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# Longest subsegment of ‘1’s formed by changing at most k ‘0’s

• Difficulty Level : Medium
• Last Updated : 25 May, 2021

Given a binary array a[] and a number k, we need to find 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
Output: Here, we can change only 2 zeros.
Maximum possible length we can get is by
changing the 3rd and 4th (or) 4th and 5th
zeros.```

We can solve this problem using two pointers technique. Let us take a subarray [l, r] which contains at most k zeroes. Let our left pointer be l and right pointer be r. We always maintain our subsegment [l, r] to contain no more than k zeroes by moving the left pointer l. Check at every step for maximum size (i.e, r-l+1).

## C++

 `// CPP program to find length of longest``// subsegment of all 1's by changing at``// most k 0's``#include ``using` `namespace` `std;` `int` `longestSubSeg(``int` `a[], ``int` `n, ``int` `k)``{``    ``int` `cnt0 = 0;``    ``int` `l = 0;``    ``int` `max_len = 0;` `    ``// i decides current ending point``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(a[i] == 0)``            ``cnt0++;` `        ``// If there are more 0's move``        ``// left point for current ending``        ``// point.``        ``while` `(cnt0 > k) {``            ``if` `(a[l] == 0)``                ``cnt0--;``            ``l++;``        ``}` `        ``max_len = max(max_len, i - l + 1);``    ``}` `    ``return` `max_len;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 1, 0, 0, 1, 0, 1, 0, 1 };``    ``int` `k = 2;``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``cout << longestSubSeg(a, n, k);``    ``return` `0;``}`

## Java

 `// Java program to find length of``// longest subsegment of all 1's``// by changing at most k 0's``import` `java.io.*;` `class` `GFG {` `static` `int` `longestSubSeg(``int` `a[], ``int` `n,``                                  ``int` `k)``{``    ``int` `cnt0 = ``0``;``    ``int` `l = ``0``;``    ``int` `max_len = ``0``;` `    ``// i decides current ending point``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `(a[i] == ``0``)``            ``cnt0++;` `        ``// If there are more 0's move``        ``// left point for current ending``        ``// point.``        ``while` `(cnt0 > k) {``            ``if` `(a[l] == ``0``)``                ``cnt0--;``            ``l++;``        ``}` `        ``max_len = Math.max(max_len, i - l + ``1``);``    ``}` `    ``return` `max_len;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `a[] = { ``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1` `};``    ``int` `k = ``2``;``    ``int` `n = a.length;``    ``System.out.println( longestSubSeg(a, n, k));``        ` `}``}` `// This code is contributed by vt_m`

## Python3

 `# Python3 program to find length``# of longest subsegment of all 1's ``# by changing at most k 0's` `def` `longestSubSeg(a, n, k):` `    ``cnt0 ``=` `0``    ``l ``=` `0``    ``max_len ``=` `0``;` `    ``# i decides current ending point``    ``for` `i ``in` `range``(``0``, n):``        ``if` `a[i] ``=``=` `0``:``            ``cnt0 ``+``=` `1` `        ``# If there are more 0's move``        ``# left point for current``        ``# ending point.``        ``while` `(cnt0 > k):``            ``if` `a[l] ``=``=` `0``:``                ``cnt0 ``-``=` `1``            ``l ``+``=` `1``        `  `        ``max_len ``=` `max``(max_len, i ``-` `l ``+` `1``);``    `  `    ``return` `max_len` `# Driver code``a ``=` `[``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``0``, ``1` `]``k ``=` `2``n ``=` `len``(a)``print``(longestSubSeg(a, n, k))` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to find length of``// longest subsegment of all 1's``// by changing at most k 0's``using` `System;` `class` `GFG {` `    ``static` `int` `longestSubSeg(``int``[] a, ``int` `n,``                                      ``int` `k)``    ``{``        ``int` `cnt0 = 0;``        ``int` `l = 0;``        ``int` `max_len = 0;` `        ``// i decides current ending point``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(a[i] == 0)``                ``cnt0++;` `            ``// If there are more 0's move``            ``// left point for current ending``            ``// point.``            ``while` `(cnt0 > k) {``                ``if` `(a[l] == 0)``                    ``cnt0--;``                ``l++;``            ``}` `            ``max_len = Math.Max(max_len, i - l + 1);``        ``}` `        ``return` `max_len;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] a = { 1, 0, 0, 1, 0, 1, 0, 1 };``        ``int` `k = 2;``        ``int` `n = a.Length;``        ``Console.WriteLine(longestSubSeg(a, n, k));``    ``}``}` `// This code is contributed by vt_m`

## PHP

 ` ``\$k``)``        ``{``            ``if` `(``\$a``[``\$l``] == 0)``                ``\$cnt0``--;``            ``\$l``++;``        ``}` `        ``\$max_len` `= max(``\$max_len``, ``\$i` `- ``\$l` `+ 1);``    ``}` `    ``return` `\$max_len``;``}` `    ``// Driver code``    ``\$a` `= ``array``(1, 0, 0, 1, 0, 1, 0, 1);``    ``\$k` `= 2;``    ``\$n` `= ``count``(``\$a``);``    ``echo` `longestSubSeg(``\$a``, ``\$n``, ``\$k``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output:

`5`

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