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 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 1sint 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 Codeint 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 approachimport java.util.LinkedList;import java.util.Queue;class GFG{// Function to Find longest subsegment of 1sstatic 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 Codepublic 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 1sdef 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 CodeN = 10K = 2arr = [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 approachusing 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)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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