Skip to content
Related Articles

Related Articles

Improve Article
Minimum operations required to convert a binary string to all 0s or all 1s
  • Last Updated : 06 May, 2021

Given a binary string str, the task is to find the minimum number of operations required to make all the characters of the string same i.e. either the resultant string contains all 0s or all 1s. In a single operation, any block of consecutive 0s can be converted to a block of consecutive 1s of the same length and vice versa.
Examples: 
 

Input: str = “000111” 
Output:
In a single operation, either change all 0s to 1s 
or change all 1s to 0s.
Input: str = “0011101010” 
Output:
All the 1s can be converted to 0s in 3 operations. 
 

 

Approach: The problem is to convert all the characters into a single one. Now since converting a whole consecutive group of characters counts as a single step. You can calculate the number of different groups as separated by each other due to the presence of other characters in between them. Now the number of steps would be simply the minimum of both the numbers. Hence, the answer will be the minimum of the count of consecutive blocks of 0s or the count of consecutive blocks of 1s.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// minimum operations required
int minOperations(string str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++) {
 
        // Increment count when consecutive
        // characters are different
        if (str[i] != str[i + 1])
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
int main()
{
    string str = "000111";
    int n = str.length();
 
    cout << minOperations(str, n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to return the count of
// minimum operations required
static int minOperations(String str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++)
    {
 
        // Increment count when consecutive
        // characters are different
        if (str.charAt(i) != str.charAt(i + 1))
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "000111";
    int n = str.length();
 
    System.out.println(minOperations(str, n));
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the approach
 
# Function to return the count of
# minimum operations required
def minOperations(str, n):
    count = 0
    for i in range(n - 1):
 
        # Increment count when consecutive
        # characters are different
        if (str[i] != str[i + 1]):
            count += 1
 
    # Answer is rounding off the
    # (count / 2) to lower
    return (count + 1) // 2
 
# Driver code
str = "000111"
n = len(str)
 
print(minOperations(str, n))
 
# This code is contributed
# by Mohit Kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count of
// minimum operations required
static int minOperations(string str, int n)
{
    int count = 0;
    for (int i = 0; i < n - 1; i++)
    {
 
        // Increment count when consecutive
        // characters are different
        if (str[(i)] != str[(i + 1)])
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
// Driver code
public static void Main()
{
    string str = "000111";
    int n = str.Length;
 
    Console.WriteLine(minOperations(str, n));
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
 //Javascript implementation of the approach
 
// Function to return the count of
// minimum operations required
function minOperations(str, n)
{
    var count = 0;
    for (var i = 0; i < n - 1; i++) {
 
        // Increment count when consecutive
        // characters are different
        if (str[i] != str[i + 1])
            count++;
    }
 
    // Answer is rounding off the
    // (count / 2) to lower
    return (count + 1) / 2;
}
 
 var str = "000111";
 var n = str.length;
document.write(minOperations(str, n));
 
// This code is contributed by SoumikMondal
</script>
Output: 
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.  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 industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :