Form lexicographically smallest string with minimum replacements having equal number of 0s, 1s and 2s
Given string str of length n (n is a multiple of 3) containing characters only from the set {0, 1, 2}. The task is to update the string such that each character has the same frequency with a minimum number of operations. In a single operation, any character of the string can be replaced with any other character (also from the same set). If there are multiple strings possible then print the lexicographically smallest one.
Examples:
Input: str = β000000β
Output: 001122
Replace 3rd and 4th β0β with β1β and 5th and 6th β0β with β2β such that given condition is satisfied and it forms lexicographically smallest stringInput: str = β211200β
Output: 211200
The string already has equal number of 0s, 1s and 2s, hence there is no need to perform any operation.
Approach: This problem can be solved using a greedy approach. We need only n / 3 number of characters of each type where n is the size of the string. Iterate over the string and count the number of characters of each type. Again, iterate over the string and now check if the current characterβs count is equal to n / 3 then there is no need to perform any operation on the current character.
However, if count(currentChar) != n / 3 then we may need to perform a replacement operation depending on the value of the character in such a way that it maintains the smallest lexicographical order as follows:
- If the current character is zero(0) and we have already processed required number of zeroes, then this character needs to be replaced with either one(1) if count[1] < (n / 3) or with two(1) if count[2] < (n / 3).
- If the current character is one(1), then we can either replace it with zero(0) if count[0] < (n / 3) or with two(2) if count[2] < (n / 3) and we have already processed required number of ones to maintain lowest lexicographical order
- If the current character is two(2), then we can either replace it with zero(0) if count[0] < (n / 3) or with two(2) if count[2] < (n / 3).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns the modified lexicographically // smallest string after performing minimum number // of given operations string formStringMinOperations(string s) { // Stores the initial frequencies of characters // 0s, 1s and 2s int count[3] = { 0 }; for ( auto & c : s) count++; // Stores number of processed characters upto that // point of each type int processed[3] = { 0 }; // Required number of characters of each type int reqd = ( int )s.size() / 3; for ( int i = 0; i < s.size(); i++) { // If the current type has already reqd // number of characters, no need to perform // any operation if (count[s[i] - '0' ] == reqd) continue ; // Process all 3 cases if (s[i] == '0' && count[0] > reqd && processed[0] >= reqd) { // Check for 1 first if (count[1] < reqd) { s[i] = '1' ; count[1]++; count[0]--; } // Else 2 else if (count[2] < reqd) { s[i] = '2' ; count[2]++; count[0]--; } } // Here we need to check processed[1] only // for 2 since 0 is less than 1 and we can // replace it anytime if (s[i] == '1' && count[1] > reqd) { if (count[0] < reqd) { s[i] = '0' ; count[0]++; count[1]--; } else if (count[2] < reqd && processed[1] >= reqd) { s[i] = '2' ; count[2]++; count[1]--; } } // Here we can replace 2 with 0 and 1 anytime if (s[i] == '2' && count[2] > reqd) { if (count[0] < reqd) { s[i] = '0' ; count[0]++; count[2]--; } else if (count[1] < reqd) { s[i] = '1' ; count[1]++; count[2]--; } } // keep count of processed characters of each // type processed[s[i] - '0' ]++; } return s; } // Driver Code int main() { string s = "011200" ; cout << formStringMinOperations(s); return 0; } |
Java
// Java implementation of the approach class GFG { // Function that returns the // modified lexicographically // smallest String after // performing minimum number // of given operations static String formStringMinOperations( char [] s) { // Stores the initial frequencies // of characters 0s, 1s and 2s int count[] = new int [ 3 ]; for ( char c : s) { count[( int )c - 48 ] += 1 ; } // Stores number of processed characters // upto that point of each type int processed[] = new int [ 3 ]; // Required number of characters of each type int reqd = ( int ) s.length / 3 ; for ( int i = 0 ; i < s.length; i++) { // If the current type has already // reqd number of characters, no // need to perform any operation if (count[s[i] - '0' ] == reqd) { continue ; } // Process all 3 cases if (s[i] == '0' && count[ 0 ] > reqd && processed[ 0 ] >= reqd) { // Check for 1 first if (count[ 1 ] < reqd) { s[i] = '1' ; count[ 1 ]++; count[ 0 ]--; } // Else 2 else if (count[ 2 ] < reqd) { s[i] = '2' ; count[ 2 ]++; count[ 0 ]--; } } // Here we need to check processed[1] only // for 2 since 0 is less than 1 and we can // replace it anytime if (s[i] == '1' && count[ 1 ] > reqd) { if (count[ 0 ] < reqd) { s[i] = '0' ; count[ 0 ]++; count[ 1 ]--; } else if (count[ 2 ] < reqd && processed[ 1 ] >= reqd) { s[i] = '2' ; count[ 2 ]++; count[ 1 ]--; } } // Here we can replace 2 with 0 and 1 anytime if (s[i] == '2' && count[ 2 ] > reqd) { if (count[ 0 ] < reqd) { s[i] = '0' ; count[ 0 ]++; count[ 2 ]--; } else if (count[ 1 ] < reqd) { s[i] = '1' ; count[ 1 ]++; count[ 2 ]--; } } // keep count of processed // characters of each type processed[s[i] - '0' ]++; } return String.valueOf(s); } // Driver Code public static void main(String[] args) { String s = "011200" ; System.out.println(formStringMinOperations(s.toCharArray())); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach import math # Function that returns the modified # lexicographically smallest string after # performing minimum number of given operations def formStringMinOperations(ss): # Stores the initial frequencies of # characters 0s, 1s and 2s count = [ 0 ] * 3 ; s = list (ss); for i in range ( len (s)): count[ ord (s[i]) - ord ( '0' )] + = 1 ; # Stores number of processed characters # upto that point of each type processed = [ 0 ] * 3 ; # Required number of characters of each type reqd = math.floor( len (s) / 3 ); for i in range ( len (s)): # If the current type has already reqd # number of characters, no need to # perform any operation if (count[ ord (s[i]) - ord ( '0' )] = = reqd): continue ; # Process all 3 cases if (s[i] = = '0' and count[ 0 ] > reqd and processed[ 0 ] > = reqd): # Check for 1 first if (count[ 1 ] < reqd): s[i] = '1' ; count[ 1 ] + = 1 ; count[ 0 ] - = 1 ; # Else 2 elif (count[ 2 ] < reqd): s[i] = '2' ; count[ 2 ] + = 1 ; count[ 0 ] - = 1 ; # Here we need to check processed[1] only # for 2 since 0 is less than 1 and we can # replace it anytime if (s[i] = = '1' and count[ 1 ] > reqd): if (count[ 0 ] < reqd): s[i] = '0' ; count[ 0 ] + = 1 ; count[ 1 ] - = 1 ; elif (count[ 2 ] < reqd and processed[ 1 ] > = reqd): s[i] = '2' ; count[ 2 ] + = 1 ; count[ 1 ] - = 1 ; # Here we can replace 2 with 0 and 1 anytime if (s[i] = = '2' and count[ 2 ] > reqd): if (count[ 0 ] < reqd): s[i] = '0' ; count[ 0 ] + = 1 ; count[ 2 ] - = 1 ; elif (count[ 1 ] < reqd): s[i] = '1' ; count[ 1 ] + = 1 ; count[ 2 ] - = 1 ; # keep count of processed characters # of each type processed[ ord (s[i]) - ord ( '0' )] + = 1 ; return ''.join(s); # Driver Code s = "011200" ; print (formStringMinOperations(s)); # This code is contributed by mits |
C#
// C# implementation of the approach using System; class GFG { // Function that returns the // modified lexicographically // smallest String after // performing minimum number // of given operations static String formStringMinOperations( char [] s) { // Stores the initial frequencies // of characters 0s, 1s and 2s int []count = new int [3]; foreach ( char c in s) { count[( int )c - 48] += 1; } // Stores number of processed characters // upto that point of each type int []processed = new int [3]; // Required number of characters of each type int reqd = ( int ) s.Length / 3; for ( int i = 0; i < s.Length; i++) { // If the current type has already // reqd number of characters, no // need to perform any operation if (count[s[i] - '0' ] == reqd) { continue ; } // Process all 3 cases if (s[i] == '0' && count[0] > reqd && processed[0] >= reqd) { // Check for 1 first if (count[1] < reqd) { s[i] = '1' ; count[1]++; count[0]--; } // Else 2 else if (count[2] < reqd) { s[i] = '2' ; count[2]++; count[0]--; } } // Here we need to check processed[1] only // for 2 since 0 is less than 1 and we can // replace it anytime if (s[i] == '1' && count[1] > reqd) { if (count[0] < reqd) { s[i] = '0' ; count[0]++; count[1]--; } else if (count[2] < reqd && processed[1] >= reqd) { s[i] = '2' ; count[2]++; count[1]--; } } // Here we can replace 2 with 0 and 1 anytime if (s[i] == '2' && count[2] > reqd) { if (count[0] < reqd) { s[i] = '0' ; count[0]++; count[2]--; } else if (count[1] < reqd) { s[i] = '1' ; count[1]++; count[2]--; } } // keep count of processed // characters of each type processed[s[i] - '0' ]++; } return String.Join( "" ,s); } // Driver Code public static void Main(String[] args) { String s = "011200" ; Console.WriteLine(formStringMinOperations(s.ToCharArray())); } } // This code is contributed by Rajput-Ji |
PHP
<?php // PHP implementation of the approach // Function that returns the modified lexicographically // smallest string after performing minimum number // of given operations function formStringMinOperations( $s ) { // Stores the initial frequencies of // characters 0s, 1s and 2s $count = array_fill (0, 3, 0); for ( $i = 0; $i < strlen ( $s ) ; $i ++) $count [ $s [ $i ] - '0' ]++; // Stores number of processed characters // upto that point of each type $processed = array_fill (0, 3, 0); // Required number of characters of each type $reqd = floor ( strlen ( $s ) / 3); for ( $i = 0; $i < strlen ( $s ); $i ++) { // If the current type has already reqd // number of characters, no need to // perform any operation if ( $count [ $s [ $i ] - '0' ] == $reqd ) continue ; // Process all 3 cases if ( $s [ $i ] == '0' && $count [0] > $reqd && $processed [0] >= $reqd ) { // Check for 1 first if ( $count [1] < $reqd ) { $s [ $i ] = '1' ; $count [1]++; $count [0]--; } // Else 2 else if ( $count [2] < $reqd ) { $s [ $i ] = '2' ; $count [2]++; $count [0]--; } } // Here we need to check processed[1] only // for 2 since 0 is less than 1 and we can // replace it anytime if ( $s [ $i ] == '1' && $count [1] > $reqd ) { if ( $count [0] < $reqd ) { $s [ $i ] = '0' ; $count [0]++; $count [1]--; } else if ( count [2] < $reqd && $processed [1] >= $reqd ) { $s [ $i ] = '2' ; $count [2]++; $count [1]--; } } // Here we can replace 2 with 0 and 1 anytime if ( $s [ $i ] == '2' && $count [2] > $reqd ) { if ( $count [0] < $reqd ) { $s [ $i ] = '0' ; $count [0]++; $count [2]--; } else if ( $count [1] < $reqd ) { $s [ $i ] = '1' ; $count [1]++; $count [2]--; } } // keep count of processed characters // of each type $processed [ $s [ $i ] - '0' ]++; } return $s ; } // Driver Code $s = "011200" ; echo formStringMinOperations( $s ); // This code is contributed by Ryuga ?> |
Javascript
<script> // JavaScript implementation of the approach // Function that returns the // modified lexicographically // smallest String after // performing minimum number // of given operations function formStringMinOperations(s) { // Stores the initial frequencies // of characters 0s, 1s and 2s var count = new Array(3).fill(0); for (const c of s) { count += 1; } // Stores number of processed characters // upto that point of each type var processed = new Array(3).fill(0); // Required number of characters of each type var reqd = parseInt(s.length / 3); for ( var i = 0; i < s.length; i++) { // If the current type has already // reqd number of characters, no // need to perform any operation if (count[s[i].charCodeAt(0) - "0" .charCodeAt(0)] === reqd) { continue ; } // Process all 3 cases if (s[i] === "0" && count[0] > reqd && processed[0] >= reqd) { // Check for 1 first if (count[1] < reqd) { s[i] = "1" ; count[1]++; count[0]--; } // Else 2 else if (count[2] < reqd) { s[i] = "2" ; count[2]++; count[0]--; } } // Here we need to check processed[1] only // for 2 since 0 is less than 1 and we can // replace it anytime if (s[i] === "1" && count[1] > reqd) { if (count[0] < reqd) { s[i] = "0" ; count[0]++; count[1]--; } else if (count[2] < reqd && processed[1] >= reqd) { s[i] = "2" ; count[2]++; count[1]--; } } // Here we can replace 2 with 0 and 1 anytime if (s[i] === "2" && count[2] > reqd) { if (count[0] < reqd) { s[i] = "0" ; count[0]++; count[2]--; } else if (count[1] < reqd) { s[i] = "1" ; count[1]++; count[2]--; } } // keep count of processed // characters of each type processed[s[i].charCodeAt(0) - "0" .charCodeAt(0)]++; } return s.join( "" ); } // Driver Code var s = "011200" ; document.write(formStringMinOperations(s.split( "" ))); </script> |
011202
Complexity Analysis:
- Time Complexity: O(n), where n is the size of the given string
- Auxiliary Space: O(1), as 3 is termed as constant.