Longest subarray having average greater than or equal to x

Given an array of integers and an integer x. Find length of maximum size subarray having average of integers greater than or equal to x.

Examples:

Input : arr[] = {-2, 1, 6, -3}, x = 3
Output : 2
Longest subarray is {1, 6} having average
3.5 greater than x = 3.

Input : arr[] = {2, -3, 3, 2, 1}, x = 2
Output : 3
Longest subarray is {3, 2, 1} having 
average 2 equal to x = 2.

A simple solution is to one by one consider each subarray and find its average. If average is greater than or equal to x, then compare length of subarray with maximum length found so far. Time complexity of this solution is O(n2).

An efficient solution is to use binary search and prefix sum. Let the required Longest subarray be arr[i..j] and its length is l = j-i+1. Thus, mathematically it can be written as:



(Σ (arr[i..j]) / l ) ≥ x
==> (Σ (arr[i..j]) / l) – x ≥ 0
==> (Σ (arr[i..j]) – x*l) / l ≥ 0 …(1)

The equation (1) is equivalent to subtracting x from each element of subarray and then taking average of resultant subarray. So the resultant subarray can be obtained by subtracting x from each element of array. Let the updated subarray is arr1. Equation (1) can be further simplified as:

Σ (arr1[i..j]) / l ≥ 0
==> Σ (arr1[i..j]) ≥ 0 …(2)

Equation (2) is simply Longest subarray having sum greater than or equal to zero. The Longest subarray having sum greater than or equal to zero can be found by method discussed in following article:
Longest subarray having sum greater than k.

The step wise algo is:
1. Subtract x from each element of array.
2. Find Longest subarray having sum greater than or equal to zero in updated array using prefix sum and binary search.

Below is the implementation of above approach:

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// CPP program to find Longest subarray
// having average greater than or equal
// to x.
#include <bits/stdc++.h>
  
using namespace std;
  
// Comparison function used to sort preSum vector.
bool compare(const pair<int, int>& a, const pair<int, int>& b)
{
    if (a.first == b.first)
        return a.second < b.second;
  
    return a.first < b.first;
}
  
// Function to find index in preSum vector upto which
// all prefix sum values are less than or equal to val.
int findInd(vector<pair<int, int> >& preSum, int n, int val)
{
  
    // Starting and ending index of search space.
    int l = 0;
    int h = n - 1;
    int mid;
  
    // To store required index value.
    int ans = -1;
  
    // If middle value is less than or equal to
    // val then index can lie in mid+1..n
    // else it lies in 0..mid-1.
    while (l <= h) {
        mid = (l + h) / 2;
        if (preSum[mid].first <= val) {
            ans = mid;
            l = mid + 1;
        }
        else
            h = mid - 1;
    }
  
    return ans;
}
  
// Function to find Longest subarray having average
// greater than or equal to x.
int LongestSub(int arr[], int n, int x)
{
    int i;
  
    // Update array by subtracting x from
    // each element.
    for (i = 0; i < n; i++)
        arr[i] -= x;
  
    // Length of Longest subarray.
    int maxlen = 0;
  
    // Vector to store pair of prefix sum
    // and corresponding ending index value.
    vector<pair<int, int> > preSum;
  
    // To store current value of prefix sum.
    int sum = 0;
  
    // To store minimum index value in range
    // 0..i of preSum vector.
    int minInd[n];
  
    // Insert values in preSum vector.
    for (i = 0; i < n; i++) {
        sum = sum + arr[i];
        preSum.push_back({ sum, i });
    }
  
    sort(preSum.begin(), preSum.end(), compare);
  
    // Update minInd array.
    minInd[0] = preSum[0].second;
  
    for (i = 1; i < n; i++) {
        minInd[i] = min(minInd[i - 1], preSum[i].second);
    }
  
    sum = 0;
    for (i = 0; i < n; i++) {
        sum = sum + arr[i];
  
        // If sum is greater than or equal to 0,
        // then answer is i+1.
        if (sum >= 0)
            maxlen = i + 1;
  
        // If sum is less than 0, then find if
        // there is a prefix array having sum
        // that needs to be added to current sum to
        // make its value greater than or equal to 0.
        // If yes, then compare length of updated
        // subarray with maximum length found so far.
        else {
            int ind = findInd(preSum, n, sum);
            if (ind != -1 && minInd[ind] < i)
                maxlen = max(maxlen, i - minInd[ind]);
        }
    }
  
    return maxlen;
}
  
// Driver code.
int main()
{
    int arr[] = { -2, 1, 6, -3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    int x = 3;
  
    cout << LongestSub(arr, n, x);
    return 0;
}

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Output:

2

Time Complexity: O(nlogn)
Auxiliary Space: O(n)

Another Approach :
Use approach of https://www.geeksforgeeks.org/largest-sum-contiguous-subarray/ just add a length variable and use the logic if the length is greater then previous saved length.

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// C program to find Longest subarray
// having average greater than or equal
// to x.
#include <stdio.h>
  
// Function to find Longest subarray having average
// greater than or equal to x.
int max_subarray(int arr[],int x,int n)
{
    int length=0 ,temp =0 ;
    int max=0,sum=0,i;
    for(i=0;i<n;i++)
    
        sum = sum+arr[i];
        temp++;
        if(sum>max && sum/temp>=x && temp>=length)
        {
            max= sum ;
            length =temp;
        }
        if(sum<0)
        {
            sum=0;
            temp=0;
        }
          
    }
  
return length; 
}
  
// Drive Code
int main() {
int arr[] = {-2, 1, 6, -3 };
int x = 2;
int n = sizeof(arr)/sizeof(int);
int length = max_subarray(arr,x,n);
printf("%d",length);
return 0;
}

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Output:

2

Time Complexity: O(n)
Auxiliary Space: O(n)



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