# Minimize deletions in a Binary String to remove all subsequences of the form “0101”

Given a binary string S of length N, the task is to find the minimum number of characters required to be deleted from the string such that no subsequence of the form “0101” exists in the string.

Examples:

Input: S = “0101101”
Output: 2
Explanation: Removing S and S modifies the string to 00111. Therefore, no subsequence of the type 0101 can be obtained from the given string.

Input: S = “0110100110”
Output: 2

Approach: Follow the steps below to solve the problem:

• The required valid string can consist of at most three blocks of the same elements i.e. the strings can be one of the following patterns “00…0”, “11…1”, “00…01…1”, “1…10..0”, “00..01…10..0”, “1…10…01…1”.
• Count frequencies of 0s and 1s of a block using partial sum.
• Fix the starting and ending indices of blocks of 0s and 1s and determine the minimum number of characters required to be deleted by the calculated partial sums.
• Therefore, check the length of the longest string that can be obtained by removing the subsequences of the given type.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find minimum characters` `// to be removed such that no subsequence` `// of the form "0101" exists in the string` `int` `findmin(string s)` `{` `    ``int` `n = s.length();` `    ``int` `i, j, maximum = 0;`   `    ``// Stores the partial sums` `    ``int` `incr[n + 1] = { 0 };`   `    ``for` `(i = 0; i < n; i++) {`   `        ``// Calculate partial sums` `        ``incr[i + 1] = incr[i];`   `        ``if` `(s[i] == ``'0'``) {` `            ``incr[i + 1]++;` `        ``}` `    ``}`   `    ``for` `(i = 0; i < n; i++) {` `        ``for` `(j = i + 1; j < n; j++) {`   `            ``// Setting endpoints and` `            ``// deleting characters indices.` `            ``maximum` `                ``= max(maximum, incr[i] + j - i + 1` `                                   ``- (incr[j + 1] - incr[i])` `                                   ``+ incr[n] - incr[j + 1]);` `        ``}` `    ``}`   `    ``// Return count of deleted characters` `    ``return` `n - maximum;` `}` `// Driver Code` `int` `main()` `{` `    ``string S = ``"0110100110"``;` `    ``int` `minimum = findmin(S);` `    ``cout << minimum << ``'\n'``;` `}`

## Java

 `// Java Program to implement` `// the above approach` `import` `java.io.*;` `class` `GFG{`   `// Function to find minimum ` `// characters to be removed ` `// such that no subsequence` `// of the form "0101" exists` `// in the string` `static` `int` `findmin(String s)` `{` `  ``int` `n = s.length();` `  ``int` `i, j, maximum = ``0``;`   `  ``// Stores the partial sums` `  ``int``[] incr = ``new` `int``[n + ``1``];`   `  ``for` `(i = ``0``; i < n; i++) ` `  ``{` `    ``// Calculate partial sums` `    ``incr[i + ``1``] = incr[i];`   `    ``if` `(s.charAt(i) == ``'0'``) ` `    ``{` `      ``incr[i + ``1``]++;` `    ``}` `  ``}`   `  ``for` `(i = ``0``; i < n; i++) ` `  ``{` `    ``for` `(j = i + ``1``; j < n; j++) ` `    ``{` `      ``// Setting endpoints and` `      ``// deleting characters indices.` `      ``maximum = Math.max(maximum, incr[i] + ` `                         ``j - i + ``1` `- ` `                        ``(incr[j + ``1``] - incr[i]) + ` `                         ``incr[n] - incr[j + ``1``]);` `    ``}` `  ``}`   `  ``// Return count of ` `  ``// deleted characters` `  ``return` `n - maximum;` `}` `  `  `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `  ``String S = ``"0110100110"``;` `  ``int` `minimum = findmin(S);` `  ``System.out.println(minimum);` `}` `}`   `// This code is contributed by akhilsaini`

## Python3

 `# Python3 Program to implement` `# the above approach`   `# Function to find minimum ` `# characters to be removed ` `# such that no subsequence` `# of the form "0101" exists ` `# in the string` `def` `findmin(s):`   `    ``n ``=` `len``(s)` `    ``maximum ``=` `0`   `    ``# Stores the partial sums` `    ``incr ``=` `[``0``] ``*` `(n ``+` `1``)` `    `  `    ``for` `i ``in` `range``(``0``, n):` `      `  `        ``# Calculate partial sums` `        ``incr[i ``+` `1``] ``=` `incr[i]`   `        ``if` `(s[i] ``=``=` `'0'``):` `            ``incr[i ``+` `1``] ``=` `incr[i ``+` `1``] ``+` `1`   `    ``for` `i ``in` `range``(``0``, n ``+` `1``):` `        ``for` `j ``in` `range``(i ``+` `1``, n):`   `            ``# Setting endpoints and` `            ``# deleting characters indices.` `            ``maximum ``=` `max``(maximum, incr[i] ``+` `                          ``j ``-` `i ``+` `1` `-` `                         ``(incr[j ``+` `1``] ``-` `incr[i]) ``+` `                          ``incr[n] ``-` `incr[j ``+` `1``])`   `    ``# Return count of` `    ``# deleted characters` `    ``return` `n ``-` `maximum`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``S ``=` `"0110100110"` `    ``minimum ``=` `findmin(S)` `    ``print``(minimum)`   `# This code is contributed by akhilsaini`

## C#

 `// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{`   `// Function to find minimum ` `// characters to be removed ` `// such that no subsequence` `// of the form "0101" exists ` `// in the string` `static` `int` `findmin(``string` `s)` `{` `  ``int` `n = s.Length;` `  ``int` `i, j, maximum = 0;`   `  ``// Stores the partial sums` `  ``int``[] incr = ``new` `int``[n + 1];`   `  ``for` `(i = 0; i < n; i++) ` `  ``{` `    ``// Calculate partial sums` `    ``incr[i + 1] = incr[i];`   `    ``if` `(s[i] == ``'0'``) ` `    ``{` `      ``incr[i + 1]++;` `    ``}` `  ``}`   `  ``for` `(i = 0; i < n; i++) ` `  ``{` `    ``for` `(j = i + 1; j < n; j++) ` `    ``{` `      ``// Setting endpoints and` `      ``// deleting characters indices.` `      ``maximum = Math.Max(maximum, incr[i] + ` `                         ``j - i + 1 - ` `                        ``(incr[j + 1] - incr[i]) + ` `                         ``incr[n] - incr[j + 1]);` `    ``}` `  ``}`   `  ``// Return count of ` `  ``// deleted characters` `  ``return` `n - maximum;` `}` `  `  `// Driver Code` `public` `static` `void` `Main()` `{` `  ``string` `S = ``"0110100110"``;` `  ``int` `minimum = findmin(S);` `  ``Console.WriteLine(minimum);` `}` `}`   `// This code is contributed by akhilsaini`

Output:

```3

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.

Improved By : akhilsaini

Article Tags :
Practice Tags :

Be the First to upvote.

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