# Length of longest consecutive ones by at most one swap in a Binary String

Given a Binary String of length . It is allowed to do at most one swap between any 0 and 1. The task is to find the length of the longest consecutive 1’s that can be achieved.

Examples:

```Input : str = "111011101"
Output : 7
We can swap 0 at 4th with 1 at 10th position

Input : str = "111000"
Output : 3
We cannot obtain more than 3 1's after
```

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

Approach:

1. Count all 1’s in the array in a variable say cnt_one.
2. Maintain two vectors or arrays storing cumulative ones from left and right.
3. Whenever there is a 0:
• if (left[i-1]+right[i+1] < cnt_one) store max_count = left[i-1] + right [i+1] + 1, as by swapping we will get one extra one in place of 0.
• else max_count = left[i-1] + right[i+1].
4. Output is the maximum value of max_count that can be achieved.

Below is the implementation of the above approach:

## C++

 `// C++ program to find length of longest consecutive ` `// ones by at most one swap in a Binary String ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the length of the ` `// longest consecutive 1's ` `int` `maximum_one(string s, ``int` `n) ` `{ ` `    ``// To count all 1's in the string ` `    ``int` `cnt_one = 0; ` `     `  `    ``int` `max_cnt = 0, temp=0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(s[i] == ``'1'``) ` `        ``{ ` `            ``cnt_one++; ` `            ``temp++; ` `        ``} ` `        ``else` `        ``{ ` `            ``max_cnt = max(temp,max_cnt); ` `            ``temp=0; ` `        ``} ` `    ``} ` `     `  `    ``max_cnt = max(max_cnt, temp); ` ` `  `    ``// To store cumulative 1's ` `    ``int` `left[n], right[n]; ` ` `  `    ``if` `(s[0] == ``'1'``) ` `        ``left[0] = 1; ` `    ``else` `        ``left[0] = 0; ` ` `  `    ``if` `(s[n - 1] == ``'1'``) ` `        ``right[n - 1] = 1; ` `    ``else` `        ``right[n - 1] = 0; ` ` `  `    ``// Counting cumulative 1's from left ` `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``if` `(s[i] == ``'1'``) ` `            ``left[i] = left[i - 1] + 1; ` ` `  `        ``// If 0 then start new cumulative ` `        ``// one from that i ` `        ``else` `            ``left[i] = 0; ` `    ``} ` ` `  `    ``for` `(``int` `i = n - 2; i >= 0; i--) { ` `        ``if` `(s[i] == ``'1'``) ` `            ``right[i] = right[i + 1] + 1; ` ` `  `        ``else` `            ``right[i] = 0; ` `    ``} ` ` `  ` `  `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` `        ``// perform step 3 of the approach ` `        ``if` `(s[i] == ``'0'``) { ` ` `  `            ``// step 3 ` `            ``int` `sum = left[i - 1] + right[i + 1]; ` ` `  `            ``if` `(sum < cnt_one) ` `                ``max_cnt = max(max_cnt, sum+1); ` ` `  `            ``else` `                ``max_cnt = max(max_cnt, sum); ` ` `  `        ``} ` `    ``} ` ` `  `    ``return` `max_cnt; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// string ` `    ``string s = ``"111011101"``; ` ` `  `    ``cout << maximum_one(s, s.length()); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java  program to find length of longest consecutive ` `// ones by at most one swap in a Binary String ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `  `  `// Function to calculate the length of the ` `// longest consecutive 1's ` ` ``static` `int` `maximum_one(String s, ``int` `n) ` `{ ` `    ``// To count all 1's in the string ` `    ``int` `cnt_one = ``0``; ` ` `  `    ``int` `max_cnt = ``0``, temp=``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(s.charAt(i) == ``'1'``) ` `        ``{ ` `            ``cnt_one++; ` `            ``temp++; ` `        ``} ` `        ``else` `        ``{ ` `            ``max_cnt = Math.max(max_cnt, temp); ` `            ``temp = ``0``; ` `        ``} ` `    ``} ` `    ``max_cnt = Math.max(max_cnt, temp); ` ` `  `    ``// To store cumulative 1's ` `    ``int` `[]left = ``new` `int``[n]; ` `     ``int` `right[]= ``new` `int``[n]; ` ` `  `    ``if` `(s.charAt(``0``) == ``'1'``) ` `        ``left[``0``] = ``1``; ` `    ``else` `        ``left[``0``] = ``0``; ` ` `  `    ``if` `(s.charAt(n - ``1``) == ``'1'``) ` `        ``right[n - ``1``] = ``1``; ` `    ``else` `        ``right[n - ``1``] = ``0``; ` ` `  `    ``// Counting cumulative 1's from left ` `    ``for` `(``int` `i = ``1``; i < n; i++) { ` `        ``if` `(s.charAt(i) == ``'1'``) ` `            ``left[i] = left[i - ``1``] + ``1``; ` ` `  `        ``// If 0 then start new cumulative ` `        ``// one from that i ` `        ``else` `            ``left[i] = ``0``; ` `    ``} ` ` `  `    ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) { ` `        ``if` `(s.charAt(i) == ``'1'``) ` `            ``right[i] = right[i + ``1``] + ``1``; ` ` `  `        ``else` `            ``right[i] = ``0``; ` `    ``} ` ` `  ` `  `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++) { ` `        ``// perform step 3 of the approach ` `        ``if` `(s.charAt(i) == ``'0'``) { ` ` `  `            ``// step 3 ` `            ``int` `sum = left[i - ``1``] + right[i + ``1``]; ` ` `  `            ``if` `(sum < cnt_one) ` `                 ``max_cnt = Math.max(max_cnt, sum+``1``); ` ` `  `            ``else` `                 ``max_cnt = Math.max(max_cnt, sum); ` `        ``} ` `    ``} ` ` `  `    ``return` `max_cnt; ` `} ` ` `  `// Driver Code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `        ``// string ` `    ``String s = ``"111011101"``; ` ` `  `    ``System.out.println( maximum_one(s, s.length())); ` `    ``} ` `} ` `// This code is contributed by anuj_67.. `

## Python3

 `# Python 3 program to find length of  ` `# longest consecutive ones by at most ` `# one swap in a Binary String ` ` `  `# Function to calculate the length  ` `# of the longest consecutive 1's ` `def` `maximum_one(s, n) : ` ` `  `    ``# To count all 1's in the string ` `    ``cnt_one ``=` `0` `     `  `    ``cnt, max_cnt ``=` `0``, ``0` ` `  `    ``for` `i ``in` `range``(n) : ` `        ``if` `(s[i] ``=``=` `'1'``) : ` `            ``cnt_one ``+``=` `1` `            ``cnt ``+``=` `1` `        ``else` `: ` `            ``max_cnt ``=` `max``(max_cnt, cnt) ` `            ``cnt ``=` `0` `             `  `    ``max_cnt ``=` `max``(max_cnt, cnt) ` `     `  `    ``# To store cumulative 1's ` `    ``left ``=` `[``0``] ``*` `n ` `    ``right ``=` `[``0``] ``*` `n ` ` `  `    ``if` `(s[``0``] ``=``=` `'1'``) : ` `        ``left[``0``] ``=` `1` `         `  `    ``else` `: ` `        ``left[``0``] ``=` `0` ` `  `    ``if` `(s[n ``-` `1``] ``=``=` `'1'``) : ` `        ``right[n ``-` `1``] ``=` `1` `         `  `    ``else` `: ` `        ``right[n ``-` `1``] ``=` `0` ` `  `    ``# Counting cumulative 1's from left ` `    ``for` `i ``in` `range``(``1``, n) : ` `        ``if` `(s[i] ``=``=` `'1'``) : ` `            ``left[i] ``=` `left[i ``-` `1``] ``+` `1` ` `  `        ``# If 0 then start new cumulative ` `        ``# one from that i ` `        ``else` `: ` `            ``left[i] ``=` `0` `     `  `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``) : ` `         `  `        ``if` `(s[i] ``=``=` `'1'``) : ` `            ``right[i] ``=` `right[i ``+` `1``] ``+` `1` ` `  `        ``else` `: ` `            ``right[i] ``=` `0` ` `  ` `  `    ``for` `i ``in` `range``(``1``, n) : ` `         `  `        ``# perform step 3 of the approach ` `        ``if` `(s[i] ``=``=` `'0'``) : ` ` `  `            ``# step 3 ` `            ``sum` `=` `left[i ``-` `1``] ``+` `right[i ``+` `1``] ` ` `  `            ``if` `(``sum` `< cnt_one) : ` `                ``max_cnt ``=` `max``(max_cnt, ``sum``+``1``) ` ` `  `            ``else` `: ` `                ``max_cnt ``=` `max``(max_cnt, ``sum``) ` ` `  ` `  `    ``return` `max_cnt ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``# string ` `    ``s ``=` `"111011101"` ` `  `    ``print``(maximum_one(s, ``len``(s))) ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# program to find length of longest consecutive ` `// ones by at most one swap in a Binary String ` `using` `System; ` ` `  `class` `GFG { ` ` `  `// Function to calculate the length of the ` `// longest consecutive 1's ` `static` `int` `maximum_one(``string` `s, ``int` `n) ` `{ ` `    ``// To count all 1's in the string ` `    ``int` `cnt_one = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(s[i] == ``'1'``) ` `            ``cnt_one++; ` `    ``} ` ` `  `    ``// To store cumulative 1's ` `    ``int` `[]left = ``new` `int``[n]; ` `    ``int` `[]right= ``new` `int``[n]; ` ` `  `    ``if` `(s[0] == ``'1'``) ` `        ``left[0] = 1; ` `    ``else` `        ``left[0] = 0; ` ` `  `    ``if` `(s[n - 1]== ``'1'``) ` `        ``right[n - 1] = 1; ` `    ``else` `        ``right[n - 1] = 0; ` ` `  `    ``// Counting cumulative 1's from left ` `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``if` `(s[i] == ``'1'``) ` `            ``left[i] = left[i - 1] + 1; ` ` `  `        ``// If 0 then start new cumulative ` `        ``// one from that i ` `        ``else` `            ``left[i] = 0; ` `    ``} ` ` `  `    ``for` `(``int` `i = n - 2; i >= 0; i--) { ` `        ``if` `(s[i] == ``'1'``) ` `            ``right[i] = right[i + 1] + 1; ` ` `  `        ``else` `            ``right[i] = 0; ` `    ``} ` ` `  `    ``int` `cnt = 0, max_cnt = 0; ` ` `  `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` `        ``// perform step 3 of the approach ` `        ``if` `(s[i] == ``'0'``) { ` ` `  `            ``// step 3 ` `            ``int` `sum = left[i - 1] + right[i + 1]; ` ` `  `            ``if` `(sum < cnt_one) ` `                ``cnt = sum + 1; ` ` `  `            ``else` `                ``cnt = sum; ` ` `  `            ``max_cnt = Math.Max(max_cnt, cnt); ` `            ``cnt = 0; ` `        ``} ` `    ``} ` ` `  `    ``return` `max_cnt; ` `} ` ` `  `// Driver Code ` ` `  `    ``public` `static` `void` `Main () { ` `        ``// string ` `    ``string` `s = ``"111011101"``; ` ` `  `    ``Console.WriteLine( maximum_one(s, s.Length)); ` `    ``} ` `} ` `// This code is contributed by inder_verma.. `

## PHP

 `= 0; ``\$i``--) ` `    ``{ ` `        ``if` `(``\$s``[``\$i``] == ``'1'``) ` `            ``\$right``[``\$i``] = ``\$right``[``\$i` `+ 1] + 1; ` ` `  `        ``else` `            ``\$right``[``\$i``] = 0; ` `    ``} ` ` `  `    ``\$cnt` `= 0; ``\$max_cnt` `= 0; ` ` `  `    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n` `- 1; ``\$i``++)  ` `    ``{ ` `        ``// perform step 3 of the approach ` `        ``if` `(``\$s``[``\$i``] == ``'0'``) ` `        ``{ ` ` `  `            ``// step 3 ` `            ``\$sum` `= ``\$left``[``\$i` `- 1] + ``\$right``[``\$i` `+ 1]; ` ` `  `            ``if` `(``\$sum` `< ``\$cnt_one``) ` `                ``\$cnt` `= ``\$sum` `+ 1; ` ` `  `            ``else` `                ``\$cnt` `= ``\$sum``; ` ` `  `            ``\$max_cnt` `= max(``\$max_cnt``, ``\$cnt``); ` `            ``\$cnt` `= 0; ` `        ``} ` `    ``} ` `    ``return` `\$max_cnt``; ` `} ` ` `  `// Driver Code ` ` `  `// string ` `\$s` `= ``"111011101"``; ` ` `  `echo` `maximum_one(``\$s``, ``strlen``(``\$s``)); ` ` `  `// This code is contributed  ` `// by Akanksha Rai ` `?> `

Output:

```7
```

Time Complexity: O(N)

My Personal Notes arrow_drop_up

Maths is the language of nature

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.