Find minimum operations to make all characters of a string identical
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.
Input: string = “xyabcdadb”
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++
#include <bits/stdc++.h>
using namespace std;
int makeequal(string s, int n)
{
int minsubarray, ans = 0;
for ( int j = 0; j < 26; j++) {
char key = 'a' + j;
int temp = 0, ma = 0;
for ( int i = 0; i < n; i++) {
if (s[i] == key) {
ma = max(ma, temp);
temp = 0;
}
else {
temp += 1;
}
if (i == n - 1) {
ma = max(ma, temp);
}
}
minsubarray = min(minsubarray, ma);
}
while (minsubarray != 0) {
ans += 1;
minsubarray = minsubarray / 2;
}
return ans;
}
int main()
{
string s = "xyabcdadb" ;
int n = s.size();
cout << makeequal(s, n);
return 0;
}
|
Java
import java.util.Arrays;
public class Main {
static int makeEqual(String s, int n) {
int minSubarray = Integer.MAX_VALUE;
int ans = 0 ;
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) {
ma = Math.max(ma, temp);
temp = 0 ;
} else {
temp += 1 ;
}
if (i == n - 1 ) {
ma = Math.max(ma, temp);
}
}
minSubarray = Math.min(minSubarray, ma);
}
while (minSubarray != 0 ) {
ans += 1 ;
minSubarray /= 2 ;
}
return ans;
}
public static void main(String[] args) {
String s = "xyabcdadb" ;
int n = s.length();
System.out.println(makeEqual(s, n));
}
}
|
Python3
def makeequal(s):
n = len (s)
min_subarray = float ( 'inf' )
ans = 0
for j in range ( 26 ):
key = chr ( ord ( 'a' ) + j)
temp = 0
ma = 0
for i in range (n):
if s[i] = = key:
ma = max (ma, temp)
temp = 0
else :
temp + = 1
if i = = n - 1 :
ma = max (ma, temp)
min_subarray = min (min_subarray, ma)
while min_subarray ! = 0 :
ans + = 1
min_subarray = min_subarray / / 2
return ans
if __name__ = = '__main__' :
s = "xyabcdadb"
print (makeequal(s))
|
C#
using System;
class Program
{
static int MakeEqual( string s, int n)
{
int minSubarray = 2147483647 , ans = 0;
for ( int j = 0; j < 26; j++)
{
char key = ( char )( 'a' + j);
int temp = 0, ma = 0;
for ( int i = 0; i < n; i++)
{
if (s[i] == key)
{
ma = Math.Max(ma, temp);
temp = 0;
}
else
{
temp += 1;
}
if (i == n - 1)
{
ma = Math.Max(ma, temp);
}
}
minSubarray = Math.Min(minSubarray, ma);
}
while (minSubarray != 0)
{
ans += 1;
minSubarray = minSubarray / 2;
}
return ans;
}
static void Main()
{
string s = "xyabcdadb" ;
int n = s.Length;
Console.WriteLine(MakeEqual(s, n));
}
}
|
Javascript
function makeequal(s) {
let n = s.length;
let min_subarray = Infinity;
let ans = 0;
for (let j = 0; j < 26; j++) {
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) {
ma = Math.max(ma, temp);
temp = 0;
} else {
temp += 1;
}
if (i === n - 1) {
ma = Math.max(ma, temp);
}
}
min_subarray = Math.min(min_subarray, ma);
}
while (min_subarray !== 0) {
ans += 1;
min_subarray = Math.floor(min_subarray / 2);
}
return ans;
}
let s = "xyabcdadb" ;
console.log(makeequal(s));
|
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 <iostream>
#include <vector>
#include <algorithm> // Include the necessary header for max_element and remove
using namespace std;
int makeEqual(string str) {
int n = str.length();
vector< int > frequency(26, 0);
for ( char ch : str) {
frequency[ch - 'a' ]++;
}
int operations = 0;
while ( true ) {
int maxFrequency = *max_element(frequency.begin(), frequency.end());
int totalOperations = n - maxFrequency;
if (totalOperations >= maxFrequency) {
operations += maxFrequency;
break ;
}
operations += totalOperations;
n -= totalOperations;
frequency.erase( remove (frequency.begin(), frequency.end(), maxFrequency), frequency.end());
}
return operations;
}
int main() {
string str = "xyabcdadb" ;
cout << makeEqual(str) << endl;
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<Integer> frequency = new ArrayList<>(Collections.nCopies( 26 , 0 ));
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;
if (totalOperations >= maxFrequency) {
operations += maxFrequency;
break ;
}
operations += totalOperations;
n -= totalOperations;
frequency.removeAll(Collections.singletonList(maxFrequency));
}
return operations;
}
public static void main(String[] args) {
String str = "xyabcdadb" ;
System.out.println(makeEqual(str));
}
}
|
Python
def make_equal(string):
n = len (string)
frequency = [ 0 ] * 26
for char in string:
frequency[ ord (char) - ord ( 'a' )] + = 1
operations = 0
while True :
max_frequency = max (frequency)
total_operations = n - max_frequency
if total_operations > = max_frequency:
operations + = max_frequency
break
operations + = total_operations
n - = total_operations
frequency = [freq for freq in frequency if freq ! = max_frequency]
return operations
string = "xyabcdadb"
print (make_equal(string))
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static int MakeEqual( string str)
{
int n = str.Length;
List< int > frequency = new List< int >(Enumerable.Repeat(0, 26));
foreach ( char ch in str)
{
frequency[ch - 'a' ]++;
}
int operations = 0;
while ( true )
{
int maxFrequency = frequency.Max();
int totalOperations = n - maxFrequency;
if (totalOperations >= maxFrequency)
{
operations += maxFrequency;
break ;
}
operations += totalOperations;
n -= totalOperations;
frequency.RemoveAll(f => f == maxFrequency);
}
return operations;
}
static void Main()
{
string str = "xyabcdadb" ;
Console.WriteLine(MakeEqual(str));
}
}
|
Javascript
const make_equal = (string) => {
const n = string.length;
const frequency = new Array(26).fill(0);
for (let char of string) {
frequency[char.charCodeAt() - 'a' .charCodeAt()] += 1;
}
let operations = 0;
while ( true ) {
const max_frequency = Math.max(...frequency);
const total_operations = n - max_frequency;
if (total_operations >= max_frequency) {
operations += max_frequency;
break ;
}
operations += total_operations;
n -= total_operations;
frequency.splice(frequency.indexOf(max_frequency), 1);
}
return operations;
}
const string = "xyabcdadb" ;
console.log(make_equal(string));
|
Time complexity: O(N)
Auxiliary space: O(1)
Last Updated :
14 Oct, 2023
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