Count negative elements present in every K-length subarray

Given an array arr[] of size N and an integer K, the task is to count the number of negative elements present in all K-length subarrays.

Example:

Input: arr[] = {-1, 2, -2, 3, 5, -7, -5}, K = 3
Output: 2 1 1 1 2
Explanation:
First Subarray: {-1, 2, -2}. Count of negative numbers = 2.
Second Subarray: {2, -2, 3}. Count of negative numbers = 1.
Third Subarray: {-2, 3, 5}. Count of negative numbers = 1.
Fourth Subarray: {3, 5, -7}. Count of negative numbers = 1.
Fifth Subarray: {5, -7, -5}. Count of negative numbers = 2.

Input: arr[] = {-1, 2, 4, 4}, K = 2
Output: 1 0 0

Naive Approach: The simplest approach is to traverse the given array considering every window of size K and find the count of negative numbers in every window.
Time Complexity: O(N*K)
Auxiliary Space: O(1)

Efficient Approach: This problem can be solved using the window sliding technique. Follow the steps below to solve the problem:

• Initialize a variable count as 0 to store the count of negative elements in a window of size K.
• Initialize two variables i and j as 0 to store the first and last index of the window respectively.
• Loop while j<N and perform the following steps:
  • If arr[j] < 0, increment count by 1.
  • If the size of the window, i.e, j-i+1 is equal to K, print the value of count, and check if arr[i] < 0, then decrement count by 1. Also, increment i by 1.
  • Increment the value of j by 1.

Below is the implementation of the above approach

Java

import java.io.*; import java.util.*;    class GFG {                       public static void countNegative(int[] arr, int k)     {                  int i = 0;         int j = 0;                     int count = 0;         int n = arr.length;                     while (j < n) {                                          if (arr[j] < 0) {                 count++;             }                             if (j - i + 1 == k) {                 System.out.print(count + " ");                                                                       if (arr[i] < 0) {                     count--;                 }                 i++;             }                j++;         }     }             public static void main(String[] args)     {                  int[] arr = { -1, 2, -2, 3, 5, -7, -5 };         int k = 3;                     countNegative(arr, k);     } }

Output:

2 1 1 1 2

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

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