First negative integer in every window of size k

Given an array and a positive integer k, find the first negative integer for each window(contiguous subarray) of size k. If a window does not contain a negative integer, then print 0 for that window.

Examples:

Input : arr[] = {-8, 2, 3, -6, 10}, k = 2
Output : -8 0 -6 -6
First negative integer for each window of size k
{-8, 2} = -8
{2, 3} = 0 (does not contain a negative integer)
{3, -6} = -6
{-6, 10} = -6

Input : arr[] = {12, -1, -7, 8, -15, 30, 16, 28} , k = 3
Output : -1 -1 -7 -15 -15 0

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Naive Approach: Run two loops. In the outer loop, take all subarrays(windows) of size k. In the inner loop, get the first negative integer of the current subarray(window).

C++

// C++ implementation to find the first negative  
// integer in every window of size k 
#include <bits/stdc++.h> 
using namespace std; 
   
// function to find the first negative  
// integer in every window of size k 
void printFirstNegativeInteger(int arr[], int n, int k) 
{ 
    // flag to check whether window contains 
    // a negative integer or not 
    bool flag; 
      
    // Loop for each subarray(window) of size k 
    for (int i = 0; i<(n-k+1); i++)            
    { 
        flag = false; 
  
        // traverse through the current window 
        for (int j = 0; j<k; j++) 
        { 
            // if a negative integer is found, then 
            // it is the first negative integer for  
            // current window. Print it, set the flag 
            // and break 
            if (arr[i+j] < 0) 
            { 
                cout << arr[i+j] << " "; 
                flag = true; 
                break; 
            } 
        } 
          
        // if the current window does not 
        // contain a negative integer 
        if (!flag) 
            cout << "0" << " "; 
    }     
} 
   
// Driver program to test above functions 
int main() 
{ 
    int arr[] = {12, -1, -7, 8, -15, 30, 16, 28}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    int k = 3; 
    printFirstNegativeInteger(arr, n, k); 
    return 0; 
} 

Java

// Java implementation to find the first negative  
// integer in every window of size k 
import java.util.*; 
  
class solution 
{ 
  
// function to find the first negative  
// integer in every window of size k 
static void printFirstNegativeInteger(int arr[], int n, int k) 
{ 
    // flag to check whether window contains 
    // a negative integer or not 
    boolean flag; 
      
    // Loop for each subarray(window) of size k 
    for (int i = 0; i<(n-k+1); i++)          
    { 
        flag = false; 
  
        // traverse through the current window 
        for (int j = 0; j<k; j++) 
        { 
            // if a negative integer is found, then 
            // it is the first negative integer for  
            // current window. Print it, set the flag 
            // and break 
            if (arr[i+j] < 0) 
            { 
                System.out.print((arr[i+j])+" "); 
                flag = true; 
                break; 
            } 
        } 
          
        // if the current window does not 
        // contain a negative integer 
        if (!flag) 
            System.out.print("0"+" "); 
    }  
} 
  
// Driver program to test above functions 
public static void main(String args[]) 
{ 
    int arr[] = {12, -1, -7, 8, -15, 30, 16, 28}; 
    int n = arr.length; 
    int k = 3; 
    printFirstNegativeInteger(arr, n, k); 
      
} 
} 
// This code is contributed by 
// Shashank_Sharma 

Python3

# Python3 implementation to find the first negative  
# integer in every window of size k 
  
# Function to find the first negative  
# integer in every window of size k 
def printFirstNegativeInteger(arr, n, k): 
      
    # Loop for each subarray(window) of size k 
    for i in range(0, (n - k + 1)): 
        flag = False
  
        # Traverse through the current window 
        for j in range(0, k): 
          
            # If a negative integer is found, then 
            # it is the first negative integer for  
            # current window. Print it, set the flag 
            # and break 
            if (arr[i + j] < 0): 
          
                print(arr[i + j], end = " ") 
                flag = True
                break
          
        # If the current window does not 
        # contain a negative integer 
        if (not(flag)): 
            print("0", end = " ") 
      
# Driver Code 
arr = [12, -1, -7, 8, -15, 30, 16, 28] 
n = len(arr) 
k = 3
printFirstNegativeInteger(arr, n, k) 
  
# This code is contributed by 'Smitha dinesh semwal' 

C#

// C# implementation to find  
// the first negative integer 
// in every window of size k  
using System; 
  
class GFG 
{  
  
// function to find the first negative  
// integer in every window of size k  
static void printFirstNegativeInteger(int []arr,  
                                    int n, int k)  
{  
    // flag to check whether window contains  
    // a negative integer or not  
    bool flag;  
      
    // Loop for each subarray(window) of size k  
    for (int i = 0; i < (n - k + 1); i++)  
    {  
        flag = false;  
  
        // traverse through the current window  
        for (int j = 0; j < k; j++)  
        {  
            // if a negative integer is found, then  
            // it is the first negative integer for  
            // current window. Print it, set the flag  
            // and break  
            if (arr[i + j] < 0)  
            {  
                Console.Write((arr[i + j]) + " ");  
                flag = true;  
                break;  
            }  
        }  
          
        // if the current window does not  
        // contain a negative integer  
        if (!flag)  
            Console.Write("0" + " ");  
    }  
}  
  
// Driver code 
public static void Main(String []args)  
{  
    int []arr = {12, -1, -7, 8, -15, 30, 16, 28};  
    int n = arr.Length;  
    int k = 3;  
    printFirstNegativeInteger(arr, n, k);  
}  
}  
  
// This code has been contributed 
// by 29AjayKumar 

Output :

-1 -1 -7 -15 -15 0

Time Complexity : The outer loop runs n-k+1 times and the inner loop runs k times for every iteration of outer loop. So time complexity is O((n-k+1)*k) which can also be written as O(nk) when k is comparitively much smaller than n, otherwise when k tends to reach n, complexity becomes O(k).

Efficient Approach: It is a variation of the problem of Sliding Window Maximum.
We create a Dequeue, Di of capacity k, that stores only useful elements of the current window of k elements. An element is useful if it is in the current window and it is a negative integer. We process all array elements one by one and maintain Di to contain useful elements of current window and these useful elements are all negative integers. For a particular window, if Di is not empty then the element at front of the Di is the first negative integer for that window, else that window does not contain a negative integer.

C++

// C++ implementation to find the first negative  
// integer in every window of size k 
#include <bits/stdc++.h> 
  
using namespace std; 
   
// function to find the first negative  
// integer in every window of size k 
void printFirstNegativeInteger(int arr[], int n, int k) 
{ 
    // A Double Ended Queue, Di that will store indexes of  
    // useful array elements for the current window of size k. 
    // The useful elements are all negative integers. 
    deque<int>  Di; 
   
    /* Process first k (or first window) elements of array */
    int i; 
    for (i = 0; i < k; i++) 
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.push_back(i); 
      
    // Process rest of the elements, i.e., from arr[k] to arr[n-1] 
    for ( ; i < n; i++) 
    { 
        // if Di is not empty then the element at the 
        // front of the queue is the first negative integer 
        // of the previous window 
        if (!Di.empty()) 
            cout << arr[Di.front()] << " "; 
          
        // else the window does not have a 
        // negative integer 
        else
            cout << "0" << " "; 
   
        // Remove the elements which are out of this window 
        while ( (!Di.empty()) && Di.front() < (i - k + 1)) 
            Di.pop_front();  // Remove from front of queue 
   
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.push_back(i); 
    } 
   
    // Print the first negative  
    // integer of last window 
    if (!Di.empty()) 
           cout << arr[Di.front()] << " "; 
    else
        cout << "0" << " ";        
      
} 
   
// Driver program to test above functions 
int main() 
{ 
    int arr[] = {12, -1, -7, 8, -15, 30, 16, 28}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    int k = 3; 
    printFirstNegativeInteger(arr, n, k); 
    return 0; 
} 

Java

// Java implementation to find the  
// first negative integer in  
// every window of size k 
import java.util.*; 
class GFG  
{  
  
// function to find the first negative  
// integer in every window of size k 
static void printFirstNegativeInteger(int arr[],  
                                      int n, int k) 
{ 
    // A Double Ended Queue, Di that will  
    // store indexes of useful array elements  
    // for the current window of size k. 
    // The useful elements are all negative integers. 
    LinkedList<Integer> Di = new LinkedList<>(); 
  
    // Process first k (or first window) 
    // elements of array 
    int i; 
    for (i = 0; i < k; i++) 
      
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.add(i); 
      
    // Process rest of the elements,  
    // i.e., from arr[k] to arr[n-1] 
    for ( ; i < n; i++) 
    { 
        // if Di is not empty then the element  
        // at the front of the queue is the first  
        // negative integer of the previous window 
        if (!Di.isEmpty()) 
            System.out.print(arr[Di.peek()] + " "); 
          
        // else the window does not have a 
        // negative integer 
        else
            System.out.print("0" + " "); 
  
        // Remove the elements which are 
        // out of this window 
        while ((!Di.isEmpty()) &&  
                 Di.peek() < (i - k + 1)) 
            Di.remove(); // Remove from front of queue 
  
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.add(i); 
    } 
  
    // Print the first negative  
    // integer of last window 
    if (!Di.isEmpty()) 
        System.out.print(arr[Di.peek()] + " "); 
    else
        System.out.print("0" + " ");      
} 
  
// Driver Code 
public static void main(String[] args)  
{ 
    int arr[] = {12, -1, -7, 8, -15, 30, 16, 28}; 
    int n = arr.length; 
    int k = 3; 
    printFirstNegativeInteger(arr, n, k); 
} 
} 
  
// This code is contributed by PrinciRaj1992  

Python3

# Python3 implementation to find the  
# first negative integer in every window  
# of size k import deque() from collections  
from collections import deque 
  
# function to find the first negative  
# integer in every window of size k 
def printFirstNegativeInteger(arr, n, k): 
      
    # A Double Ended Queue, Di that will store 
    # indexes of useful array elements for the  
    # current window of size k. The useful  
    # elements are all negative integers. 
    Di = deque() 
  
    # Process first k (or first window) 
    # elements of array  
    for i in range(k): 
          
        # Add current element at the rear of Di 
        # if it is a negative integer 
        if (arr[i] < 0): 
            Di.append(i); 
      
    # Process rest of the elements, i.e.,  
    # from arr[k] to arr[n-1] 
    for i in range(k, n): 
          
        # if the window does not have 
        # a negative integer 
        if (not Di): 
            print(0, end = ' ') 
          
        # if Di is not empty then the element  
        # at the front of the queue is the first  
        # negative integer of the previous window 
        else: 
            print(arr[Di[0]], end = ' '); 
  
        # Remove the elements which are 
        # out of this window 
        while Di and Di[0] <= (i - k): 
            Di.popleft() # Remove from front of queue 
  
        # Add current element at the rear of Di 
        # if it is a negative integer 
        if (arr[i] < 0): 
            Di.append(i); 
  
    # Print the first negative  
    # integer of last window 
    if not Di: 
        print(0) 
    else: 
        print(arr[Di[0]], end = " ")  
      
# Driver Code 
if __name__ =="__main__": 
    arr = [12, -1, -7, 8, -15, 30, 16, 28] 
    n = len(arr) 
    k = 3
    printFirstNegativeInteger(arr, n, k); 
  
# This code is contributed by 
# chaudhary_19 (Mayank Chaudhary) 

C#

// C# implementation to find the  
// first negative integer in  
// every window of size k 
using System; 
using System.Collections.Generic; 
  
class GFG  
{  
  
// function to find the first negative  
// integer in every window of size k 
static void printFirstNegativeint(int []arr,  
                                  int n, int k) 
{ 
    // A Double Ended Queue, Di that will  
    // store indexes of useful array elements  
    // for the current window of size k. 
    // The useful elements are all  
    // negative integers. 
    List<int> Di = new List<int>(); 
  
    // Process first k (or first window) 
    // elements of array 
    int i; 
    for (i = 0; i < k; i++) 
      
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.Add(i); 
      
    // Process rest of the elements,  
    // i.e., from arr[k] to arr[n-1] 
    for ( ; i < n; i++) 
    { 
        // if Di is not empty then the element  
        // at the front of the queue is the first  
        // negative integer of the previous window 
        if (Di.Count != 0) 
            Console.Write(arr[Di[0]] + " "); 
          
        // else the window does not have a 
        // negative integer 
        else
            Console.Write("0" + " "); 
  
        // Remove the elements which are 
        // out of this window 
        while ((Di.Count != 0) &&  
                Di[0] < (i - k + 1)) 
                  
            // Remove from front of queue 
            Di.RemoveAt(0);  
  
        // Add current element at the rear of Di 
        // if it is a negative integer 
        if (arr[i] < 0) 
            Di.Add(i); 
    } 
  
    // Print the first negative  
    // integer of last window 
    if (Di.Count!=0) 
        Console.Write(arr[Di[0]] + " "); 
    else
        Console.Write("0" + " ");      
} 
  
// Driver Code 
public static void Main(String[] args)  
{ 
    int []arr = {12, -1, -7, 8, -15, 30, 16, 28}; 
    int n = arr.Length; 
    int k = 3; 
    printFirstNegativeint(arr, n, k); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Output:

-1 -1 -7 -15 -15 0

Time Complexity: O(n)
Auxiliary Space: O(k)

Optimized Approach:: It is also possible to accomplish this with constant space. The idea is to have a variable firstNegativeIndex to keep track of the first negative element in the k sized window. At every iteration, we skip the elements which no longer fall under the current k size window (firstNegativeIndex <= i – k) as well as the positive elements.
Below is the solution based upon this approach.

Python3

def printFirstNegativeInteger(arr, k): 
    firstNegativeIndex = 0
  
    for i in range(k - 1, len(arr)): 
  
        # skip out of window and positive elements 
        while firstNegativeIndex < i and (firstNegativeIndex <= i - k or arr[firstNegativeIndex] > 0): 
            firstNegativeIndex += 1
  
        # check if a negative element is found, otherwise use 0 
        firstNegativeElement = arr[firstNegativeIndex] if arr[firstNegativeIndex] < 0 else 0
        print(firstNegativeElement, end=' ') 
  
  
if __name__ == "__main__": 
    arr = [12, -1, -7, 8, -15, 30, 16, 28] 
    k = 3
    printFirstNegativeInteger(arr, k) 
  
# contributed by Arjun Lather 

Output:

-1 -1 -7 -15 -15 0

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

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeek’s main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

Python3

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