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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