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Minimum number of replacements to make the binary string alternating | Set 2

  • Difficulty Level : Easy
  • Last Updated : 19 Mar, 2021

Given a binary string str, the task is to find the minimum number of characters in the string that have to be replaced in order to make the string alternating (i.e. of the form 01010101… or 10101010…).
Examples: 
 

Input: str = “1100” 
Output:
Replace 2nd character with ‘0’ and 3rd character with ‘1’
Input: str = “1010” 
Output:
The string is already alternating. 
 

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We have discussed one approach in Number of flips to make binary string alternate. In this post a better approach is discussed.
Approach: For the string str, there can be two possible solutions. Either the resultant string can be 
 

  1. 010101… or
  2. 101010…

In order to find the minimum replacements, count the number of replacements to convert the string in type 1 and store it in count then minimum replacement will be min(count, len – count) where len is the length of the string. len – count is the number of replacements to convert the string in type 2.
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 minimum number of
// characters of the given binary string
// to be replaced to make the string alternating
int minReplacement(string s, int len)
{
    int ans = 0;
    for (int i = 0; i < len; i++) {
 
        // If there is 1 at even index positions
        if (i % 2 == 0 && s[i] == '1')
            ans++;
 
        // If there is 0 at odd index positions
        if (i % 2 == 1 && s[i] == '0')
            ans++;
    }
    return min(ans, len - ans);
}
 
// Driver code
int main()
{
    string s = "1100";
    int len = s.size();
    cout << minReplacement(s, len);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return the minimum number of
    // characters of the given binary string
    // to be replaced to make the string alternating
    static int minReplacement(String s, int len)
    {
        int ans = 0;
        for (int i = 0; i < len; i++) {
 
            // If there is 1 at even index positions
            if (i % 2 == 0 && s.charAt(i) == '1')
                ans++;
 
            // If there is 0 at odd index positions
            if (i % 2 == 1 && s.charAt(i) == '0')
                ans++;
        }
        return Math.min(ans, len - ans);
    }
 
    // Driver code
    public static void main(String args[])
    {
        String s = "1100";
        int len = s.length();
        System.out.print(minReplacement(s, len));
    }
}

Python3




# Python3 implementation of the approach.
 
# Function to return the minimum number of
# characters of the given binary string
# to be replaced to make the string alternating
def minReplacement(s, length):
 
    ans = 0
    for i in range(0, length):
 
        # If there is 1 at even index positions
        if i % 2 == 0 and s[i] == '1':
            ans += 1
 
        # If there is 0 at odd index positions
        if i % 2 == 1 and s[i] == '0':
            ans += 1
     
    return min(ans, length - ans)
 
# Driver code
if __name__ == "__main__":
 
    s = "1100"
    length = len(s)
    print(minReplacement(s, length))
         
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the minimum number of
    // characters of the given binary string
    // to be replaced to make the string alternating
    static int minReplacement(String s, int len)
    {
        int ans = 0;
        for (int i = 0; i < len; i++)
        {
 
            // If there is 1 at even index positions
            if (i % 2 == 0 && s[i] == '1')
                ans++;
 
            // If there is 0 at odd index positions
            if (i % 2 == 1 && s[i] == '0')
                ans++;
        }
        return Math.Min(ans, len - ans);
    }
 
    // Driver code
    public static void Main(String []args)
    {
        String s = "1100";
        int len = s.Length;
        Console.Write(minReplacement(s, len));
    }
}
 
// This code contributed by Rajput-Ji

PHP




<?php
// PHP implementation of the approach
 
// Function to return the minimum number of
// characters of the given binary string
// to be replaced to make the string alternating
function minReplacement($s, $len)
{
    $ans = 0;
    for ($i = 0; $i < $len; $i++)
    {
 
        // If there is 1 at even index positions
        if ($i % 2 == 0 && $s[$i] == '1')
            $ans++;
 
        // If there is 0 at odd index positions
        if ($i % 2 == 1 && $s[$i] == '0')
            $ans++;
    }
    return min($ans, $len - $ans);
}
 
// Driver code
$s = "1100";
$len = strlen($s);
echo minReplacement($s, $len);
 
// This code is contributed by chandan_jnu
?>

Javascript




<script>
 
 
// Javascript implementation of the approach
 
// Function to return the minimum number of
// characters of the given binary string
// to be replaced to make the string alternating
function minReplacement(s, len)
{
    var ans = 0;
    for (var i = 0; i < len; i++) {
 
        // If there is 1 at even index positions
        if (i % 2 == 0 && s[i] == '1')
            ans++;
 
        // If there is 0 at odd index positions
        if (i % 2 == 1 && s[i] == '0')
            ans++;
    }
    return Math.min(ans, len - ans);
}
 
// Driver code
var s = "1100";
var len = s.length;
document.write(minReplacement(s, len));
 
 
</script>
Output: 
2

 

Time Complexity: O(len) where len is the length of the given string.
 




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