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

Given a binary array a[] and a number k, we need to find length of he 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. ` `?> `

Output:

```5
```

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