每个大小为 k 的窗口中的第一个负整数

给定一个数组和一个正整数 k，找到每个大小为 k 的窗口（连续子数组）的第一个负整数。如果窗口不包含负整数，则为该窗口打印 0。

例子：

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。

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

天真的方法：运行两个循环。在外循环中，取所有大小为 k 的子数组（窗口）。在内循环中，获取当前子数组（窗口）的第一个负整数。

C++

// C++ implementation to find the first negative
// integer in every window of size k
#include 
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

---

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

---

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

---

输出 ：

-1 -1 -7 -15 -15 0

时间复杂度：外循环运行 n-k+1 次，内循环每次外循环迭代运行 k 次。所以时间复杂度是 O((n-k+1)*k) 当 k 远小于 n 时也可以写成 O(nk)，否则当 k 趋于达到 n 时，复杂度变为 O(k)。

高效方法：它是滑动窗口最大值问题的变体。
我们创建了一个容量为 k 的出队队列 Di，它只存储当前 k 个元素窗口的有用元素。如果元素在当前窗口中并且是负整数，则该元素很有用。我们将所有数组元素一一处理，并维护 Di 以包含当前窗口的有用元素，这些有用元素都是负整数。对于特定窗口，如果 Di 不为空，则 Di 前面的元素是该窗口的第一个负整数，否则该窗口不包含负整数。

C++

// C++ implementation to find the first negative
// integer in every window of size k
#include 
 
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  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 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

蟒蛇3

# 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 Di = new List();
 
    // 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

输出：

-1 -1 -7 -15 -15 0

时间复杂度： O(n)
辅助空间： O(k)

优化方法： : 也可以使用恒定空间来实现这一点。这个想法是有一个变量 firstNegativeIndex 来跟踪 k 大小窗口中的第一个负元素。在每次迭代中，我们跳过不再属于当前 k 大小窗口（firstNegativeIndex <= i – k）的元素以及正元素。

下面是基于这种方法的解决方案。

C++

// C++ code for First negative integer
// in every window of size k
#include 
using namespace std;
 
void printFirstNegativeInteger(int arr[], int k, int n)
{
    int firstNegativeIndex = 0;
    int firstNegativeElement;
     
    
    for(int i = k - 1; i < n; i++)
    {
       
        // skip out of window and positive elements
        while((firstNegativeIndex < i ) &&
              (firstNegativeIndex <= i - k ||
               arr[firstNegativeIndex] > 0))
        {
            firstNegativeIndex ++;
        }
         
        // check if a negative element is found, otherwise use 0
        if(arr[firstNegativeIndex] < 0)
        {
            firstNegativeElement = arr[firstNegativeIndex];
        }
        else
        {
            firstNegativeElement = 0;
        }
        cout<

Java

// Java code for First negative integer
// in every window of size k
import java.util.*;
 
class GFG{
     
static void printFirstNegativeInteger(int arr[],
                                      int k, int n)
{
    int firstNegativeIndex = 0;
    int firstNegativeElement;
     
    for(int i = k - 1; i < n; i++)
    {
         
        // Skip out of window and positive elements
        while ((firstNegativeIndex < i ) &&
               (firstNegativeIndex <= i - k ||
            arr[firstNegativeIndex] > 0))
        {
            firstNegativeIndex ++;
        }
         
        // Check if a negative element is
        // found, otherwise use 0
        if (arr[firstNegativeIndex] < 0)
        {
            firstNegativeElement = arr[firstNegativeIndex];
        }
        else
        {
            firstNegativeElement = 0;
        }
        System.out.print(firstNegativeElement + " ");
    }
}
 
// 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, k, n);   
}
}
 
// This code is contributed by amreshkumar3

蟒蛇3

# Python3 code for First negative integer
# in every window of size k
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

C＃

// C# code for First negative integer
// in every window of size k
using System;
 
class GFG{
 
static void printFirstNegativeInteger(int[] arr,
                                      int k, int n)
{
    int firstNegativeIndex = 0;
    int firstNegativeElement;
     
    for(int i = k - 1; i < n; i++)
    {
         
        // Skip out of window and positive elements
        while ((firstNegativeIndex < i ) &&
               (firstNegativeIndex <= i - k ||
            arr[firstNegativeIndex] > 0))
        {
            firstNegativeIndex ++;
        }
         
        // Check if a negative element is
        // found, otherwise use 0
        if (arr[firstNegativeIndex] < 0)
        {
            firstNegativeElement = arr[firstNegativeIndex];
        }
        else
        {
            firstNegativeElement = 0;
        }
        Console.Write(firstNegativeElement + " ");
    }   
}
 
// Driver code
static public void Main()
{
    int[] arr = { 12, -1, -7, 8, -15, 30, 16, 28 };
    int n = arr.Length;
    int k = 3;
     
    printFirstNegativeInteger(arr, k, n);
}
}
 
// This code is contributed by rag2127

Javascript

        // JavaScript Program for the above approach
 
 
 
        function printFirstNegativeInteger(arr, k, n) {
            let firstNegativeIndex = 0;
            let firstNegativeElement;
 
 
            for (let i = k - 1; i < n; i++) {
 
                // skip out of window and positive elements
                while ((firstNegativeIndex < i) &&
                    (firstNegativeIndex <= i - k ||
                        arr[firstNegativeIndex] > 0)) {
                    firstNegativeIndex++;
                }
 
                // check if a negative element is found, otherwise use 0
                if (arr[firstNegativeIndex] < 0) {
                    firstNegativeElement = arr[firstNegativeIndex];
                }
                else {
                    firstNegativeElement = 0;
                }
                document.write(firstNegativeElement + " ");
            }
 
        }
 
        // Driver code
 
        let arr = [12, -1, -7, 8, -15, 30, 16, 28];
        let n = arr.length;
        let k = 3;
        printFirstNegativeInteger(arr, k, n);
 
    // This code is contributed by Potta Lokesh
    

输出：

-1 -1 -7 -15 -15 0

时间复杂度： O(n)
辅助空间： O(1)