C++ Program for Mean of range in array

Given an array of n integers. You are given q queries. Write a program to print the floor value of mean in range l to r for each query in a new line.

Examples :

Input : arr[] = {1, 2, 3, 4, 5}
        q = 3
        0 2
        1 3
        0 4
Output : 2
         3
         3
Here for 0 to 2 (1 + 2 + 3) / 3 = 2

Input : arr[] = {6, 7, 8, 10}
        q = 2
        0 3
        1 2
Output : 7
         7

Naive Approach: We can run loop for each query l to r and find sum and number of elements in range. After this we can print floor of mean for each query.  

C++

// CPP program to find floor value
// of mean in range l to r
#include <bits/stdc++.h>
using namespace std;

// To find mean of range in l to r
int findMean(int arr[], int l, int r)
{
    // Both sum and count are
    // initialize to 0
    int sum = 0, count = 0;

    // To calculate sum and number
    // of elements in range l to r
    for (int i = l; i <= r; i++) {
        sum += arr[i];
        count++;
    }

    // Calculate floor value of mean
    int mean = floor(sum / count);

    // Returns mean of array
    // in range l to r
    return mean;
}

// Driver program to test findMean()
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    cout << findMean(arr, 0, 2) << endl;
    cout << findMean(arr, 1, 3) << endl;
    cout << findMean(arr, 0, 4) << endl;
    return 0;
}

Output :

2
3
3

Time complexity: O(n*q) where q is the number of queries and n is the size of the array. Here in the above code q is 3 as the findMean function is used 3 times.
Auxiliary Space: O(1)

Efficient Approach: We can find sum of numbers using numbers using prefix sum. The prefixSum[i] denotes the sum of first i elements. So sum of numbers in range l to r will be prefixSum[r] – prefixSum[l-1]. Number of elements in range l to r will be r – l + 1. So we can now print mean of range l to r in O(1). 

C++

// CPP program to find floor value
// of mean in range l to r
#include <bits/stdc++.h>
#define MAX 1000005
using namespace std;

int prefixSum[MAX];

// To calculate prefixSum of array
void calculatePrefixSum(int arr[], int n)
{
    // Calculate prefix sum of array
    prefixSum[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefixSum[i] = prefixSum[i - 1] + arr[i];
}

// To return floor of mean
// in range l to r
int findMean(int l, int r)
{
    if (l == 0)
      return floor(prefixSum[r]/(r+1));

    // Sum of elements in range l to
    // r is prefixSum[r] - prefixSum[l-1]
    // Number of elements in range
    // l to r is r - l + 1
    return floor((prefixSum[r] -
          prefixSum[l - 1]) / (r - l + 1));
}

// Driver program to test above functions
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    calculatePrefixSum(arr, n);
    cout << findMean(0, 2) << endl;
    cout << findMean(1, 3) << endl;
    cout << findMean(0, 4) << endl;
    return 0;
}

Output:

2
3
3

Time complexity: O(n+q) where q is the number of queries and n is the size of the array. Here in the above code q is 3 as the findMean function is used 3 times.
Auxiliary Space: O(k) where k=1000005.

Please refer complete article on Mean of range in array for more details!