Related Articles
Minimum removal of subsequences of distinct consecutive characters required to empty a given string
• Last Updated : 03 Dec, 2020

Given a binary string, str, the task is to empty the given string by minimum number of removals of a single character or a subsequence containing distinct consecutive characters from str.

Examples:

Input: str = “0100100111”
Output:
Explanation:
Removing the subsequence “010101” from the string modifies str to “0011”.
Removing the subsequence “01” from the string modifies str to “01”.
Removing the subsequence “01” from the string modifies str to “” which is an empty string.
Therefore, the required output is 3.

Input: str = “010110”
Output: 2

Naive Approach: The simplest approach to solve this problem is to traverse the string repetitively and remove the longest subsequence of distinct consecutive characters from the string and increment the count after every removal. Finally, print the count.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

• Initialize two variables, say cntOne and cntZero, to store the count of 1s and 0s.
• Traverse the string using variable i and check the following conditions:
• If str[i] == ‘0’, then increment the value of cntZero and check if the value of cntOne is greater than 0 or not. If found to be true, then decrement the value of cntOne.
• If str[i] == ‘1’, then increment the value of cntOne and check if the value of cntZero is greater than 0 or not. If found to be true, then decrement the value of cntZero.
• Finally, print the value of (cntZero + cntOne).

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `//Function to count minimum operations required` `// to make the string an empty string` `int` `findMinOperationsReqEmpStr(string str)` `{` `    ``// Stores count of 1s by removing` `    ``// consecutive distinct subsequence` `    ``int` `cntOne = 0;`   `    ``// Stores count of 0s by removing` `    ``// consecutive distinct subsequence` `    ``int` `cntZero = 0 ;`   `    ``// Stores length of str` `    ``int` `N = str.length();`   `    ``// Traverse the string` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// If current character` `        ``// is 0` `        ``if``(str[i] == ``'0'``){` `            ``if` `(cntOne) {`   `                ``// Update cntOne` `                ``cntOne--;` `            ``}`   `            ``// Update cntZero` `            ``cntZero++;` `        ``}`   `        ``// If current character` `        ``// is 1` `        ``else``{`   `            ``//Update cntZero` `            ``if` `(cntZero) {` `                ``cntZero--;` `            ``}` `            `  `            `  `            ``// Update cntOne` `            ``cntOne++;` `        ``}` `    ``}` `    `  `    ``return` `(cntOne + cntZero);` `}`     `// Driver Code` `int` `main()` `{` `    ``string str = ``"0100100111"``;` `    ``cout<< findMinOperationsReqEmpStr(str);` `}`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.util.*;` `import` `java.lang.*;`   `class` `GFG{`   `//Function to count minimum operations required ` `// to make the string an empty string ` `static` `int` `findMinOperationsReqEmpStr(String str)` `{` `    `  `    ``// Stores count of 1s by removing ` `    ``// consecutive distinct subsequence ` `    ``int` `cntOne = ``0``; `   `    ``// Stores count of 0s by removing ` `    ``// consecutive distinct subsequence ` `    ``int` `cntZero = ``0``; `   `    ``// Stores length of str ` `    ``int` `N = str.length(); `   `    ``// Traverse the string ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        `  `        ``// If current character ` `        ``// is 0 ` `        ``if``(str.charAt(i) == ``'0'``)` `        ``{ ` `            ``if` `(cntOne != ``0``)` `            ``{ ` `                `  `                ``// Update cntOne ` `                ``cntOne--; ` `            ``} `   `            ``// Update cntZero ` `            ``cntZero++; ` `        ``} `   `        ``// If current character ` `        ``// is 1 ` `        ``else` `        ``{ ` `            `  `            ``// Update cntZero ` `            ``if` `(cntZero != ``0``) ` `            ``{` `                ``cntZero--; ` `            ``} ` `            `  `            ``// Update cntOne ` `            ``cntOne++; ` `        ``} ` `    ``} ` `    ``return` `(cntOne + cntZero); ` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String str = ``"0100100111"``; ` `    `  `    ``System.out.print(findMinOperationsReqEmpStr(str)); ` `}` `}`   `// This code is contributed by ajaykr00kj`

## Python3

 `# Python3 program to implement ` `# the above approach `   `# Function to count minimum operations ` `# required to make the string an empty ` `# string ` `def` `findMinOperationsReqEmpStr(``str``):` `    `  `    ``# Stores count of 1s by removing ` `    ``# consecutive distinct subsequence ` `    ``cntOne ``=` `0` `  `  `    ``# Stores count of 0s by removing ` `    ``# consecutive distinct subsequence ` `    ``cntZero ``=` `0` `    `  `    ``# Traverse the string ` `    ``for` `element ``in` `str``: ` `        `  `        ``# If current character ` `        ``# is 0 ` `        ``if` `element ``=``=` `'0'``: ` `            ``if` `cntOne > ``0``:  ` `  `  `                ``# Update cntOne ` `                ``cntOne ``=` `cntOne ``-` `1` `                `  `            ``# Update cntZero ` `            ``cntZero ``=` `cntZero ``+` `1` `            `  `        ``# If current character ` `        ``# is 1 ` `        ``else``:` `            `  `            ``# Update cntZero ` `            ``if` `cntZero > ``0``: ` `                ``cntZero ``=` `cntZero ``-` `1` `             `  `            ``# Update cntOne ` `            ``cntOne ``=` `cntOne ``+` `1` `    `  `    ``return` `cntOne ``+` `cntZero   ` `    `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    `  `    ``str` `=` `"0100100111"` `    `  `    ``print``(findMinOperationsReqEmpStr(``str``)) `   `# This code is contributed by ajaykr00kj`

## C#

 `// C# program to implement ` `// the above approach ` `using` `System;`   `class` `GFG` `{`   `// Function to count minimum operations required ` `// to make the string an empty string ` `static` `int` `findMinOperationsReqEmpStr(String str)` `{` `    `  `    ``// Stores count of 1s by removing ` `    ``// consecutive distinct subsequence ` `    ``int` `cntOne = 0; `   `    ``// Stores count of 0s by removing ` `    ``// consecutive distinct subsequence ` `    ``int` `cntZero = 0; `   `    ``// Stores length of str ` `    ``int` `N = str.Length; `   `    ``// Traverse the string ` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{ ` `        `  `        ``// If current character ` `        ``// is 0 ` `        ``if``(str[i] == ``'0'``)` `        ``{ ` `            ``if` `(cntOne != 0)` `            ``{ ` `                `  `                ``// Update cntOne ` `                ``cntOne--; ` `            ``} `   `            ``// Update cntZero ` `            ``cntZero++; ` `        ``} `   `        ``// If current character ` `        ``// is 1 ` `        ``else` `        ``{ ` `            `  `            ``// Update cntZero ` `            ``if` `(cntZero != 0) ` `            ``{` `                ``cntZero--; ` `            ``} ` `            `  `            ``// Update cntOne ` `            ``cntOne++; ` `        ``} ` `    ``} ` `    ``return` `(cntOne + cntZero); ` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``String str = ``"0100100111"``; ` `    `  `    ``Console.Write(findMinOperationsReqEmpStr(str)); ` `}` `}`   `// This code is contributed by 29AjayKumar `

Output:

`3`

Time Complexity: O(N)
Auxiliary Space: O(1)

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
Recommended Articles
Page :