Find the Subarray of given length with Least Average

Difficulty Level Easy
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Problem Statement

In the “Find the Subarray of given length with Least Average” problem we have given an array and an input integer X. Write a program to find the subarray of length X with least/minimum average. Prints the starting and ending indexes of the subarray which has the least average.

Input Format

The first line containing an integer value n.

Second-line containing n space-separated integers.

The third line containing an integer value X.

Output Format

The first and only one line containing two integer values with space-separated. The first integer represents the starting and the second integer represents the ending indexes of the subarray which has the least average.

Constraints

  • 1<=N<=10^5
  • 1<=a[i]<=10^9
  • 1<=X<=N

Example

5
3 5 1 7 6
2
1 2

Explanation: The subarray is between index 1 and 2. We can see clearly that the sum of 5 and  1 is the least.

Algorithm to Find the Subarray of given length with Least Average

In this method, we use the idea of the sliding window of size X.

1. Initialize index =0, which is the starting index of the subarray with the least average

2. Find the sum of the first X elements and store it in sum variable

3. Initialize least_sum to the above sum variable

4. Traverse the array from (X+1)th index till the end of the array

  • For every element arr[i], calculate sum = sum + arr[i] -arr[i-X]
  • If sum < least_sum, then make index = (i-X+1) and least_sum =sum.

5. Print the index and the index + X -1 as the starting and ending of the subarray with least average.

Implementation

C++ Program to Find the Subarray of given length with Least Average

#include <bits/stdc++.h> 
using namespace std; 

int main() 
{ 
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
  int k;
  cin>>k;
  if(n>=k)
  {
      int ans = 0; 
    	int sum = 0; 
    	for(int i=0;i<k;i++) 
    	{
    		sum+=a[i];
    	}
    	int min_sum=sum; 
    	for (int i=k;i<n;i++) 
    	{ 
    		sum+=a[i]-a[i-k]; 
    		if(sum<min_sum) 
    		{ 
    			min_sum = sum; 
    			ans = (i-k+1); 
    		} 
    	} 
    	cout<<ans<<" "<<ans+k-1<<endl;
  }
  else
  {
     cout<<-1<<endl;
  }
  return 0; 
}

Java Program to Find the Subarray of given length with Least Average

import java.util.Scanner;
class sum
{
    public static void main(String[] args)
    {
        Scanner sr = new Scanner(System.in);
        int n=sr.nextInt();
        int a[]= new int[n];
        for(int i=0;i<n;i++)
        {
            a[i]=sr.nextInt();
        }
        int k = sr.nextInt();
        if(n>=k)
        {
            int ans = 0; 
            int sum = 0; 
            for(int i=0;i<k;i++) 
            {
                    sum+=a[i];
            }
            int min_sum=sum; 
            for (int i=k;i<n;i++) 
            { 
                    sum+=a[i]-a[i-k]; 
                    if(sum<min_sum) 
                    { 
                            min_sum = sum; 
                            ans = (i-k+1); 
                    } 
            } 
            System.out.println(ans +" "+(ans+k-1));
        }
        else
        {
            System.out.println(-1);
        } 
    }
}

10
1 4 3 6 23 76 43 1 2 89
4
0 3

Complexity Analysis to Find the Subarray of given length with Least Average

Time Complexity

O(n) where n is the size of the given array a[]. Here we use the concept of the sliding window which take linear time.

Space Complexity

O(1) because we don’t use any auxiliary space at here.

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