# Find minimum operations to make all characters of a string identical

Last Updated : 14 Oct, 2023

Given a string of size N, the task is to find the minimum operation required in a string such that all elements of a string are the same. In one operation, you can choose at least one character from the string and remove all chosen characters from the string, but you can’t choose an adjacent character from the string.

Examples:

Input: string = “geeks”
Output: 2
Explanation

• In one operation, you will remove elements at index 1 and 4. Now the string is “ees”,
• In the second operation, you will remove the element at index 3.

Output: 2
Explanation

• In one operation, you will remove elements at indexes 1, 4, 6, and 8. Now the string is “yacab”,
• In the second operation, you will remove elements at index 1, 3, and 5.

Approach: To solve the problem follow the below idea:

We will iterate from key = ‘a’ to ‘z’ and we will erase all elements except the key. For every key, we will find the length of the maximum subarray that doesn’t contain the current key. we will take the minimum from all maximum subarray lengths from a to z. Now our aim is to remove all elements from the subarray because we remove all elements from the maximum subarray for a particular key, then we can easily remove all subarray that length is less than or equal to the maximum subarray length in the same operation. Now, we will choose an element from the min length subarray such that no adjacent element is chosen and remove it and it will take a total Log2x+1 operation, where x is the length of the minimum subarray. We can find the answer also by dividing x by 2 until it is not zero and increasing the answer by 1 when we divide.

Below are the steps to implement the above idea:

• We will iterate the key from char- ‘a’ to ‘z’.
• Now we will find the maximum subarray length that doesn’t contain the current key for every key from ‘a’ to ‘z’.
• Then, we will take the minimum subarray length from all maximum subarray for a key from ‘a’ to ‘z’.
• Now, we have to remove all elements from the minimum length subarray.
• Then, we will divide the subarray length by 2 until it becomes zero and increase the answer by 1 when we divide.
• Finally, print the final answer.

Below is the implementation of the above approach:

## C++

 // C++ code for the above approach:   #include using namespace std;   // Function to find minimum operation // required in a string such that all // elements become same after performing // operation in a string int makeequal(string s, int n) {     int minsubarray, ans = 0;       // Iterating all lower case alphabets     for (int j = 0; j < 26; j++) {           // Current lower case alphabet         char key = 'a' + j;         int temp = 0, ma = 0;           for (int i = 0; i < n; i++) {             if (s[i] == key) {                   // Updating maximum                 // length of subarray                 ma = max(ma, temp);                   // That doesn't contain                 // current key                 temp = 0;             }             else {                   temp += 1;             }             if (i == n - 1) {                 ma = max(ma, temp);             }         }           minsubarray = min(minsubarray, ma);           // Taking minimum length subarray         // from maximum subarray length         // for each key from 'a' to 'z'.     }       // Dividing minsubarray by 2     // until it is not zero     while (minsubarray != 0) {         ans += 1;         minsubarray = minsubarray / 2;     }       // Return the answer     return ans; }   // Drive code int main() {     string s = "xyabcdadb";     int n = s.size();       // Function call     cout << makeequal(s, n);       return 0; }

## Java

 import java.util.Arrays;   public class Main {       // Function to find minimum operation     // required in a string such that all     // elements become same after performing     // operation in a string     static int makeEqual(String s, int n) {         int minSubarray = Integer.MAX_VALUE;         int ans = 0;           // Iterating all lower case alphabets         for (char key = 'a'; key <= 'z'; key++) {             int temp = 0;             int ma = 0;               for (int i = 0; i < n; i++) {                 if (s.charAt(i) == key) {                       // Updating maximum                     // length of subarray                     ma = Math.max(ma, temp);                       // That doesn't contain                     // current key                     temp = 0;                 } else {                     temp += 1;                 }                   if (i == n - 1) {                     ma = Math.max(ma, temp);                 }             }               minSubarray = Math.min(minSubarray, ma);               // Taking minimum length subarray             // from maximum subarray length             // for each key from 'a' to 'z'.         }           // Dividing minSubarray by 2         // until it is not zero         while (minSubarray != 0) {             ans += 1;             minSubarray /= 2;         }           // Return the answer         return ans;     }       // Drive code     public static void main(String[] args) {         String s = "xyabcdadb";         int n = s.length();           // Function call         System.out.println(makeEqual(s, n));     } } //This code is Contributed by chinmaya121221

## Python3

 # Python code for the above approach   # Function to find minimum operation # required in a string such that all # elements become same after performing # operation in a string def makeequal(s):     n = len(s)     min_subarray = float('inf')     ans = 0       # Iterating all lower case alphabets     for j in range(26):         # Current lower case alphabet         key = chr(ord('a') + j)         temp = 0         ma = 0           for i in range(n):             if s[i] == key:                 # Updating maximum length of subarray                 # that doesn't contain current key                 ma = max(ma, temp)                 temp = 0             else:                 temp += 1             if i == n - 1:                 ma = max(ma, temp)           min_subarray = min(min_subarray, ma)           # Taking minimum length subarray         # from maximum subarray length         # for each key from 'a' to 'z'.       # Dividing min_subarray by 2     # until it is not zero     while min_subarray != 0:         ans += 1         min_subarray = min_subarray // 2       # Return the answer     return ans   # Driver code if __name__ == '__main__':     s = "xyabcdadb"       # Function call     print(makeequal(s))

## C#

 using System;   class Program {       // Function to find minimum operation     // required in a string such that all     // elements become same after performing     // operation in a string     static int MakeEqual(string s, int n)     {         int minSubarray = 2147483647 , ans = 0;                     // Iterating all lower case alphabets         for (int j = 0; j < 26; j++)         {                           // Current lower case alphabet                char key = (char)('a' + j);             int temp = 0, ma = 0;                           for (int i = 0; i < n; i++)             {                 if (s[i] == key)                 {                         // Updating maximum                     // length of subarray                                      ma = Math.Max(ma, temp);                                           // That doesn't contain                    // current key                     temp = 0;                 }                 else                 {                     temp += 1;                 }                                   if (i == n - 1)                 {                     ma = Math.Max(ma, temp);                 }             }                           minSubarray = Math.Min(minSubarray, ma);                           // Taking minimum length subarray             // from maximum subarray length             // for each key from 'a' to 'z'.         }                     // Dividing minsubarray by 2         // until it is not zero         while (minSubarray != 0)         {             ans += 1;             minSubarray = minSubarray / 2;         }                     // Return the answer             return ans;     }         // Drive code     static void Main()     {         string s = "xyabcdadb";         int n = s.Length;                   Console.WriteLine(MakeEqual(s, n));     } }

## Javascript

 // Function to find minimum operation // required in a string such that all // elements become same after performing // operation in a string function makeequal(s) {     let n = s.length;     let min_subarray = Infinity;     let ans = 0;      // Iterating all lower case alphabets    for (let j = 0; j < 26; j++) {                 // Current lower case alphabet            let key = String.fromCharCode('a'.charCodeAt(0) + j);         let temp = 0;         let ma = 0;         for (let i = 0; i < n; i++) {             if (s[i] === key) {                                  // Updating maximum                 // length of subarray                    ma = Math.max(ma, temp);                                   // That doesn't contain                 // current key                 temp = 0;             } else {                 temp += 1;             }             if (i === n - 1) {                 ma = Math.max(ma, temp);             }         }                   // Taking minimum length subarray         // from maximum subarray length         // for each key from 'a' to 'z'.         min_subarray = Math.min(min_subarray, ma);     }       // Dividing minsubarray by 2     // until it is not zero     while (min_subarray !== 0) {         ans += 1;         min_subarray = Math.floor(min_subarray / 2);     }           // Return the answer     return ans; }   // Test the function let s = "xyabcdadb"; console.log(makeequal(s));

Output

2

Time Complexity: O(26*N)
Auxiliary Space: O(1)

### Approach 2

Another approach to solve the problem of finding the minimum operations to make all characters of a string identical is to use a greedy algorithm. Here’s an alternative approach

## C++

 #include #include #include // Include the necessary header for max_element and remove using namespace std;   int makeEqual(string str) {     int n = str.length();     vector frequency(26, 0); // Frequency array for lowercase letters       // Count the occurrences of each character     for (char ch : str) {         frequency[ch - 'a']++;     }       int operations = 0;       while (true) {         int maxFrequency = *max_element(frequency.begin(), frequency.end());         int totalOperations = n - maxFrequency; // Number of operations to remove all other characters           if (totalOperations >= maxFrequency) {             // All other characters can be removed in one operation             operations += maxFrequency;             break;         }           // Remove characters with maximum frequency from the string         operations += totalOperations;         n -= totalOperations;           // Update the frequency array for the remaining characters         frequency.erase(remove(frequency.begin(), frequency.end(), maxFrequency), frequency.end());     }       return operations; }   // Test the function int main() {     string str = "xyabcdadb";     cout << makeEqual(str) << endl; // Output: 2       return 0; }

## Java

 import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.List;   public class Main {       static int makeEqual(String str) {         int n = str.length();         List frequency = new ArrayList<>(Collections.nCopies(26, 0)); // Frequency list for lowercase letters           // Count the occurrences of each character         for (char ch : str.toCharArray()) {             frequency.set(ch - 'a', frequency.get(ch - 'a') + 1);         }           int operations = 0;           while (true) {             int maxFrequency = Collections.max(frequency);             int totalOperations = n - maxFrequency; // Number of operations to remove all other characters               if (totalOperations >= maxFrequency) {                 // All other characters can be removed in one operation                 operations += maxFrequency;                 break;             }               // Remove characters with maximum frequency from the string             operations += totalOperations;             n -= totalOperations;               // Update the frequency list for the remaining characters             frequency.removeAll(Collections.singletonList(maxFrequency));         }           return operations;     }       // Test the function     public static void main(String[] args) {         String str = "xyabcdadb";         System.out.println(makeEqual(str)); // Output: 2     } } //This code is Contrbuted by chinmaya121221

## Python

 def make_equal(string):     n = len(string)     frequency = [0] * 26  # Frequency array for lowercase letters       # Count the occurrences of each character     for char in string:         frequency[ord(char) - ord('a')] += 1       operations = 0       while True:         max_frequency = max(frequency)         total_operations = n - max_frequency  # Number of operations to remove all other characters           if total_operations >= max_frequency:             # All other characters can be removed in one operation             operations += max_frequency             break           # Remove characters with maximum frequency from the string         operations += total_operations         n -= total_operations           # Update the frequency array for the remaining characters         frequency = [freq for freq in frequency if freq != max_frequency]       return operations   # Test the function string = "xyabcdadb" print(make_equal(string))  # Output: 2

## C#

 using System; using System.Collections.Generic; using System.Linq;   class Program {     static int MakeEqual(string str)     {         int n = str.Length;         List frequency = new List(Enumerable.Repeat(0, 26)); // Frequency array for lowercase letters           // Count the occurrences of each character         foreach (char ch in str)         {             frequency[ch - 'a']++;         }           int operations = 0;           while (true)         {             int maxFrequency = frequency.Max();             int totalOperations = n - maxFrequency; // Number of operations to remove all other characters               if (totalOperations >= maxFrequency)             {                 // All other characters can be removed in one operation                 operations += maxFrequency;                 break;             }               // Remove characters with maximum frequency from the string             operations += totalOperations;             n -= totalOperations;               // Update the frequency array for the remaining characters             frequency.RemoveAll(f => f == maxFrequency);         }           return operations;     }       // Test the function     static void Main()     {         string str = "xyabcdadb";         Console.WriteLine(MakeEqual(str)); // Output: 2     } }

## Javascript

 const make_equal = (string) => {     const n = string.length;     // Frequency array for lowercase letters     const frequency = new Array(26).fill(0);          // Count the occurrences of each character     for (let char of string) {         frequency[char.charCodeAt() - 'a'.charCodeAt()] += 1;     }       let operations = 0;     while (true) {         const max_frequency = Math.max(...frequency);                   // Number of operations to remove all other characters         const total_operations = n - max_frequency;           if (total_operations >= max_frequency) {             // All other characters can be removed in one operation             operations += max_frequency;             break;         }           // Remove characters with maximum frequency from the string         operations += total_operations;         n -= total_operations;               // Update the frequency array for the remaining characters         frequency.splice(frequency.indexOf(max_frequency), 1);     }       return operations; }   // Test the function const string = "xyabcdadb"; console.log(make_equal(string));

Output

2

Time complexity: O(N)
Auxiliary space: O(1)