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:
Java
// Java program of the// above approachclass GFG{ // Function to find the minimum// characters to be removed from// the stringpublic 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 occurring 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 occurring 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 codepublic static void main(String[] args){ String str = "geeksgeeks"; System.out.print(getMinimumRemoval(str));}}// This code is contributed by divyeshrabadiya07 |
6
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1), as we are not using any extra space.
Please refer complete article on Minimize characters to be changed to make the left and right rotation of a string same for more details!
