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 stringint 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 occurring 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 occurring 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 Codeint main(){ string str = "neveropenneveropen"; 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 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 = "neveropenneveropen"; 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 stringdef 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 occurring # 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 occurring # 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 Codestr = "neveropenneveropen"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 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[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 = "neveropenneveropen"; 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 = "neveropenneveropen"; document.write(getMinimumRemoval(str)); </script> |
Output:
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.
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