Find the subarray with least average (original) (raw)

Last Updated : 16 Jul, 2022

Given an array arr[] of size n and integer k such that k <= n.

Examples :

Input: arr[] = {3, 7, 90, 20, 10, 50, 40}, k = 3 Output: Subarray between indexes 3 and 5 The subarray {20, 10, 50} has the least average among all subarrays of size 3.

Input: arr[] = {3, 7, 5, 20, -10, 0, 12}, k = 2 Output: Subarray between [4, 5] has minimum average

We strongly recommend that you click here and practice it, before moving on to the solution.

A Simple Solution is to consider every element as beginning of subarray of size k and compute sum of subarray starting with this element. Time complexity of this solution is O(nk).

Try It Yourself

Java `

// java program code of Naive approach to // find subarray with minimum average import java.util.*;

class GFG {

static void findsubarrayleast(int arr[], int k, int n)
{
    int min = Integer.MAX_VALUE;
    int minindex = 0;
    for (int i = 0; i <= n - k; i++) {
        int sum = 0;
        for (int j = i; j < i + k; j++) {
            sum += arr[j];
        }
        if (sum < min) {
            min = sum;
            minindex = i;
        }
    }
    // printing the desired subarray
    System.out.println(
        "subarray with minimum average is: ");
    for (int i = minindex; i < minindex + k; i++) {
        System.out.print(arr[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    // Test Case 1
    int arr[] = {20, 3, 13, 5, 10, 14, 8, 5, 11, 9, 1, 11};
    int n = arr.length;
    int k = 9;

    // function call
    findsubarrayleast(arr, k, n);
}

}

// This code is contributed by Aarti_Rathi

C++

//C++ code of Naive approach to //find subarray with minimum average

#include<bits/stdc++.h> using namespace std; //function to find subarray void findsubarrayleast(int arr[],int k,int n){ int min=INT_MAX,minindex; for (int i = 0; i <= n-k; i++) { int sum=0; for (int j = i; j < i+k; j++) { sum+=arr[j]; } if(sum<min){ min=sum; minindex=i; }

}

//printing the desired subarray cout<<"subarray with minimum average is: "; for (int i = minindex; i < minindex+k; i++) { cout<<arr[i]<<" "; }

} //driver code int main() { int arr[]={3, 7, 90, 20, 10, 50, 40}; int n=sizeof(arr)/sizeof(arr[0]),k=3; //function call findsubarrayleast(arr,k,n); return 0; }

//this code is contributed by Machhaliya Muhammad

Python3

Python3 program to find

minimum average subarray

Prints beginning and ending

indexes of subarray of size k

with minimum average

def findsubarrayleast(arr, k, n): min = 999999 minindex = 0 for i in range(n-k+1): sum = 0 j = i for j in range(i, i+k): sum += arr[j] if sum < min: min = sum minindex = i

# printing the desired subarray
print("subarray with minimum average is: ", end='')
for i in range(minindex, minindex+k):
    print(arr[i], end=" ")

Driver Code

arr = [20, 3, 13, 5, 10, 14, 8, 5, 11, 9, 1, 11] k = 9 # Subarray size n = len(arr) findsubarrayleast(arr, k, n)

This code is contributed by Aarti_Rathi

C#

// C# program code of Naive approach to //find subarray with minimum average using System;

class GFG {

static void findsubarrayleast(int []arr,int k,int n){ int min = Int32.MaxValue; int minindex = 0; for (int i = 0; i <= n-k; i++) { int sum = 0; for (int j = i; j < i+k; j++) { sum += arr[j]; } if(sum < min){ min = sum; minindex = i; }

}
//printing the desired subarray
Console.Write("subarray with minimum average is: ");
for (int i = minindex; i < minindex+k; i++)
{
  Console.Write(arr[i]+" ");
}

}

// Driver Code static public void Main() { // Test Case 1 int[] arr = { 3, 7, 90, 20, 10, 50, 40}; int n = arr.Length; int k=3;

// function call
findsubarrayleast(arr,k,n);

} }

// This code is contributed by Aarti_Rathi

JavaScript

// javascript program code of Naive approach to // find subarray with minimum average

function findsubarrayleast(arr, k, n) { var min = Number.MAX_VALUE; var minindex = 0; for (var i = 0; i < n - k; i++) { var sum = 0; for (var j=i; j < i + k; j++) { sum += arr[j]; } if (sum < min) { min = sum; minindex = i; } } // printing the desired subarray console.log("subarray with minimum average is: "); for (var i=minindex; i < minindex + k; i++) { console.log(arr[i] + " "); } }

// Driver Code var arr = [3, 7, 90, 20, 10, 50, 40]; var n = arr.length; var k = 3;

// function call findsubarrayleast(arr, k, n);

// This code is contributed by Aarti_Rathi

`

Output

subarray with minimum average is: 20 10 50

Time Complexity: O(n*k) where n is the size of array.
Auxiliary Space: O(1)

An Efficient Solution is to solve the above problem in O(n) time and O(1) extra space. The idea is to use sliding window of size k. Keep track of sum of current k elements. To compute sum of current window, remove first element of previous window and add current element (last element of current window).

1. Initialize res_index = 0 // Beginning of result index
2. Find sum of first k elements. Let this sum be 'curr_sum'
3. Initialize min_sum = sum
4. Iterate from (k+1)'th to n'th element, do following for every element arr[i] a) curr_sum = curr_sum + arr[i] - arr[i-k] b) If curr_sum < min_sum res_index = (i-k+1)
5. Print res_index and res_index+k-1 as beginning and ending indexes of resultant subarray.

Below is the implementation of above algorithm.

C++ `

// A Simple C++ program to find minimum average subarray #include <bits/stdc++.h> using namespace std;

// Prints beginning and ending indexes of subarray // of size k with minimum average void findMinAvgSubarray(int arr[], int n, int k) { // k must be smaller than or equal to n if (n < k) return;

// Initialize  beginning index of result
int res_index = 0;

// Compute sum of first subarray of size k
int curr_sum = 0;
for (int i = 0; i < k; i++)
    curr_sum += arr[i];

// Initialize minimum sum as current sum
int min_sum = curr_sum;

// Traverse from (k+1)'th element to n'th element
for (int i = k; i < n; i++) {
    // Add current item and remove first item of
    // previous subarray
    curr_sum += arr[i] - arr[i - k];

    // Update result if needed
    if (curr_sum < min_sum) {
        min_sum = curr_sum;
        res_index = (i - k + 1);
    }
}

cout << "Subarray between [" << res_index << ", "
     << res_index + k - 1 << "] has minimum average";

}

// Driver program int main() { int arr[] = { 3, 7, 90, 20, 10, 50, 40 }; int k = 3; // Subarray size int n = sizeof arr / sizeof arr[0]; findMinAvgSubarray(arr, n, k); return 0; }

Java

// A Simple Java program to find // minimum average subarray

class Test {

static int arr[] = new int[] { 3, 7, 90, 20, 10, 50, 40 };

// Prints beginning and ending indexes of subarray
// of size k with minimum average
static void findMinAvgSubarray(int n, int k)
{
    // k must be smaller than or equal to n
    if (n < k)
        return;

    // Initialize beginning index of result
    int res_index = 0;

    // Compute sum of first subarray of size k
    int curr_sum = 0;
    for (int i = 0; i < k; i++)
        curr_sum += arr[i];

    // Initialize minimum sum as current sum
    int min_sum = curr_sum;

    // Traverse from (k+1)'th element to n'th element
    for (int i = k; i < n; i++) 
    {
        // Add current item and remove first
        // item of previous subarray
        curr_sum += arr[i] - arr[i - k];

        // Update result if needed
        if (curr_sum < min_sum) {
            min_sum = curr_sum;
            res_index = (i - k + 1);
        }
    }

    System.out.println("Subarray between [" +
                        res_index + ", " + (res_index + k - 1) +
                        "] has minimum average");
}

// Driver method to test the above function
public static void main(String[] args)
{
    int k = 3; // Subarray size
    findMinAvgSubarray(arr.length, k);
}

}

Python3

Python3 program to find

minimum average subarray

Prints beginning and ending

indexes of subarray of size k

with minimum average

def findMinAvgSubarray(arr, n, k):

# k must be smaller than or equal to n
if (n < k): return 0

# Initialize beginning index of result
res_index = 0

# Compute sum of first subarray of size k
curr_sum = 0
for i in range(k):
    curr_sum += arr[i]

# Initialize minimum sum as current sum
min_sum = curr_sum

# Traverse from (k + 1)'th
# element to n'th element
for i in range(k, n):

    # Add current item and remove first 
    # item of previous subarray
    curr_sum += arr[i] - arr[i-k]

    # Update result if needed
    if (curr_sum < min_sum):
    
        min_sum = curr_sum
        res_index = (i - k + 1)
    
print("Subarray between [", res_index,
      ", ", (res_index + k - 1),
      "] has minimum average")

Driver Code

arr = [3, 7, 90, 20, 10, 50, 40] k = 3 # Subarray size n = len(arr) findMinAvgSubarray(arr, n, k)

This code is contributed by Anant Agarwal.

C#

// A Simple C# program to find // minimum average subarray using System;

class Test {

static int[] arr = new int[] { 3, 7, 90, 20, 10, 50, 40 };

// Prints beginning and ending indexes of subarray
// of size k with minimum average
static void findMinAvgSubarray(int n, int k)
{
    // k must be smaller than or equal to n
    if (n < k)
        return;

    // Initialize beginning index of result
    int res_index = 0;

    // Compute sum of first subarray of size k
    int curr_sum = 0;
    for (int i = 0; i < k; i++)
        curr_sum += arr[i];

    // Initialize minimum sum as current sum
    int min_sum = curr_sum;

    // Traverse from (k+1)'th element to n'th element
    for (int i = k; i < n; i++) 
    {
        // Add current item and remove first item of
        // previous subarray
        curr_sum += arr[i] - arr[i - k];

        // Update result if needed
        if (curr_sum < min_sum) {
            min_sum = curr_sum;
            res_index = (i - k + 1);
        }
    }

    Console.Write("Subarray between [" + res_index + ", " +
                 (res_index + k - 1) + 
                 "] has minimum average");
}

// Driver method to test the above function
public static void Main()
{
    int k = 3; // Subarray size
    findMinAvgSubarray(arr.Length, k);
}

}

// This code is contributed by nitin mittal.

PHP

JavaScript

`

Output

Subarray between [3, 5] has minimum average

Time Complexity: O(n)
Auxiliary Space: O(1)