Write(Math.Min(mp[X].Item1, mp[X].Item2) + " ");
}
}
// Driver code
static void Main() {
int[] arr = { 2, 3, 6, 7, 4, 5, 30 };
int[] queries = { 6, 5 };
int N = arr.Length;
int M = queries.Length;
calculateQuery(arr, N, queries, M);
}
}
// This code is contributed by divyeshrabadiya07.
<script>
// Javascript implementation
// of the above approach
// Function to find the minimum sum
// for the given queries
function calculateQuery(arr, N, query, M)
{
// Stores prefix and suffix sums
var prefix = 0, suffix = 0;
// Stores pairs of prefix and suffix sums
var mp = new Map();
// Traverse the array
for(var i = 0; i < N; i++)
{
// Add element to prefix
prefix += arr[i];
// Store prefix for each element
if (!mp.has(arr[i]))
{
mp.set(arr[i], [0, 0]);
}
var tmp = mp.get(arr[i]);
tmp[0] = prefix;
mp.set(arr[i], tmp);
}
// Traverse the array in reverse
for(var i = N - 1; i >= 0; i--)
{
// Add element to suffix
suffix += arr[i];
// Storing suffix for each element
if (!mp.has(arr[i]))
{
mp.set(arr[i], [0, 0]);
}
var tmp = mp.get(arr[i]);
tmp[1] = suffix;
mp.set(arr[i], tmp);
}
// Traverse the array queries[]
for(var i = 0; i < M; i++)
{
var X = query[i];
var tmp = mp.get(X);
// Minimum of suffix
// and prefix sums
document.write(Math.min(tmp[0], tmp[1]) + " ");
}
}
// Driver Code
var arr = [ 2, 3, 6, 7, 4, 5, 30 ];
var queries = [ 6, 5 ];
var N = arr.length;
var M = queries.length;
calculateQuery(arr, N, queries, M);
// This code is contributed by rutvik_56
</script>
Output:
11 27
Time Complexity: O(M + N)
Auxiliary Space: O(N)
