Given an array, arr[] of size N, two positive integers K and S, the task is to find the length of the smallest subarray of size greater than K, whose sum is greater than S.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 1, S = 8
Output: 2
Explanation:
Smallest subarray with sum greater than S(=8) is {4, 5}
Therefore, the required output is 2.Input: arr[] = {1, 3, 5, 1, 8, 2, 4}, K= 2, S= 13
Output: 3
Naive Approach
The idea is to find all subarrays and in that choose those subarrays whose sum is greater than S and whose length is greater than K. After that from those subarrays, choose that subarray whose length is minimum. Then print the length of that subarray.
Steps that were to follow the above approach:
- Make a variable “ans” and initialize it with the maximum value
- After that run two nested loops to find all subarrays
- While finding all subarray calculate their size and sum of all elements of that subarray
- If the sum of all elements is greater than S and its size is greater than K, then update answer with minimum of answer and length of the subarray
Below is the code to implement the above approach:
C++
// C++ program to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to find the length of the smallest subarray of// size > K with sum greater than Sint smallestSubarray(int K, int S, int arr[], int N){ //To store answer int ans=INT_MAX; //Traverse the array for(int i=0;i<N;i++){ //To store size of subarray int size=0; //To store sum of all elements of subarray int sum=0; for(int j=i;j<N;j++){ size++; sum+=arr[j]; //when size of subarray is greater than k and sum of all element //is greater than S then if size is less than previously stored answer //then update answer with size if(size>K && sum>S){ if(size<ans){ans=size;} } } } return ans;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = sizeof(arr) / sizeof(arr[0]); cout << smallestSubarray(K, S, arr, N);} |
Java
import java.util.*;public class Main{ // Function to find the length of the smallest subarray of // size > K with sum greater than S static int smallestSubarray(int K, int S, int[] arr, int N) { // To store answer int ans = Integer.MAX_VALUE; // Traverse the array for (int i = 0; i < N; i++) { // To store size of subarray int size = 0; // To store sum of all elements of subarray int sum = 0; for (int j = i; j < N; j++) { size++; sum += arr[j]; // when size of subarray is greater than k and sum of all element // is greater than S then if size is less than previously stored answer // then update answer with size if (size > K && sum > S) { if (size < ans) { ans = size; } } } } return ans; } // Driver Code public static void main(String[] args) { int[] arr = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = arr.length; System.out.println(smallestSubarray(K, S, arr, N)); }} |
Python3
# Function to find the length of the smallest subarray of# size > K with sum greater than Sdef smallestSubarray(K, S, arr, N): # To store answer ans = float('inf') # Traverse the array for i in range(N): # To store size of subarray size = 0 # To store sum of all elements of subarray sum = 0 for j in range(i, N): size += 1 sum += arr[j] # when size of subarray is greater than k and sum of all element # is greater than S then if size is less than previously stored answer # then update answer with size if size > K and sum > S: if size < ans: ans = size return ans# Driver Codearr = [1, 2, 3, 4, 5]K = 1S = 8N = len(arr)print(smallestSubarray(K, S, arr, N)) |
C#
// C# program to implement the above approachusing System;class GFG { // Function to find the length of the smallest subarray // of size > K with sum greater than S static int SmallestSubarray(int K, int S, int[] arr, int N) { // To store the answer int ans = int.MaxValue; // Traverse the array for (int i = 0; i < N; i++) { // To store the size of the subarray int size = 0; // To store the sum of all elements of the // subarray int sum = 0; for (int j = i; j < N; j++) { size++; sum += arr[j]; // When the size of the subarray is greater // than K and the sum of all elements is // greater than S, then if the size is less // than the previously stored answer update // the answer with the size if (size > K && sum > S) { if (size < ans) { ans = size; } } } } return ans; } // Driver Code public static void Main(string[] args) { int[] arr = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = arr.Length; Console.WriteLine(SmallestSubarray(K, S, arr, N)); }}// This code is contributed by shivamgupta0987654321 |
Javascript
// Function to find the length of the smallest subarray of// size > K with sum greater than Sfunction smallestSubarray(K, S, arr, N) { // To store answer let ans = Infinity; // Traverse the array for (let i = 0; i < N; i++) { // To store size of subarray let size = 0; // To store sum of all elements of subarray let sum = 0; for (let j = i; j < N; j++) { size += 1; sum += arr[j]; // when size of subarray is greater than k and sum of all element // is greater than S then if size is less than previously stored answer // then update answer with size if (size > K && sum > S) { if (size < ans) { ans = size; } } } } return ans;}// Driver Codelet arr = [1, 2, 3, 4, 5];let K = 1;let S = 8;let N = arr.length;console.log(smallestSubarray(K, S, arr, N)); |
Output-
2
Time Complexity: O(N2), because of two nested for loops
Auxiliary Space:O(1) , because no extra space has been used
Approach: The problem can be solved using Sliding Window Technique. Follow the steps below to solve the problem:
- Initialize two variables say, i = 0 and j = 0 both pointing to the start of array i.e index 0.
- Initialize a variable sum to store the sum of the subArray currently being processed.
- Traverse the array, arr[] and by incrementing j and adding arr[j]
- Take our the window length or the length of the current subArray which is given by j-i+1 (+1 because the indexes start from zero) .
- Firstly, check if the size of the current subArray i.e winLen here is greater than K. if this is not the case increment the j value and continue the loop.
- Else , we get that the size of the current subArray is greater than K, now we have to check if we meet the second condition i.e sum of the current Subarray is greater than S.
- If this is the case, we update minLength variable which stores the minimum length of the subArray satisfying the above conditions.
- At this time , we check if the size of the subArray can be reduced (by incrementing i ) such that it still is greater than K and sum is also greater than S. We constantly remove the ith element of the array from the sum to reduce the subArray size in the While loop and then increment j such that we move to the next element in the array .the
- Finally, print the minimum length of required subarray obtained that satisfies the conditions.
Below is the implementation of the above approach:
C++
// C++ program to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to find the length of the smallest subarray of// size > K with sum greater than Sint smallestSubarray(int K, int S, int arr[], int N){ // Store the first index of the current subarray int start = 0; // Store the last index of the current subarray int end = 0; // Store the sum of the current subarray int currSum = arr[0]; // Store the length of the smallest subarray int res = INT_MAX; while (end < N - 1) { // If sum of the current subarray <= S or length of // current subarray <= K if (currSum <= S || (end - start + 1) <= K) { // Increase the subarray sum and size currSum += arr[++end]; } else { // Update to store the minimum size of subarray // obtained res = min(res, end - start + 1); // Decrement current subarray size by removing // first element currSum -= arr[start++]; } } // Check if it is possible to reduce the length of the // current window while (start < N) { if (currSum > S && (end - start + 1) > K) res = min(res, (end - start + 1)); currSum -= arr[start++]; } return res;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = sizeof(arr) / sizeof(arr[0]); cout << smallestSubarray(K, S, arr, N);}// This code is contributed by Sania Kumari Gupta |
C
// C program to implement the above approach#include <limits.h>#include <stdio.h>// Find minimum between two numbers.int min(int num1, int num2){ return (num1 > num2) ? num2 : num1;}// Function to find the length of the smallest subarray of// size > K with sum greater than Sint smallestSubarray(int K, int S, int arr[], int N){ // Store the first index of the current subarray int start = 0; // Store the last index of the current subarray int end = 0; // Store the sum of the current subarray int currSum = arr[0]; // Store the length of the smallest subarray int res = INT_MAX; while (end < N - 1) { // If sum of the current subarray <= S or length of // current subarray <= K if (currSum <= S || (end - start + 1) <= K) { // Increase the subarray sum and size currSum += arr[++end]; } else { // Update to store the minimum size of subarray // obtained res = min(res, end - start + 1); // Decrement current subarray size by removing // first element currSum -= arr[start++]; } } // Check if it is possible to reduce the length of the // current window while (start < N) { if (currSum > S && (end - start + 1) > K) res = min(res, (end - start + 1)); currSum -= arr[start++]; } return res;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = sizeof(arr) / sizeof(arr[0]); printf("%d", smallestSubarray(K, S, arr, N));}// This code is contributed by Sania Kumari Gupta |
Java
// Java program to implement// the above approachimport java.io.*;class GFG{// Function to find the length of the// smallest subarray of size > K with// sum greater than Spublic static int smallestSubarray(int k, int s, int[] array, int N){ int i=0; int j=0; int minLen = Integer.MAX_VALUE; int sum = 0; while(j < N) { sum += array[j]; int winLen = j-i+1; if(winLen <= k) j++; else{ if(sum > s) { minLen = Math.min(minLen,winLen); while(sum > s) { sum -= array[i]; i++; } j++; } } } return minLen;}// Driver Codepublic static void main(String[] args){ int[] arr = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = arr.length; System.out.print(smallestSubarray(K, S, arr, N));}}// This code is contributed by akhilsaini |
Python3
# Python3 program to implement# the above approachimport sys# Function to find the length of the# smallest subarray of size > K with# sum greater than Sdef smallestSubarray(K, S, arr, N): # Store the first index of # the current subarray start = 0 # Store the last index of # the current subarray end = 0 # Store the sum of the # current subarray currSum = arr[0] # Store the length of # the smallest subarray res = sys.maxsize while end < N - 1: # If sum of the current subarray <= S # or length of current subarray <= K if ((currSum <= S) or ((end - start + 1) <= K)): # Increase the subarray # sum and size end = end + 1; currSum += arr[end] # Otherwise else: # Update to store the minimum # size of subarray obtained res = min(res, end - start + 1) # Decrement current subarray # size by removing first element currSum -= arr[start] start = start + 1 # Check if it is possible to reduce # the length of the current window while start < N: if ((currSum > S) and ((end - start + 1) > K)): res = min(res, (end - start + 1)) currSum -= arr[start] start = start + 1 return res;# Driver Codeif __name__ == "__main__": arr = [ 1, 2, 3, 4, 5 ] K = 1 S = 8 N = len(arr) print(smallestSubarray(K, S, arr, N))# This code is contributed by akhilsaini |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to find the length of the// smallest subarray of size > K with// sum greater than Sstatic int smallestSubarray(int K, int S, int[] arr, int N){ // Store the first index of // the current subarray int start = 0; // Store the last index of // the current subarray int end = 0; // Store the sum of the // current subarray int currSum = arr[0]; // Store the length of // the smallest subarray int res = int.MaxValue; while (end < N - 1) { // If sum of the current subarray <= S // or length of current subarray <= K if (currSum <= S || (end - start + 1) <= K) { // Increase the subarray // sum and size currSum += arr[++end]; } // Otherwise else { // Update to store the minimum // size of subarray obtained res = Math.Min(res, end - start + 1); // Decrement current subarray // size by removing first element currSum -= arr[start++]; } } // Check if it is possible to reduce // the length of the current window while (start < N) { if (currSum > S && (end - start + 1) > K) res = Math.Min(res, (end - start + 1)); currSum -= arr[start++]; } return res;}// Driver Codestatic public void Main(){ int[] arr = { 1, 2, 3, 4, 5 }; int K = 1, S = 8; int N = arr.Length; Console.Write(smallestSubarray(K, S, arr, N));}}// This code is contributed by akhilsaini |
Javascript
<script>// JavaScript program to implement // the above approach // Function to find the length of the // smallest subarray of size > K with // sum greater than S function smallestSubarray(K, S, arr, N) { // Store the first index of // the current subarray let start = 0; // Store the last index of // the current subarray let end = 0; // Store the sum of the // current subarray let currSum = arr[0]; // Store the length of // the smallest subarray let res = Number.MAX_SAFE_INTEGER; while (end < N - 1) { // If sum of the current subarray <= S // or length of current subarray <= K if (currSum <= S || (end - start + 1) <= K) { // Increase the subarray // sum and size currSum += arr[++end]; } // Otherwise else { // Update to store the minimum // size of subarray obtained res = Math.min(res, end - start + 1); // Decrement current subarray // size by removing first element currSum -= arr[start++]; } } // Check if it is possible to reduce // the length of the current window while (start < N) { if (currSum > S && (end - start + 1) > K) res = Math.min(res, (end - start + 1)); currSum -= arr[start++]; } return res; } // Driver Code let arr = [ 1, 2, 3, 4, 5 ]; let K = 1, S = 8; let N = arr.length; document.write(smallestSubarray(K, S, arr, N)); // This code is contributed by Surbhi tyagi.</script> |
Output
2
Time Complexity: O(N)
Auxiliary Space:O(1)
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