Minimum replacements such that the difference between the index of the same characters is divisible by 3

Given a string of ‘0’, ‘1’ and ‘2’. The task is to find the minimum replacements in the string such that the differences between the indexes of the same characters is divisible by 3. 

Examples:

Input: s = “2101200” 
Output: 3 
1201201 or 2102102 can be the resultant string 
which has 3 replacements. 

Input: s = “012” 
Output: 0 

Approach: There can be 6 different strings such that the difference between the index of similar characters is divisible by 3. Hence generate all 6 different strings, and compare the replacements done with the original string. Store the string which has a minimal number of replacements. The different strings can be generated using next_permutation in C++. 

Below is the implementation of the above approach: 

C++

// C++ program to find minimum replacements
// such that the difference between the
// index of the same characters
// is divisible by 3
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// minimal replacements
int countMinimalReplacements(string s)
{

    int n = s.length();

    int mini = INT_MAX;
    string dup = "012";

    // Generate all permutations
    do {

        // Count the replacements
        int dif = 0;
        for (int i = 0; i < n; i++)
            if (s[i] != dup[i % 3])
                dif++;

        mini = min(mini, dif);
    } while (next_permutation(dup.begin(), dup.end()));

    // Return the replacements
    return mini;
}

// Driver Code
int main()
{
    string s = "2101200";
    cout << countMinimalReplacements(s);
    return 0;
}

Java

// Java program to find minimum replacements
// such that the difference between the
// index of the same characters
// is divisible by 3
class GFG {

    // Function to count the number of
    // minimal replacements
    static int countMinimalReplacements(String s)
    {

        int n = s.length();

        int mini = Integer.MAX_VALUE;
        char[] dup = "012".toCharArray();

        // Generate all permutations
        do {

            // Count the replacements
            int dif = 0;
            for (int i = 0; i < n; i++) {
                if (s.charAt(i) != dup[i % 3]) {
                    dif++;
                }
            }

            mini = Math.min(mini, dif);
        } while (next_permutation(dup));

        // Return the replacements
        return mini;
    }

    static boolean next_permutation(char[] p)
    {
        for (int a = p.length - 2; a >= 0; --a) {
            if (p[a] < p[a + 1]) {
                for (int b = p.length - 1;; --b) {
                    if (p[b] > p[a]) {
                        char t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
                }
            }
        }
        return false;
    }

    // Driver Code
    public static void main(String args[])
    {
        String s = "2101200";
        System.out.println(countMinimalReplacements(s));
    }
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python program to find minimum replacements
# such that the difference between the
# index of the same characters
# is divisible by 3

import sys

# Function to count the number of
# minimal replacements
def countMinimalReplacements(s):

    n = len(s)

    mini = sys.maxsize
    dup = "012"

    # Generate all permutations
    for i in range(6):
        # Count the replacements
        dif = 0
        for j in range(n):
            if s[j] != dup[j % 3]:
                dif += 1

        mini = min(mini, dif)
        dup = dup[1:] + dup[0]

    # Return the replacements
    return mini

# Driver Code
if __name__ == '__main__':
    s = "2101200"
    print(countMinimalReplacements(s))

C#

// C# program to find minimum replacements
// such that the difference between the
// index of the same characters
// is divisible by 3
using System;

class GFG {

    // Function to count the number of
    // minimal replacements
    static int countMinimalReplacements(String s)
    {

        int n = s.Length;

        int mini = int.MaxValue;
        char[] dup = "012".ToCharArray();

        // Generate all permutations
        do {

            // Count the replacements
            int dif = 0;
            for (int i = 0; i < n; i++) {
                if (s[i] != dup[i % 3]) {
                    dif++;
                }
            }

            mini = Math.Min(mini, dif);
        } while (next_permutation(dup));

        // Return the replacements
        return mini;
    }

    static bool next_permutation(char[] p)
    {
        for (int a = p.Length - 2; a >= 0; --a) {
            if (p[a] < p[a + 1]) {
                for (int b = p.Length - 1;; --b) {
                    if (p[b] > p[a]) {
                        char t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.Length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
                }
            }
        }
        return false;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String s = "2101200";
        Console.WriteLine(countMinimalReplacements(s));
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>
      // JavaScript program to find minimum replacements
      // such that the difference between the
      // index of the same characters
      // is divisible by 3
      // Function to count the number of
      // minimal replacements
      function countMinimalReplacements(s) {
        var n = s.length;

        //Max Integer Value
        var mini = 2147483647;
        var dup = "012".split("");

        // Generate all permutations
        do {
          // Count the replacements
          var dif = 0;
          for (var i = 0; i < n; i++) {
            if (s[i] !== dup[i % 3]) {
              dif++;
            }
          }

          mini = Math.min(mini, dif);
        } while (next_permutation(dup));

        // Return the replacements
        return mini;
      }

      function next_permutation(p) {
        for (var a = p.length - 2; a >= 0; --a) {
          if (p[a] < p[a + 1]) {
            for (var b = p.length - 1; ; --b) {
              if (p[b] > p[a]) {
                var t = p[a];
                p[a] = p[b];
                p[b] = t;
                for (++a, b = p.length - 1; a < b; ++a, --b) {
                  t = p[a];
                  p[a] = p[b];
                  p[b] = t;
                }
                return true;
              }
            }
          }
        }
        return false;
      }

      // Driver Code
      var s = "2101200";
      document.write(countMinimalReplacements(s));
    </script>

Output

3

Complexity Analysis:

• Time Complexity: O(3*N), as we are a loop to traverse N times. Where n is the length of the string.
• Auxiliary Space: O(1), as we are not using any extra space.