Given an array arr[] of n integers and q queries represented by an array queries[][], where queries[i][0] = l and queries[i][1] = r. For each query, the task is to calculate the mean of elements in the range l to r and return its floor value.
Examples:
Input: arr[] = [3, 7, 2, 8, 5] queries[][] = [[1, 3], [2, 5]]
Output: 4 5
Explanation: For query [1, 3] - Elements in range are 3, 7, 2
Mean is (3+7+2)/3 = 4, Floor value is 4
For query [2, 5] - Elements in range are 7, 2, 8, 5,
Mean is (7+2+8+5)/4 = 5.5, Floor value is 5
Input: arr[] = [10, 20, 30, 40, 50, 60], queries[][] = [[4, 6]]
Output: 50
Explanation: For query [4, 6] - Elements in range are 40, 50, 60
Mean is (40+50+60)/3 = 50, Floor value is 50
[Naive Approach] Compute for Each Query Separately - O(n*q) Time and O(1) Space
The idea is to process each query separately by iterating over the given range l to r and computing the sum and count of elements. Since the mean is sum/count, we directly compute its floor value using integer division. The approach ensures that each query is handled independently.
C++
// C++ Code to find the mean in an array
// for each query using Naive Approach
#include <bits/stdc++.h>
using namespace std;
// Function to compute floor of mean for each query
vector<int> findMean(vector<int> &arr,
vector<vector<int>> &queries) {
int n = arr.size();
int q = queries.size();
vector<int> result;
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i][0] - 1;
// Convert to 0-based index
int r = queries[i][1] - 1;
int sum = 0, count = 0;
// Calculate sum of elements in
// range l to r
for (int j = l; j <= r; j++) {
sum += arr[j];
count++;
}
// Store floor of mean in result
result.push_back(sum / count);
}
return result;
}
// Function to print the result
void printArr(vector<int> &arr) {
for (int num : arr) {
cout << num << " ";
}
cout << endl;
}
// Driver code
int main() {
vector<int> arr = {3, 7, 2, 8, 5};
vector<vector<int>> queries = {{1, 3}, {2, 5}};
vector<int> result = findMean(arr, queries);
printArr(result);
return 0;
}
// Java Code to find the mean in an array
// for each query using Naive Approach
import java.util.*;
class GfG {
// Function to compute floor of mean for each query
public static int[] findMean(int[] arr, int[][] queries) {
int n = arr.length;
int q = queries.length;
int[] result = new int[q];
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i][0] - 1;
// Convert to 0-based index
int r = queries[i][1] - 1;
int sum = 0, count = 0;
// Calculate sum of elements in
// range l to r
for (int j = l; j <= r; j++) {
sum += arr[j];
count++;
}
// Store floor of mean in result
result[i] = sum / count;
}
return result;
}
// Function to print the result
public static void printArr(int[] arr) {
for (int num : arr) {
System.out.print(num + " ");
}
System.out.println();
}
// Driver code
public static void main(String[] args) {
int[] arr = {3, 7, 2, 8, 5};
int[][] queries = { {1, 3}, {2, 5} };
int[] result = findMean(arr, queries);
printArr(result);
}
}
# Python Code to find the mean in an array
# for each query using Naive Approach
# Function to compute floor of mean for each query
def findMean(arr, queries):
n = len(arr)
q = len(queries)
result = []
# Iterate through each query
for i in range(q):
# Convert to 0-based index
l = queries[i][0] - 1
# Convert to 0-based index
r = queries[i][1] - 1
sum_val = 0
count = 0
# Calculate sum of elements in
# range l to r
for j in range(l, r + 1):
sum_val += arr[j]
count += 1
# Store floor of mean in result
result.append(sum_val // count)
return result
# Function to print the result
def printArr(arr):
print(*arr)
# Driver code
if __name__ == "__main__":
arr = [3, 7, 2, 8, 5]
queries = [[1, 3], [2, 5]]
result = findMean(arr, queries)
printArr(result)
// C# Code to find the mean in an array
// for each query using Naive Approach
using System;
using System.Collections.Generic;
class GfG {
// Function to compute floor of mean for each query
public static int[] findMean(int[] arr, int[,] queries) {
int n = arr.Length;
int q = queries.GetLength(0);
int[] result = new int[q];
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i, 0] - 1;
// Convert to 0-based index
int r = queries[i, 1] - 1;
int sum = 0, count = 0;
// Calculate sum of elements in
// range l to r
for (int j = l; j <= r; j++) {
sum += arr[j];
count++;
}
// Store floor of mean in result
result[i] = sum / count;
}
return result;
}
// Function to print the result
public static void printArr(int[] arr) {
foreach (int num in arr) {
Console.Write(num + " ");
}
Console.WriteLine();
}
// Driver code
public static void Main() {
int[] arr = {3, 7, 2, 8, 5};
int[,] queries = { {1, 3}, {2, 5} };
int[] result = findMean(arr, queries);
printArr(result);
}
}
// JavaScript Code to find the mean in an array
// for each query using Naive Approach
// Function to compute floor of mean for each query
function findMean(arr, queries) {
let n = arr.length;
let q = queries.length;
let result = [];
// Iterate through each query
for (let i = 0; i < q; i++) {
// Convert to 0-based index
let l = queries[i][0] - 1;
// Convert to 0-based index
let r = queries[i][1] - 1;
let sum = 0, count = 0;
// Calculate sum of elements in
// range l to r
for (let j = l; j <= r; j++) {
sum += arr[j];
count++;
}
// Store floor of mean in result
result.push(Math.floor(sum / count));
}
return result;
}
// Function to print the result
function printArr(arr) {
console.log(arr.join(" "));
}
// Driver code
let arr = [3, 7, 2, 8, 5];
let queries = [[1, 3], [2, 5]];
let result = findMean(arr, queries);
printArr(result);
Time Complexity: O(n * q), for each of the q queries, we iterate over the range l to r, taking O(n) time.
Space Complexity: O(1), as we use only a few extra variables for storing sum, count, and result, ignoring output storage.
[Expected Approach] Using Prefix Sum - O(n+q) Time and O(n) Space
The idea is to optimize the sum calculation for each query by using Prefix Sum. Instead of iterating through the range (l to r) for each query, we precompute the sum of elements up to each index in prefixSum. This allows us to find the sum of any subarray in O(1) time using the formula prefixSum[r + 1] - prefixSum[l]. This reduces query processing time from O(n) to O(1).
Steps to implement the above idea:
- Initialize prefixSum array of size n+1 with 0 to store cumulative sums for efficient range sum calculation.
- Compute prefixSum by iterating through arr[] and storing the sum of elements up to each index for quick access.
- Iterate through queries[][] to process each range query efficiently without looping through elements in the given range.
- Extract l and r from queries[][], converting them to 0-based indexing for correct array access.
- Calculate sum efficiently using prefixSum[r+1] - prefixSum[l] instead of iterating through elements in the range.
- Compute floor of mean by dividing the sum by the count of elements in the range and store the result.
- Return and print result, using a helper function to display all floor values of means for given queries.
Below is the implementation of the above approach:
C++
// C++ Code to find the mean in an array
// for each query using Prefix Sum
#include <bits/stdc++.h>
using namespace std;
// Function to compute floor of mean for each query
vector<int> findMean(vector<int> &arr,
vector<vector<int>> &queries) {
int n = arr.size();
int q = queries.size();
vector<int> prefixSum(n + 1, 0);
vector<int> result;
// Compute prefix sum
for (int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i - 1];
}
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i][0] - 1;
// Convert to 0-based index
int r = queries[i][1] - 1;
// Calculate sum using prefix sum
int sum = prefixSum[r + 1] - prefixSum[l];
int count = (r - l + 1);
// Store floor of mean in result
result.push_back(sum / count);
}
return result;
}
// Function to print the result
void printArr(vector<int> &arr) {
for (int num : arr) {
cout << num << " ";
}
cout << endl;
}
// Driver code
int main() {
vector<int> arr = {3, 7, 2, 8, 5};
vector<vector<int>> queries = {{1, 3}, {2, 5}};
vector<int> result = findMean(arr, queries);
printArr(result);
return 0;
}
// Java Code to find the mean in an array
// for each query using Prefix Sum
import java.util.*;
class GfG {
// Function to compute floor of mean for each query
static int[] findMean(int[] arr, int[][] queries) {
int n = arr.length;
int q = queries.length;
int[] prefixSum = new int[n + 1];
int[] result = new int[q];
// Compute prefix sum
for (int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i - 1];
}
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i][0] - 1;
// Convert to 0-based index
int r = queries[i][1] - 1;
// Calculate sum using prefix sum
int sum = prefixSum[r + 1] - prefixSum[l];
int count = (r - l + 1);
// Store floor of mean in result
result[i] = sum / count;
}
return result;
}
// Function to print the result
static void printArr(int[] arr) {
for (int num : arr) {
System.out.print(num + " ");
}
System.out.println();
}
// Driver code
public static void main(String[] args) {
int[] arr = {3, 7, 2, 8, 5};
int[][] queries = {{1, 3}, {2, 5}};
int[] result = findMean(arr, queries);
printArr(result);
}
}
# Python Code to find the mean in an array
# for each query using Prefix Sum
# Function to compute floor of mean for each query
def findMean(arr, queries):
n = len(arr)
q = len(queries)
prefixSum = [0] * (n + 1)
result = []
# Compute prefix sum
for i in range(1, n + 1):
prefixSum[i] = prefixSum[i - 1] + arr[i - 1]
# Iterate through each query
for i in range(q):
# Convert to 0-based index
l = queries[i][0] - 1
# Convert to 0-based index
r = queries[i][1] - 1
# Calculate sum using prefix sum
total = prefixSum[r + 1] - prefixSum[l]
count = (r - l + 1)
# Store floor of mean in result
result.append(total // count)
return result
# Function to print the result
def printArr(arr):
print(*arr)
# Driver code
if __name__ == "__main__":
arr = [3, 7, 2, 8, 5]
queries = [[1, 3], [2, 5]]
result = findMean(arr, queries)
printArr(result)
// C# Code to find the mean in an array
// for each query using Prefix Sum
using System;
class GfG {
// Function to compute floor of mean for each query
static int[] findMean(int[] arr, int[][] queries) {
int n = arr.Length;
int q = queries.Length;
int[] prefixSum = new int[n + 1];
int[] result = new int[q];
// Compute prefix sum
for (int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i - 1];
}
// Iterate through each query
for (int i = 0; i < q; i++) {
// Convert to 0-based index
int l = queries[i][0] - 1;
// Convert to 0-based index
int r = queries[i][1] - 1;
// Calculate sum using prefix sum
int sum = prefixSum[r + 1] - prefixSum[l];
int count = (r - l + 1);
// Store floor of mean in result
result[i] = sum / count;
}
return result;
}
// Function to print the result
static void printArr(int[] arr) {
foreach (int num in arr) {
Console.Write(num + " ");
}
Console.WriteLine();
}
// Driver code
public static void Main() {
int[] arr = {3, 7, 2, 8, 5};
int[][] queries = {new int[] {1, 3}, new int[] {2, 5}};
int[] result = findMean(arr, queries);
printArr(result);
}
}
// JavaScript Code to find the mean in an array
// for each query using Prefix Sum
// Function to compute floor of mean for each query
function findMean(arr, queries) {
let n = arr.length;
let q = queries.length;
let prefixSum = new Array(n + 1).fill(0);
let result = [];
// Compute prefix sum
for (let i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i - 1];
}
// Iterate through each query
for (let i = 0; i < q; i++) {
// Convert to 0-based index
let l = queries[i][0] - 1;
// Convert to 0-based index
let r = queries[i][1] - 1;
// Calculate sum using prefix sum
let sum = prefixSum[r + 1] - prefixSum[l];
let count = (r - l + 1);
// Store floor of mean in result
result.push(Math.floor(sum / count));
}
return result;
}
// Function to print the result
function printArr(arr) {
console.log(arr.join(" "));
}
// Driver code
let arr = [3, 7, 2, 8, 5];
let queries = [[1, 3], [2, 5]];
let result = findMean(arr, queries);
printArr(result);
Time Complexity: O(n + q), as the method efficiently computes prefix sum in O(n) and processes each query in O(1).
Space Complexity: O(n), as the method uses an extra array of size n+1 to store prefix sums.
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