Find the subarray with least average
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
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).
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++ 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 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# 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 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
Outputsubarray 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;
}
// 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 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.
// 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
// A Simple PHP program to find
// minimum average subarray
// Prints beginning and ending
// indexes of subarray of size
// k with minimum average
function findMinAvgSubarray($arr, $n, $k)
{
// k must be smaller
// than or equal to n
if ($n < $k)
return;
// Initialize beginning
// index of result
$res_index = 0;
// Compute sum of first
// subarray of size k
$curr_sum = 0;
for ($i = 0; $i < $k; $i++)
$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 = $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 - $k + 1);
}
}
echo "Subarray between [" ,$res_index
, ", " ,$res_index + $k - 1, "] has minimum average";
}
// Driver Code
$arr = array(3, 7, 90, 20, 10, 50, 40);
// Subarray size
$k = 3;
$n = sizeof ($arr) / sizeof ($arr[0]);
findMinAvgSubarray($arr, $n, $k);
return 0;
// This code is contributed by nitin mittal.
?>
<script>
// A Simple JavaScript program to find
// minimum average subarray
// Prints beginning and ending indexes
// of subarray of size k with minimum average
function findMinAvgSubarray(arr, n, k)
{
// k must be smaller than or equal to n
if (n < k)
return;
// Initialize beginning index of result
let res_index = 0;
// Compute sum of first subarray of size k
let curr_sum = 0;
for(let i = 0; i < k; i++)
curr_sum += arr[i];
// Initialize minimum sum as current sum
let min_sum = curr_sum;
// Traverse from (k+1)'th element
// to n'th element
for(let 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);
}
}
document.write("Subarray between [" + res_index +
", " + (res_index + k - 1) +
"] has minimum average");
}
// Driver code
let arr = [ 3, 7, 90, 20, 10, 50, 40 ];
// Subarray size
let k = 3;
let n = arr.length;
findMinAvgSubarray(arr, n, k);
// This code is contributed by Surbhi Tyagi.
</script>
OutputSubarray between [3, 5] has minimum average
Time Complexity: O(n)
Auxiliary Space: O(1)
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