Count negative elements present in every K-length subarray

Difficulty Level : Basic
Last Updated : 30 Jun, 2021

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

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// negative elements in every window
// of size K
void countNegative(vector<int> arr, int k)
{
     
    // Initialize the window pointers
    int i = 0;
    int j = 0;
 
    // Store the count of negative numbers
    int count = 0;
    int n = arr.size();
 
    // Traverse the array, arr[]
    while (j < n)
    {
         
        // Increase the count
        // if element is less then 0
        if (arr[j] < 0)
        {
            count++;
        }
 
        // If size of the window equal to k
        if (j - i + 1 == k)
        {
            cout << count << " ";
 
            // If the first element of
            // the window is less than 0,
            // decrement count by 1
            if (arr[i] < 0)
            {
                count--;
            }
            i++;
        }
        j++;
    }
}
 
// Driver Code
int main()
{
     
    // Given Input
    vector<int> arr{ -1, 2, -2, 3, 5, -7, -5 };
    int k = 3;
 
    // Function Call
    countNegative(arr, k);
}
     
// This code is contributed by bgangwar59

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to count the number of
    // negative elements in every window
    // of size K
    public static void countNegative(int[] arr, int k)
    {
        // Initialize the window pointers
        int i = 0;
        int j = 0;
 
        // Store the count of negative numbers
        int count = 0;
        int n = arr.length;
 
        // Traverse the array, arr[]
        while (j < n) {
 
            // Increase the count
            // if element is less then 0
            if (arr[j] < 0) {
                count++;
            }
 
            // If size of the window equal to k
            if (j - i + 1 == k) {
                System.out.print(count + " ");
 
                // If the first element of
                // the window is less than 0,
                // decrement count by 1
                if (arr[i] < 0) {
                    count--;
                }
                i++;
            }
 
            j++;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given Input
        int[] arr = { -1, 2, -2, 3, 5, -7, -5 };
        int k = 3;
 
        // Function Call
        countNegative(arr, k);
    }
}

Python3

# Function to count the number of
# negative elements in every window
# of size K
def countNegative(arr,k):
     
    # Initialize the window pointers
    i = 0
    j = 0
     
    # Store the count of negative numbers
    count = 0
    n = len(arr)
     
    while(j < n):
         
        # Increase the count
        # if element is less then 0
         
        if (arr[j] < 0):
            count = count + 1
             
        # If size of the window equal to k  
        if (j - i + 1 == k):
            print(count,end=" ")
             
            # If the first element of
            # the window is less than 0,
            # decrement count by 1
             
            if(arr[i] < 0):
                count = count - 1
             
            i = i+1
        j = j+1
         
# Driver Code
 
# Given Input
arr = [-1, 2, -2, 3, 5, -7, -5]
k = 3
countNegative(arr, k)
 
# This code is contributed by abhinavjain194.

C#

// C#  program for the above approach
using System;
 
class GFG{
 
// Function to count the number of
// negative elements in every window
// of size K
public static void countNegative(int[] arr, int k)
{
     
    // Initialize the window pointers
    int i = 0;
    int j = 0;
 
    // Store the count of negative numbers
    int count = 0;
    int n = arr.Length;
 
    // Traverse the array, arr[]
    while (j < n)
    {
         
        // Increase the count
        // if element is less then 0
        if (arr[j] < 0)
        {
            count++;
        }
 
        // If size of the window equal to k
        if (j - i + 1 == k)
        {
            Console.Write(count + " ");
 
            // If the first element of
            // the window is less than 0,
            // decrement count by 1
            if (arr[i] < 0)
            {
                count--;
            }
            i++;
        }
        j++;
    }
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Input
    int[] arr = { -1, 2, -2, 3, 5, -7, -5 };
    int k = 3;
 
    // Function Call
    countNegative(arr, k);
}
}    
 
// This code is contributed by ukasp

Javascript

<script>
// Javascript program for the above approach
 
// Function to count the number of
// negative elements in every window
// of size K
function countNegative(arr, k)
{
     
    // Initialize the window pointers
    var i = 0;
    var j = 0;
 
    // Store the count of negative numbers
    var count = 0;
    var n = arr.length;
 
    // Traverse the array, arr[]
    while (j < n)
    {
         
        // Increase the count
        // if element is less then 0
        if (arr[j] < 0)
        {
            count++;
        }
 
        // If size of the window equal to k
        if (j - i + 1 == k)
        {
            document.write( count + " ");
 
            // If the first element of
            // the window is less than 0,
            // decrement count by 1
            if (arr[i] < 0)
            {
                count--;
            }
            i++;
        }
        j++;
    }
}
 
var arr = [ -1, 2, -2, 3, 5, -7, -5 ];
var k = 3;
 
// Function Call
countNegative(arr, k);
 
//This code is contributed by SoumikMondal
</script>

Output:

2 1 1 1 2

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

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