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Minimum replacements in a string to make adjacent characters unequal

  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2021

Given a lowercase character string str of size N. In one operation any character can be changed into some other character. The task is to find the minimum number of operations such that no two adjacent characters are equal.
Examples:

Input: Str = “caaab” 
Output:
Explanation: 
Change the second a to any other character, let’s change it to b. So the string becomes “cabab”. and no two adjacent characters are equal. So minimum number of operations is 1.
Input: Str = “xxxxxxx” 
Output:
Explanation: 
Replace ‘x’ at index 1, 3 and 5 to ‘a’, ‘b’, and ‘c’ respectively.

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Approach:  The idea is similar to implement sliding window technique. In this, we need to find the non-overlapping substrings that have all the characters the same. Then the minimum operations will be the sum of the floor of half the length of each substring.

  1. There is no need to change a character directly. Instead, consider all substring started from any index having only one character.
  2. Now consider any substring of length l such that all the characters of that substring are equal then change floor ( l / 2) characters of this substring to some other character.
  3. So just iterate over all the characters of the string from any character ch find out the maximal length of the substring such that all the characters in that substring are equal to the character ch.
  4. Find the length l of this substring and add floor ( l / 2) to the ans.
  5. After that start from the character just next to the end of the above substring.

C++14




// C++ program to find minimum
// replacements in a string to
// make adjacent characters unequal
 
#include <bits/stdc++.h>
using namespace std;
 
// Function which counts the minimum
// number of required operations
void count_minimum(string s)
{
    // n stores the length of the string s
    int n = s.length();
 
    // ans will store the required ans
    int ans = 0;
 
    // i is the current index in the string
    int i = 0;
 
    while (i < n) {
 
        int j = i;
 
        // Move j until characters s[i] & s[j]
        // are equal or the end of the
        // string is reached
        while (s[j] == s[i] && j < n) {
            j++;
        }
 
        // diff stores the length of the
        // substring such that all the
        // characters are equal in it
        int diff = j - i;
 
        // We need atleast diff/2 operations
        // for this substring
        ans += diff / 2;
        i = j;
    }
 
    cout << ans << endl;
}
 
// Driver code
int main()
{
    string str = "caaab";
    count_minimum(str);
    return 0;
}

Java




// Java program to find minimum
// replacements in a string to
// make adjacent characters unequal
import java.util.*;
 
class GFG{
 
// Function which counts the minimum
// number of required operations
static void count_minimum(String s)
{
     
    // n stores the length of the string s
    int n = s.length();
 
    // ans will store the required ans
    int ans = 0;
 
    // i is the current index in the string
    int i = 0;
 
    while (i < n)
    {
        int j = i;
 
        // Move j until characters s[i] & s[j]
        // are equal or the end of the
        // string is reached
        while (j < n && s.charAt(j) ==
                        s.charAt(i))
        {
            j++;
        }
 
        // diff stores the length of the
        // substring such that all the
        // characters are equal in it
        int diff = j - i;
 
        // We need atleast diff/2 operations
        // for this substring
        ans += diff / 2;
        i = j;
    }
    System.out.println(ans);
}
 
// Driver code
public static void main(String[] args)
{
    String str = "caaab";
 
    count_minimum(str);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program to find minimum
# replacements in a string to
# make adjacent characters unequal
 
# Function which counts the minimum
# number of required operations
def count_minimum(s):
     
    # n stores the length of the string s
    n = len(s)
 
    # ans will store the required ans
    ans = 0
     
    # i is the current index in the string
    i = 0
     
    while i < n:
        j = i
         
        # Move j until characters s[i] & s[j]
        # are equal or the end of the
        # string is reached
        while j < n and (s[j] == s[i]):
            j += 1
         
        # diff stores the length of the
        # substring such that all the
        # characters are equal in it
        diff = j - i
         
        # We need atleast diff/2 operations
        # for this substring
        ans += diff // 2
        i = j
         
    print(ans)
     
# Driver code
if __name__=="__main__":
     
    str = "caaab"
    count_minimum(str)
     
# This code is contributed by rutvik_56

C#




// C# program to find minimum
// replacements in a string to
// make adjacent characters unequal
using System;
 
class GFG{
     
// Function which counts the minimum
// number of required operations
static void count_minimum(string s)
{
     
    // n stores the length of the string s
    int n = s.Length;
 
    // ans will store the required ans
    int ans = 0;
 
    // i is the current index in the string
    int i = 0;
 
    while (i < n)
    {
        int j = i;
 
        // Move j until characters s[i] & s[j]
        // are equal or the end of the
        // string is reached
        while (j < n && s[j] == s[i])
        {
            j++;
        }
 
        // diff stores the length of the
        // substring such that all the
        // characters are equal in it
        int diff = j - i;
 
        // We need atleast diff/2 operations
        // for this substring
        ans += diff / 2;
        i = j;
    }
    Console.WriteLine(ans);
}
 
// Driver code
static void Main()
{
    string str = "caaab";
     
    count_minimum(str);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
      // JavaScript program to find minimum
      // replacements in a string to
      // make adjacent characters unequal
 
      // Function which counts the minimum
      // number of required operations
      function count_minimum(s) {
        // n stores the length of the string s
        var n = s.length;
 
        // ans will store the required ans
        var ans = 0;
 
        // i is the current index in the string
        var i = 0;
 
        while (i < n) {
          var j = i;
 
          // Move j until characters s[i] & s[j]
          // are equal or the end of the
          // string is reached
          while (s[j] === s[i] && j < n) {
            j++;
          }
 
          // diff stores the length of the
          // substring such that all the
          // characters are equal in it
          var diff = j - i;
 
          // We need atleast diff/2 operations
          // for this substring
          ans += parseInt(diff / 2);
          i = j;
        }
 
        document.write(ans + "<br>");
      }
 
      // Driver code
      var str = "caaab";
      count_minimum(str);
    </script>
Output
1

Time Complexity: O (N)

Auxiliary Space: O (1)




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