# Longest subsequence of the form 0*1*0* in a binary string

Given a binary string, find the longest subsequence of the form (0)*(1)*(0)* in it. Basically we need to divide the string into 3 non overlapping strings (these strings might be empty) without changing the order of letters. **First and third** strings are made up of **only 0** and the second string is made up of only 1. These strings could be made **by deleting** some characters in original string. What is the **maximum size** of string, we can get?

**Examples:**

Input : 000011100000 Output : 12 Explanation : First part from 1 to 4. Second part 5 to 7. Third part from 8 to 12 Input : 100001100 Output : 8 Explanation : Delete the first letter. First part from 2 to 4. Second part from 5 to 6. Last part from 7. Input : 00000 Output : 5 Explanation : Special Case of Only 0 Input : 111111 Output : 6 Explanation : Special Case of Only 1 Input : 0000001111011011110000 Output : 20 Explanation : Second part is from 7 to 18. Remove all the 0 between indices 7 to 18.

A **simple solution** is to generate all subsequences of given sequence. For every subsequence, check if it is in given form. If yes, compare it with result so far and update the result if needed.

This problem can be **efficiently** solved by pre-computing below arrays in **O(n^2) time**.

Let `pre_count_0[i]`

be the count of letter 0 in the prefix of string till index i.

Let `pre_count_1[i]`

be the count of letter 1 in the prefix of string till length i.

Let `post_count_0[i]`

be the count of letter 0 in the suffix string from index i till index n (here n is the size of string).

Now we fix two two positions i and j, 1 <=i <= j <=n. We will remove all 0 from substring which starts at index i and ends in index j. Thus this makes the second substring of only 1. In the prefix before the index i and in the postfix after the index j we will delete all 1 and thus it will make first and third part of the string.

then the maximum length of string attainable is max of

` pre_count_0[i-1] + (pre_count_1[j]-pre_count_1[i-1]) + pre_count_1[j+1]`

**Special cases :** When String is made of only 0s or 1s, ans is n where n is length of string.

## C++

`// CPP program to find longest subsequence ` `// of the form 0*1*0* in a binary string ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns length of the longest subsequence ` `// of the form 0*1*0* ` `int` `longestSubseq(string s) ` `{ ` ` ` `int` `n = s.length(); ` ` ` ` ` `// Precomputing values in three arrays ` ` ` `// pre_count_0[i] is going to store count ` ` ` `// of 0s in prefix str[0..i-1] ` ` ` `// pre_count_1[i] is going to store count ` ` ` `// of 1s in prefix str[0..i-1] ` ` ` `// post_count_0[i] is going to store count ` ` ` `// of 0s in suffix str[i-1..n-1] ` ` ` `int` `pre_count_0[n + 2]; ` ` ` `int` `pre_count_1[n + 1]; ` ` ` `int` `post_count_0[n + 1]; ` ` ` `pre_count_0[0] = 0; ` ` ` `post_count_0[n + 1] = 0; ` ` ` `pre_count_1[0] = 0; ` ` ` `for` `(` `int` `j = 1; j <= n; j++) ` ` ` `{ ` ` ` `pre_count_0[j] = pre_count_0[j - 1]; ` ` ` `pre_count_1[j] = pre_count_1[j - 1]; ` ` ` `post_count_0[n - j + 1] = post_count_0[n - j + 2]; ` ` ` ` ` `if` `(s[j - 1] == ` `'0'` `) ` ` ` `pre_count_0[j]++; ` ` ` `else` ` ` `pre_count_1[j]++; ` ` ` `if` `(s[n - j] == ` `'0'` `) ` ` ` `post_count_0[n - j + 1]++; ` ` ` `} ` ` ` ` ` `// string is made up of all 0s or all 1s ` ` ` `if` `(pre_count_0[n] == n || ` ` ` `pre_count_0[n] == 0) ` ` ` `return` `n; ` ` ` ` ` ` ` `// Compute result using precomputed values ` ` ` `int` `ans = 0; ` ` ` `for` `(` `int` `i = 1; i <= n; i++) ` ` ` `for` `(` `int` `j = i; j <= n; j++) ` ` ` `ans = max(pre_count_0[i - 1] ` ` ` `+ pre_count_1[j] ` ` ` `- pre_count_1[i - 1] ` ` ` `+ post_count_0[j + 1], ` ` ` `ans); ` ` ` `return` `ans; ` `} ` ` ` `// driver program ` `int` `main() ` `{ ` ` ` `string s = ` `"000011100000"` `; ` ` ` `cout << longestSubseq(s); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find longest subsequence ` `// of the form 0*1*0* in a binary string ` `class` `GFG ` `{ ` ` ` ` ` `// Returns length of the longest subsequence ` ` ` `// of the form 0*1*0* ` ` ` `public` `static` `int` `longestSubseq(String s) ` ` ` `{ ` ` ` `int` `n = s.length(); ` ` ` ` ` `// Precomputing values in three arrays ` ` ` `// pre_count_0[i] is going to store count ` ` ` `// of 0s in prefix str[0..i-1] ` ` ` `// pre_count_1[i] is going to store count ` ` ` `// of 1s in prefix str[0..i-1] ` ` ` `// post_count_0[i] is going to store count ` ` ` `// of 0s in suffix str[i-1..n-1] ` ` ` `int` `[] pre_count_0 = ` `new` `int` `[n + ` `2` `]; ` ` ` `int` `[] pre_count_1 = ` `new` `int` `[n + ` `1` `]; ` ` ` `int` `[] post_count_0 = ` `new` `int` `[n + ` `2` `]; ` ` ` `pre_count_0[` `0` `] = ` `0` `; ` ` ` `post_count_0[n + ` `1` `] = ` `0` `; ` ` ` `pre_count_1[` `0` `] = ` `0` `; ` ` ` `for` `(` `int` `j = ` `1` `; j <= n; j++) ` ` ` `{ ` ` ` `pre_count_0[j] = pre_count_0[j - ` `1` `]; ` ` ` `pre_count_1[j] = pre_count_1[j - ` `1` `]; ` ` ` `post_count_0[n - j + ` `1` `] = post_count_0[n - j + ` `2` `]; ` ` ` ` ` `if` `(s.charAt(j - ` `1` `) == ` `'0'` `) ` ` ` `pre_count_0[j]++; ` ` ` `else` ` ` `pre_count_1[j]++; ` ` ` ` ` `if` `(s.charAt(n - j) == ` `'0'` `) ` ` ` `post_count_0[n - j + ` `1` `]++; ` ` ` `} ` ` ` ` ` `// string is made up of all 0s or all 1s ` ` ` `if` `(pre_count_0[n] == n || ` ` ` `pre_count_0[n] == ` `0` `) ` ` ` `return` `n; ` ` ` ` ` `// Compute result using precomputed values ` ` ` `int` `ans = ` `0` `; ` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) ` ` ` `for` `(` `int` `j = i; j <= n; j++) ` ` ` `ans = Math.max(pre_count_0[i - ` `1` `] + ` ` ` `pre_count_1[j] - ` ` ` `pre_count_1[i - ` `1` `] + ` ` ` `post_count_0[j + ` `1` `], ans); ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `String s = ` `"000011100000"` `; ` ` ` `System.out.println(longestSubseq(s)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ` `// sanjeev2552 ` |

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## Python3

`# Python 3 program to find longest subsequence ` `# of the form 0*1*0* in a binary string ` ` ` `# Returns length of the longest subsequence ` `# of the form 0*1*0* ` `def` `longestSubseq(s): ` ` ` `n ` `=` `len` `(s) ` ` ` ` ` `# Precomputing values in three arrays ` ` ` `# pre_count_0[i] is going to store count ` ` ` `# of 0s in prefix str[0..i-1] ` ` ` `# pre_count_1[i] is going to store count ` ` ` `# of 1s in prefix str[0..i-1] ` ` ` `# post_count_0[i] is going to store count ` ` ` `# of 0s in suffix str[i-1..n-1] ` ` ` `pre_count_0 ` `=` `[` `0` `for` `i ` `in` `range` `(n ` `+` `2` `)] ` ` ` `pre_count_1 ` `=` `[` `0` `for` `i ` `in` `range` `(n ` `+` `1` `)] ` ` ` `post_count_0 ` `=` `[` `0` `for` `i ` `in` `range` `(n ` `+` `2` `)] ` ` ` `pre_count_0[` `0` `] ` `=` `0` ` ` `post_count_0[n ` `+` `1` `] ` `=` `0` ` ` `pre_count_1[` `0` `] ` `=` `0` ` ` `for` `j ` `in` `range` `(` `1` `, n ` `+` `1` `): ` ` ` `pre_count_0[j] ` `=` `pre_count_0[j ` `-` `1` `] ` ` ` `pre_count_1[j] ` `=` `pre_count_1[j ` `-` `1` `] ` ` ` `post_count_0[n ` `-` `j ` `+` `1` `] ` `=` `post_count_0[n ` `-` `j ` `+` `2` `] ` ` ` ` ` `if` `(s[j ` `-` `1` `] ` `=` `=` `'0'` `): ` ` ` `pre_count_0[j] ` `+` `=` `1` ` ` `else` `: ` ` ` `pre_count_1[j] ` `+` `=` `1` ` ` `if` `(s[n ` `-` `j] ` `=` `=` `'0'` `): ` ` ` `post_count_0[n ` `-` `j ` `+` `1` `] ` `+` `=` `1` ` ` ` ` `# string is made up of all 0s or all 1s ` ` ` `if` `(pre_count_0[n] ` `=` `=` `n ` `or` ` ` `pre_count_0[n] ` `=` `=` `0` `): ` ` ` `return` `n ` ` ` ` ` `# Compute result using precomputed values ` ` ` `ans ` `=` `0` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `): ` ` ` `for` `j ` `in` `range` `(i, n ` `+` `1` `, ` `1` `): ` ` ` `ans ` `=` `max` `(pre_count_0[i ` `-` `1` `] ` `+` ` ` `pre_count_1[j] ` `-` ` ` `pre_count_1[i ` `-` `1` `] ` `+` ` ` `post_count_0[j ` `+` `1` `], ans) ` ` ` `return` `ans ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `s ` `=` `"000011100000"` ` ` `print` `(longestSubseq(s)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

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## C#

`// C# program to find longest subsequence ` `// of the form 0*1*0* in a binary string ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Returns length of the longest subsequence ` ` ` `// of the form 0*1*0* ` ` ` `public` `static` `int` `longestSubseq(String s) ` ` ` `{ ` ` ` `int` `n = s.Length; ` ` ` ` ` `// Precomputing values in three arrays ` ` ` `// pre_count_0[i] is going to store count ` ` ` `// of 0s in prefix str[0..i-1] ` ` ` `// pre_count_1[i] is going to store count ` ` ` `// of 1s in prefix str[0..i-1] ` ` ` `// post_count_0[i] is going to store count ` ` ` `// of 0s in suffix str[i-1..n-1] ` ` ` `int` `[] pre_count_0 = ` `new` `int` `[n + 2]; ` ` ` `int` `[] pre_count_1 = ` `new` `int` `[n + 1]; ` ` ` `int` `[] post_count_0 = ` `new` `int` `[n + 2]; ` ` ` `pre_count_0[0] = 0; ` ` ` `post_count_0[n + 1] = 0; ` ` ` `pre_count_1[0] = 0; ` ` ` `for` `(` `int` `j = 1; j <= n; j++) ` ` ` `{ ` ` ` `pre_count_0[j] = pre_count_0[j - 1]; ` ` ` `pre_count_1[j] = pre_count_1[j - 1]; ` ` ` `post_count_0[n - j + 1] = post_count_0[n - j + 2]; ` ` ` ` ` `if` `(s[j - 1] == ` `'0'` `) ` ` ` `pre_count_0[j]++; ` ` ` `else` ` ` `pre_count_1[j]++; ` ` ` ` ` `if` `(s[n - j] == ` `'0'` `) ` ` ` `post_count_0[n - j + 1]++; ` ` ` `} ` ` ` ` ` `// string is made up of all 0s or all 1s ` ` ` `if` `(pre_count_0[n] == n || ` ` ` `pre_count_0[n] == 0) ` ` ` `return` `n; ` ` ` ` ` `// Compute result using precomputed values ` ` ` `int` `ans = 0; ` ` ` `for` `(` `int` `i = 1; i <= n; i++) ` ` ` `for` `(` `int` `j = i; j <= n; j++) ` ` ` `ans = Math.Max(pre_count_0[i - 1] + ` ` ` `pre_count_1[j] - ` ` ` `pre_count_1[i - 1] + ` ` ` `post_count_0[j + 1], ans); ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main(String[] args) ` ` ` `{ ` ` ` `String s = ` `"000011100000"` `; ` ` ` `Console.WriteLine(longestSubseq(s)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Princi Singh ` |

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**Output :**

12

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