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Longest subsegment of ‘1’s formed by changing at most k ‘0’s | Set 2 (Using Queue)

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  • Difficulty Level : Hard
  • Last Updated : 07 Jan, 2022

Given a binary array a[] and a number k, we need to find the length of the longest subsegment of ‘1’s possible by changing at most k ‘0’s.

Examples: 

Input: a[] = {1, 0, 0, 1, 1, 0, 1},  k = 1
Output: 4
Explanation: Here, we should only change 1 zero(0). Maximum possible length we can get is by changing the 3rd zero in the array, we get a[] = {1, 0, 0, 1, 1, 1, 1}

Input: a[] = {1, 0, 0, 1, 0, 1, 0, 1, 0, 1}, k = 2
Output: 5 

 

Two Pointer Approach: Refer the Set 1 of this article for the implementation of Two-pointer approach.
 

Queue Approach: The task can be solved with the help of a queue. Store the indices of 0s encountered so far in a queue. For each 0, check if the value of K is greater than 0 or not, if it is non-zero, flip it to 1, and maximize the subsegment length correspondingly, else shift the left pointer (initially at the start index of the string) to the index of first zero (queue’s front) + 1.
Follow the below steps to solve the problem:

  • Declare a queue for storing Indices of 0s Visited.
  • Iterate over the string and if  If the current character is 0 and some spells are left i.e. (k != 0) then use the spell i.e. (decrement k). Also, store the index of “0” occurred.
  • If k = 0, Take out the front of the queue and store it in a variable.
  • Store the length as max between i-low and that of the previous answer.
  • Shift low to index of first “0” + 1 and increment k.
  • Finally, return the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to Find longest subsegment of 1s
int get(int n, int k, int arr[])
{
    // Queue for storing indices of 0s
    queue<int> q;
 
    int low = 0;
    int ans = INT_MIN;
 
    int p = k;
    int i = 0;
 
    while (i < n) {
        // If the current character is 1
        // then increment i by 1
        if (arr[i] == 1) {
            i++;
        }
 
        // If the current character is 0
        // and some spells are
        // left then use them
        else if (arr[i] == 0 && k != 0) {
            q.push(i);
            k--;
            i++;
        }
        // If k == 0
        else {
            // Take out the index where
            // the first "0" was found
            int x = q.front();
            q.pop();
 
            // Store the length as max
            // between i-low and that
            // of the previous answer
            ans = max(ans, i - low);
 
            // Shift low to index
            // of first "O" + 1
            low = x + 1;
 
            // Increase spell by 1
            k++;
        }
 
        // Store the length between
        // the i-low and that of
        // previous answer
        ans = max(ans, i - low);
    }
    return ans;
}
 
// Driver Code
int main()
{
    int N = 10;
    int K = 2;
    int arr[] = { 1, 0, 0, 1, 0,
                  1, 0, 1, 0, 1 };
 
    cout << get(N, K, arr) << endl;
    return 0;
}


Java




// Java program for the above approach
import java.util.LinkedList;
import java.util.Queue;
 
class GFG{
 
// Function to Find longest subsegment of 1s
static int get(int n, int k, int arr[])
{
     
    // Queue for storing indices of 0s
    Queue<Integer> q = new LinkedList<Integer>();
 
    int low = 0;
    int ans = Integer.MIN_VALUE;
    int i = 0;
 
    while (i < n)
    {
         
        // If the current character is 1
        // then increment i by 1
        if (arr[i] == 1)
        {
            i++;
        }
 
        // If the current character is 0
        // and some spells are
        // left then use them
        else if (arr[i] == 0 && k != 0)
        {
            q.add(i);
            k--;
            i++;
        }
         
        // If k == 0
        else
        {
             
            // Take out the index where
            // the first "0" was found
            int x = q.peek();
            q.remove();
 
            // Store the length as max
            // between i-low and that
            // of the previous answer
            ans = Math.max(ans, i - low);
 
            // Shift low to index
            // of first "O" + 1
            low = x + 1;
 
            // Increase spell by 1
            k++;
        }
 
        // Store the length between
        // the i-low and that of
        // previous answer
        ans = Math.max(ans, i - low);
    }
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 10;
    int K = 2;
    int arr[] = { 1, 0, 0, 1, 0,
                  1, 0, 1, 0, 1 };
 
    System.out.println(get(N, K, arr));
}
}
 
// This code is contributed by gfking


Python3




# Python code for the above approach
 
# Function to Find longest subsegment of 1s
def get(n, k, arr):
   
    # Queue for storing indices of 0s
    q = []
 
    low = 0
    ans = 10 ** -9
 
    p = k
    i = 0
 
    while (i < n):
       
        # If the current character is 1
        # then increment i by 1
        if (arr[i] == 1):
            i += 1
 
        # If the current character is 0
        # and some spells are
        # left then use them
        elif (arr[i] == 0 and k != 0):
            q.append(i)
            k -= 1
            i += 1
        # If k == 0
        else:
            # Take out the index where
            # the first "0" was found
            x = q[0]
            q.pop(0)
 
            # Store the length as max
            # between i-low and that
            # of the previous answer
            ans = max(ans, i - low)
 
            # Shift low to index
            # of first "O" + 1
            low = x + 1
 
            # Increase spell by 1
            k += 1
 
        # Store the length between
        # the i-low and that of
        # previous answer
        ans = max(ans, i - low)
     
    return ans
 
# Driver Code
N = 10
K = 2
arr = [1, 0, 0, 1, 0, 1, 0, 1, 0, 1]
print(get(N, K, arr))
 
# This code is contributed by Saurabh Jaiswal


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Function to Find longest subsegment of 1s
  static int get(int n, int k, int[] arr)
  {
 
    // Queue for storing indices of 0s
    Queue<int> q = new Queue<int>();
 
    int low = 0;
    int ans = Int32.MinValue;
    int i = 0;
 
    while (i < n) {
 
      // If the current character is 1
      // then increment i by 1
      if (arr[i] == 1) {
        i++;
      }
 
      // If the current character is 0
      // and some spells are
      // left then use them
      else if (arr[i] == 0 && k != 0) {
        q.Enqueue(i);
        k--;
        i++;
      }
 
      // If k == 0
      else {
 
        // Take out the index where
        // the first "0" was found
        int x = q.Peek();
        q.Dequeue();
 
        // Store the length as max
        // between i-low and that
        // of the previous answer
        ans = Math.Max(ans, i - low);
 
        // Shift low to index
        // of first "O" + 1
        low = x + 1;
 
        // Increase spell by 1
        k++;
      }
 
      // Store the length between
      // the i-low and that of
      // previous answer
      ans = Math.Max(ans, i - low);
    }
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 10;
    int K = 2;
    int[] arr = { 1, 0, 0, 1, 0, 1, 0, 1, 0, 1 };
 
    Console.WriteLine(get(N, K, arr));
  }
}
 
// This code is contributed by ukasp.


Javascript




<script>
        // JavaScript code for the above approach
 
 
        // Function to Find longest subsegment of 1s
        function get(n, k, arr) {
            // Queue for storing indices of 0s
            let q = [];
 
            let low = 0;
            let ans = Number.MIN_VALUE;
 
            let p = k;
            let i = 0;
 
            while (i < n) {
                // If the current character is 1
                // then increment i by 1
                if (arr[i] == 1) {
                    i++;
                }
 
                // If the current character is 0
                // and some spells are
                // left then use them
                else if (arr[i] == 0 && k != 0) {
                    q.push(i);
                    k--;
                    i++;
                }
                // If k == 0
                else {
                    // Take out the index where
                    // the first "0" was found
                    let x = q[0];
                    q.shift();
 
                    // Store the length as max
                    // between i-low and that
                    // of the previous answer
                    ans = Math.max(ans, i - low);
 
                    // Shift low to index
                    // of first "O" + 1
                    low = x + 1;
 
                    // Increase spell by 1
                    k++;
                }
 
                // Store the length between
                // the i-low and that of
                // previous answer
                ans = Math.max(ans, i - low);
            }
            return ans;
        }
 
        // Driver Code
 
        let N = 10;
        let K = 2;
        let arr = [1, 0, 0, 1, 0,
            1, 0, 1, 0, 1];
 
        document.write(get(N, K, arr) + '<br>');
 
  // This code is contributed by Potta Lokesh
    </script>


Output

5

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


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