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Minimize characters to be changed to make the left and right rotation of a string same

  • Difficulty Level : Expert
  • Last Updated : 18 Jun, 2021

Given a string S of lowercase English alphabets, the task is to find the minimum number of characters to be changed such that the left and right rotation of the string are the same.

Examples:

Input: S = “abcd”
Output: 2
Explanation:
String after the left shift: “bcda”
String after the right shift: “dabc”
Changing the character at position 3 to ‘a’ and character at position 4 to ‘b’, the string is modified to “abab”.
Therefore, both the left and right rotations becomes “baba”.

Input: S = “gfg”
Output: 1
Explanation:
After updating the character at position 1 to ‘g’, the string becomes “ggg”.
Therefore, the left and right rotation are equal.

Approach: The key observation to solve the problem is that when the length of the string is even, then all the characters at even index and characters at odd index must be the same for the left and right rotations to be the same. For strings of odd length, all the characters must be equal. Follow the steps below to solve the problem:

  • Check if the length of the string is even, then the minimum number of characters to be changed is the length of the string excluding the frequency of the most occurring element at the even indices and odd indices.
  • Otherwise, if the length of the string is odd, then the minimum number of characters to be changed is the length of the string excluding the frequency of the most occurring character in the string.
  • Print the final count obtained.

Below is the implementation of the above approach:

C++




// C++ Program of the
// above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum
// characters to be removed from
// the string
int getMinimumRemoval(string str)
{
    int n = str.length();
 
    // Initialize answer by N
    int ans = n;
 
    // If length is even
    if (n % 2 == 0) {
 
        // Frequency array for odd
        // and even indices
        vector<int> freqEven(128);
        vector<int> freqOdd(128);
 
        // Store the frequency of the
        // characters at even and odd
        // indices
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                freqEven[str[i]]++;
            }
            else {
                freqOdd[str[i]]++;
            }
        }
 
        // Stores the most occuring frequency
        // for even and odd indices
        int evenMax = 0, oddMax = 0;
 
        for (char chr = 'a'; chr <= 'z'; chr++) {
 
            evenMax = max(evenMax, freqEven[chr]);
            oddMax = max(oddMax, freqOdd[chr]);
        }
 
        // Update the answer
        ans = ans - evenMax - oddMax;
    }
 
    // If length is odd
    else {
        // Stores the frequency of the
        // characters of the string
        vector<int> freq(128);
        for (int i = 0; i < n; i++) {
            freq[str[i]]++;
        }
 
        // Stores the most occuring character
        // in the string
        int strMax = 0;
        for (char chr = 'a'; chr <= 'z'; chr++) {
            strMax = max(strMax, freq[chr]);
        }
 
        // Update the answer
        ans = ans - strMax;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    string str = "geeksgeeks";
 
    cout << getMinimumRemoval(str);
}

Java




// Java program of the
// above approach
class GFG{
     
// Function to find the minimum
// characters to be removed from
// the string
public static int getMinimumRemoval(String str)
{
    int n = str.length();
 
    // Initialize answer by N
    int ans = n;
 
    // If length is even
    if (n % 2 == 0)
    {
 
        // Frequency array for odd
        // and even indices
        int[] freqEven = new int[128];
        int[] freqOdd = new int[128];
 
        // Store the frequency of the
        // characters at even and odd
        // indices
        for(int i = 0; i < n; i++)
        {
            if (i % 2 == 0)
            {
                freqEven[str.charAt(i)]++;
            }
            else
            {
                freqOdd[str.charAt(i)]++;
            }
        }
 
        // Stores the most occuring frequency
        // for even and odd indices
        int evenMax = 0, oddMax = 0;
 
        for(char chr = 'a'; chr <= 'z'; chr++)
        {
            evenMax = Math.max(evenMax,
                               freqEven[chr]);
            oddMax = Math.max(oddMax,
                              freqOdd[chr]);
        }
 
        // Update the answer
        ans = ans - evenMax - oddMax;
    }
 
    // If length is odd
    else
    {
         
        // Stores the frequency of the
        // characters of the string
        int[] freq = new int[128];
        for(int i = 0; i < n; i++)
        {
            freq[str.charAt(i)]++;
        }
 
        // Stores the most occuring character
        // in the string
        int strMax = 0;
        for(char chr = 'a'; chr <= 'z'; chr++)
        {
            strMax = Math.max(strMax, freq[chr]);
        }
         
        // Update the answer
        ans = ans - strMax;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String str = "geeksgeeks";
     
    System.out.print(getMinimumRemoval(str));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 Program of the
# above approach
 
# Function to find the minimum
# characters to be removed from
# the string
def getMinimumRemoval(str):
   
    n = len(str)
 
    # Initialize answer by N
    ans = n
 
    # If length is even
    if(n % 2 == 0):
 
        # Frequency array for odd
        # and even indices
        freqEven = {}
        freqOdd = {}
        for ch in range(ord('a'),
                        ord('z') + 1):
            freqEven[chr(ch)] = 0
            freqOdd[chr(ch)] = 0
 
        # Store the frequency of the
        # characters at even and odd
        # indices
        for i in range(n):
            if(i % 2 == 0):
                if str[i] in freqEven:
                    freqEven[str[i]] += 1
            else:
                if str[i] in freqOdd:
                    freqOdd[str[i]] += 1
 
        # Stores the most occuring
        # frequency for even and
        # odd indices
        evenMax = 0
        oddMax = 0
 
        for ch in range(ord('a'),
                        ord('z') + 1):
            evenMax = max(evenMax,
                      freqEven[chr(ch)])
            oddMax = max(oddMax,
                     freqOdd[chr(ch)])
 
        # Update the answer
        ans = ans - evenMax - oddMax
 
    # If length is odd
    else:
 
        # Stores the frequency of the
        # characters of the string
        freq = {}
         
        for ch in range('a','z'):
            freq[chr(ch)] = 0
        for i in range(n):
            if str[i] in freq:
                freq[str[i]] += 1
 
        # Stores the most occuring
        # characterin the string
        strMax = 0
         
        for ch in range('a','z'):
            strMax = max(strMax,
                     freq[chr(ch)])
 
        # Update the answer
        ans = ans - strMax
 
    return ans
 
# Driver Code
str = "geeksgeeks"
print(getMinimumRemoval(str))
 
# This code is contributed by avanitrachhadiya2155

C#




// C# program of the
// above approach
using System;
class GFG{
     
// Function to find the minimum
// characters to be removed from
// the string
public static int getMinimumRemoval(String str)
{
  int n = str.Length;
 
  // Initialize answer by N
  int ans = n;
 
  // If length is even
  if (n % 2 == 0)
  {
    // Frequency array for odd
    // and even indices
    int[] freqEven = new int[128];
    int[] freqOdd = new int[128];
 
    // Store the frequency of the
    // characters at even and odd
    // indices
    for(int i = 0; i < n; i++)
    {
      if (i % 2 == 0)
      {
        freqEven[str[i]]++;
      }
      else
      {
        freqOdd[str[i]]++;
      }
    }
 
    // Stores the most occuring frequency
    // for even and odd indices
    int evenMax = 0, oddMax = 0;
 
    for(char chr = 'a'; chr <= 'z'; chr++)
    {
      evenMax = Math.Max(evenMax,
                         freqEven[chr]);
      oddMax = Math.Max(oddMax,
                        freqOdd[chr]);
    }
 
    // Update the answer
    ans = ans - evenMax - oddMax;
  }
 
  // If length is odd
  else
  {
    // Stores the frequency of the
    // characters of the string
    int[] freq = new int[128];
    for(int i = 0; i < n; i++)
    {
      freq[str[i]]++;
    }
 
    // Stores the most occuring character
    // in the string
    int strMax = 0;
    for(char chr = 'a'; chr <= 'z'; chr++)
    {
      strMax = Math.Max(strMax, freq[chr]);
    }
 
    // Update the answer
    ans = ans - strMax;
  }
  return ans;
}
 
// Driver code
public static void Main(String[] args)
{
  String str = "geeksgeeks";
  Console.Write(getMinimumRemoval(str));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
      // JavaScript program of the
      // above approach
      // Function to find the minimum
      // characters to be removed from
      // the string
      function getMinimumRemoval(str) {
        var n = str.length;
 
        // Initialize answer by N
        var ans = n;
 
        // If length is even
        if (n % 2 === 0) {
          // Frequency array for odd
          // and even indices
          var freqEven = new Array(128).fill(0);
          var freqOdd = new Array(128).fill(0);
 
          // Store the frequency of the
          // characters at even and odd
          // indices
          for (var i = 0; i < n; i++) {
            if (i % 2 === 0) {
              freqEven[str[i].charCodeAt(0)]++;
            } else {
              freqOdd[str[i].charCodeAt(0)]++;
            }
          }
 
          // Stores the most occuring frequency
          // for even and odd indices
          var evenMax = 0,
            oddMax = 0;
 
          for (var chr = "a".charCodeAt(0); chr <= "z".charCodeAt(0); chr++) {
            evenMax = Math.max(evenMax, freqEven[chr]);
            oddMax = Math.max(oddMax, freqOdd[chr]);
          }
 
          // Update the answer
          ans = ans - evenMax - oddMax;
        }
 
        // If length is odd
        else {
          // Stores the frequency of the
          // characters of the string
          var freq = new Array(128).fill(0);
          for (var i = 0; i < n; i++) {
            freq[str[i].charCodeAt(0)]++;
          }
 
          // Stores the most occuring character
          // in the string
          var strMax = 0;
          for (var chr = "a".charCodeAt(0); chr <= "z".charCodeAt(0); chr++) {
            strMax = Math.max(strMax, freq[chr]);
          }
 
          // Update the answer
          ans = ans - strMax;
        }
        return ans;
      }
 
      // Driver code
      var str = "geeksgeeks";
      document.write(getMinimumRemoval(str));
    </script>
Output: 
6

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


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