Minimum removal of consecutive similar characters required to empty a Binary String

Given a binary string S of length N, the task is to find the minimum number of removal of adjacent similar characters required to empty the given binary string.

Examples:

Input: S = “1100011“
Output: 2
Explanation:
Operation 1: Removal of all 0s modifies S to “1111“.
Operation 2: Removal of all remaining 1s makes S empty.
Therefore, the minimum number of operations required is 2.

Input: S = “0010100“
Output: 3
Explanation:
Operation 1: Removal of all 1s modifes S to “000100“.
Operation 2: Removal of all 1s modifies S = “00000“.
Operation 3: Removal of all remaining 0s makes S empty.
Therefore, the minimum number of operations required is 3.

Approach: The given problem can be solved using Greedy Approach. The idea is to delete the consecutive occurrences of the character with higher frequency. Follow the steps below to solve the problem:



  • Traverse the given string S and generate a new string, say newString, by removing consecutive occurrences of the character with higher frequency.
  • Finally, print (sizeof(newString) + 1)/2 as the required answer

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum steps
// to make the string empty
int minSteps(string S)
{
    // Stores the modified string
    string new_str;
 
    // Size of string
    int N = S.length();
 
    int i = 0;
 
    while (i < N) {
 
        new_str += S[i];
 
        // Remving substring of same
        // character from modified string
        int j = i;
        while (i < N && S[i] == S[j])
            ++i;
    }
 
    // Print the minimum steps required
    cout << ceil((new_str.size() + 1) / 2.0);
}
 
// Driver Code
int main()
{
    // Given string S
    string S = "0010100";
 
    // Function Call
    minSteps(S);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find minimum steps
// to make the String empty
static void minSteps(String S)
{
     
    // Stores the modified String
    String new_str = "";
 
    // Size of String
    int N = S.length();
 
    int i = 0;
 
    while (i < N)
    {
        new_str += S.charAt(i);
         
        // Remving subString of same
        // character from modified String
        int j = i;
        while (i < N && S.charAt(i) == S.charAt(j))
            ++i;
    }
 
    // Print the minimum steps required
    System.out.print((int)Math.ceil(
        (new_str.length() + 1) / 2.0));
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String S
    String S = "0010100";
 
    // Function Call
    minSteps(S);
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above approach
from math import ceil
 
# Function to find minimum steps
# to make the empty
def minSteps(S):
     
    # Stores the modified string
    new_str = ""
 
    # Size of string
    N = len(S)
 
    i = 0
 
    while (i < N):
        new_str += S[i]
 
        # Removing substring of same character
        # from modified string
        j = i
        while (i < N and S[i] == S[j]):
            i += 1
 
    # Print the minimum steps required
    print(ceil((len(new_str) + 1) / 2))
 
# Driver Code
if __name__ == '__main__':
     
    # Given S
    S = "0010100"
 
    # Function Call
    minSteps(S)
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
 
class GFG{
 
// Function to find minimum steps
// to make the string empty
static void minSteps(string S)
{
     
    // Stores the modified string
    string new_str = "";
     
    // Size of string
    int N = S.Length;
 
    int i = 0;
 
    while (i < N)
    {
        new_str += S[i];
         
        // Remving substring of same
        // character from modified string
        int j = i;
         
        while (i < N && S[i] == S[j])
            ++i;
    }
 
    // Print the minimum steps required
    Console.Write((int)Math.Ceiling(
        (new_str.Length + 1) / 2.0));
}
 
// Driver Code
public static void Main()
{
     
    // Given string S
    string S = "0010100";
 
    // Function Call
    minSteps(S);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

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Output: 

3










 

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

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