# Longest subsequence possible that starts and ends with 1 and filled with 0 in the middle

Given a binary string s, the task is to find the length of longest subsequence that can be divided into three substrings such that the first and third substrings are either empty or filled with 1 and the substring at the middle is either empty or filled with 0.

Examples:

Input: s = “1001”
Output: 4
Explanation:
The entire string can be divided into the desired three parts: “1”, “00”, “1”.

Input: s = “010”
Output: 2
Explanation:
The subsequence “00”, “01” and “10” can be split into desired three parts {“”, “00”, “”}, {“”, “0”, “1”} and {“1”, “0”, “”}

Approach:
To solve the problem, we need to follow the steps given below:

• Firstly, pre-compute and store in prefix arrays, the occurences of ‘1’ and ‘0’ respectively.
• Initialize two integers i and j, where i will be the point of partition between first and second string and j will be the point of partition between second and third string.
• Iterate over all possible values of i & j (0 <= i < j <=n) and find the maximum possible length of the subsequence possible which satisfies the given condition.

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the ` `// longest subsequence possible ` `// that starts and ends with 1 ` `// and filled with 0 in the middle ` ` `  `#include ` `using` `namespace` `std; ` ` `  `int` `longestSubseq(string s, ``int` `length) ` `{ ` `    ``// Prefix array to store the ` `    ``// occurences of '1' and '0' ` `    ``int` `ones[length + 1], zeroes[length + 1]; ` ` `  `    ``// Initialise prefix arrays with 0 ` `    ``memset``(ones, 0, ``sizeof``(ones)); ` `    ``memset``(zeroes, 0, ``sizeof``(zeroes)); ` ` `  `    ``// Iterate over the length of the string ` `    ``for` `(``int` `i = 0; i < length; i++) { ` ` `  `        ``// If current character is '1' ` `        ``if` `(s[i] == ``'1'``) { ` `            ``ones[i + 1] = ones[i] + 1; ` `            ``zeroes[i + 1] = zeroes[i]; ` `        ``} ` ` `  `        ``// If current character is '0' ` `        ``else` `{ ` `            ``zeroes[i + 1] = zeroes[i] + 1; ` `            ``ones[i + 1] = ones[i]; ` `        ``} ` `    ``} ` ` `  `    ``int` `answer = INT_MIN; ` `    ``int` `x = 0; ` ` `  `    ``for` `(``int` `i = 0; i <= length; i++) { ` `        ``for` `(``int` `j = i; j <= length; j++) { ` `            ``// Add '1' available for ` `            ``// the first string ` `            ``x += ones[i]; ` ` `  `            ``// Add '0' available for ` `            ``// the second string ` `            ``x += (zeroes[j] - zeroes[i]); ` ` `  `            ``// Add '1' available for ` `            ``// the third string ` `            ``x += (ones[length] - ones[j]); ` ` `  `            ``// Update answer ` `            ``answer = max(answer, x); ` ` `  `            ``x = 0; ` `        ``} ` `    ``} ` ` `  `    ``// Print the final result ` `    ``cout << answer << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``string s = ``"10010010111100101"``; ` ` `  `    ``int` `length = s.length(); ` ` `  `    ``longestSubseq(s, length); ` ` `  `    ``return` `0; ` `} `

Output:

```12
```

Time Complexity: O(N2)
Auxillary Space: O(N)

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