Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Minimize insertions required to make all characters of a given string equal

  • Last Updated : 29 Apr, 2021

Given a binary string S of length N, the task is to find the minimum number of characters required to be inserted such that all the characters in the string becomes the same based on the condition that:

If ‘1’ is inserted into the string, then all the ‘0’s nearest to the inserted ‘1’ is flipped or vice-versa.

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Examples:



Input: S = “11100”
Output: 1
Explanation:
Operation 1: Inserting ‘1’ at the last of the given string modifies S to “111001”. Adding ‘1’ to the last flips all the nearest ‘0’s to the inserted ‘1’. Therefore, the resultant string is “111111”.
After completing the above operation, all the characters of the string are the same. Therefore count of operations is 1.

Input: S = “0101010101”
Output: 9

 

Approach: The idea is to solve this problem by Greedy  Approach based on the following observations:

  • It can be seen that inverting one continuous section of ‘1’s or ‘0’s reduces the number of sections by one in this operation. Therefore, it is sufficient to repeat this operation to make it all into one section. The number of operations required is equal to sections – 1.
  • In simpler terms, count the total number of non-equal adjacent pair of characters, so that inverting one of them can convert the whole substring into similar substrings.

Follow the steps below to solve the problem:

  • Initialize a variable, say count, that stores the count of different adjacent characters.
  • Traverse the string and check if the current and the next characters are different, then increment the value of count.
  • After completing the above steps, print the value of count as the minimum required operations.

Below is the implementation of the above approach:

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the minimum
// number of operations required to make
// all characters of the string same
int minOperations(string& S)
{
    // Stores count of operations
    int count = 0;
 
    // Traverse the string
    for (int i = 1; i < S.length(); i++) {
 
        // Check if adjacent
        // characters are same or not
        if (S[i] != S[i - 1]) {
 
            // Increment count
            count += 1;
        }
    }
 
    // Print the count obtained
    cout << count;
}
 
// Driver Code
int main()
{
    string S = "0101010101";
    minOperations(S);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG
{
   
  // Function to calculate the minimum
  // number of operations required to make
  // all characters of the string same
  static void minOperations(String S)
  {
 
    // Stores count of operations
    int count = 0;
 
    // Traverse the string
    for (int i = 1; i < S.length(); i++)
    {
 
      // Check if adjacent
      // characters are same or not
      if (S.charAt(i) != S.charAt(i - 1))
      {
 
        // Increment count
        count += 1;
      }
    }
 
    // Print the count obtained
     System.out.print(count);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String S = "0101010101";
    minOperations(S);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Python3




# Python program to implement
# the above approach
 
# Function to calculate the minimum
# number of operations required to make
# all characters of the string same
def minOperations(S):
   
  # Stores count of operations
    count = 0;
 
    # Traverse the string
    for i in range(1, len(S)):
 
        # Check if adjacent
        # characters are same or not
        if (S[i] != S[i - 1]):
           
            # Increment count
            count += 1;
 
    # Prthe count obtained
    print(count);
 
# Driver Code
if __name__ == '__main__':
    S = "0101010101";
    minOperations(S);
 
# This code is contributed by 29AjayKumar

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to calculate the minimum
  // number of operations required to make
  // all characters of the string same
  static void minOperations(string S)
  {
 
    // Stores count of operations
    int count = 0;
 
    // Traverse the string
    for (int i = 1; i < S.Length; i++)
    {
 
      // Check if adjacent
      // characters are same or not
      if (S[i] != S[i - 1])
      {
 
        // Increment count
        count += 1;
      }
    }
 
    // Print the count obtained
    Console.Write(count);
  }
 
  // Driver Code
  public static void Main()
  {
    string S = "0101010101";
    minOperations(S);
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
      // JavaScript program for the above approach
 
      // Function to calculate the minimum
      // number of operations required to make
      // all characters of the string same
      function minOperations(S) {
        // Stores count of operations
        var count = 0;
 
        // Traverse the string
        for (var i = 1; i < S.length; i++) {
          // Check if adjacent
          // characters are same or not
          if (S[i] !== S[i - 1]) {
            // Increment count
            count += 1;
          }
        }
 
        // Print the count obtained
        document.write(count);
      }
 
      // Driver Code
      var S = "0101010101";
      minOperations(S);
    </script>
Output: 
9

 

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!