Swap all occurrences of two characters to get lexicographically smallest string
Given string str of lower case English alphabets. One can choose any two characters in the string and replace all the occurrences of the first character with the second character and replace all the occurrences of the second character with the first character. Find the lexicographically smallest string that can be obtained by doing this operation at most once. Examples:
Input: str = “ccad”
Output: aacd
Swap all the occurrences of ‘c’ with ‘a’ and all the occurrences of ‘a’ with ‘c’ to get “aacd” which is the lexicographically smallest string that we can get.Input: str = “abba”
Output: abba
The only possible operation will convert the given string to “baab” which is not lexicographically smallest.
Approach:
- First, we store the first appearance of every character in a string in a hash array chk[].
- In order to find the lexicographically smaller string, the leftmost character must be replaced with some character that is smaller than it. This will only happen if the smaller character appears after it in the array.
- So, start traversing the string from the left and for every character, find the smallest character (even smaller than the current character) that appears after swapping all of their occurrences to get the required string.
- If no such character pair is found in the previous string then print the given string as it is the smallest string possible.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; #define MAX 26 // Function to return the lexicographically // smallest string after swapping all the // occurrences of any two characters string smallestStr(string str, int n) { int i, j; // To store the first index of // every character of str int chk[MAX]; for (i = 0; i < MAX; i++) chk[i] = -1; // Store the first occurring // index every character for (i = 0; i < n; i++) { // If current character is appearing // for the first time in str if (chk[str[i] - 'a' ] == -1) chk[str[i] - 'a' ] = i; } // Starting from the leftmost character for (i = 0; i < n; i++) { bool flag = false ; // For every character smaller than str[i] for (j = 0; j < str[i] - 'a' ; j++) { // If there is a character in str which is // smaller than str[i] and appears after it if (chk[j] > chk[str[i] - 'a' ]) { flag = true ; break ; } } // If the required character pair is found if (flag) break ; } // If swapping is possible if (i < n-1) { // Characters to be swapped char ch1 = str[i]; char ch2 = char (j + 'a' ); // For every character for (i = 0; i < n; i++) { // Replace every ch1 with ch2 // and every ch2 with ch1 if (str[i] == ch1) str[i] = ch2; else if (str[i] == ch2) str[i] = ch1; } } return str; } // Driver code int main() { string str = "ccad" ; int n = str.length(); cout << smallestStr(str, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { static int MAX = 26 ; // Function to return the lexicographically // smallest string after swapping all the // occurrences of any two characters static String smallestStr( char []str, int n) { int i, j = 0 ; // To store the first index of // every character of str int []chk = new int [MAX]; for (i = 0 ; i < MAX; i++) chk[i] = - 1 ; // Store the first occurring // index every character for (i = 0 ; i < n; i++) { // If current character is appearing // for the first time in str if (chk[str[i] - 'a' ] == - 1 ) chk[str[i] - 'a' ] = i; } // Starting from the leftmost character for (i = 0 ; i < n; i++) { boolean flag = false ; // For every character smaller than str[i] for (j = 0 ; j < str[i] - 'a' ; j++) { // If there is a character in str which is // smaller than str[i] and appears after it if (chk[j] > chk[str[i] - 'a' ]) { flag = true ; break ; } } // If the required character pair is found if (flag) break ; } // If swapping is possible if (i < n- 1 ) { // Characters to be swapped char ch1 = str[i]; char ch2 = ( char ) (j + 'a' ); // For every character for (i = 0 ; i < n; i++) { // Replace every ch1 with ch2 // and every ch2 with ch1 if (str[i] == ch1) str[i] = ch2; else if (str[i] == ch2) str[i] = ch1; } } return String.valueOf(str); } // Driver code public static void main(String[] args) { String str = "ccad" ; int n = str.length(); System.out.println(smallestStr( str.toCharArray(), n)); } } |
Python
# python3 implementation of the approach MAX = 256 # Function to return the lexicographically # smallest after swapping all the # occurrences of any two characters def smallestStr( str , n): i, j = 0 , 0 # To store the first index of # every character of str chk = [ 0 for i in range ( MAX )] for i in range ( MAX ): chk[i] = - 1 # Store the first occurring # index every character for i in range (n): # If current character is appearing # for the first time in str if (chk[ ord ( str [i])] = = - 1 ): chk[ ord ( str [i])] = i # Starting from the leftmost character for i in range (n): flag = False # For every character smaller than ord(str[i]) for j in range ( ord ( str [i])): # If there is a character in str which is # smaller than ord(str[i]) and appears after it if (chk[j] > chk[ ord ( str [i])]): flag = True break # If the required character pair is found if (flag): break # If swapping is possible if (i < n - 1 ): # Characters to be swapped ch1 = ( str [i]) ch2 = chr (j) # For every character for i in range (n): # Replace every ch1 with ch2 # and every ch2 with ch1 if ( str [i] = = ch1): str [i] = ch2 elif ( str [i] = = ch2): str [i] = ch1 return "".join( str ) # Driver code st = "ccad" str = [i for i in st] n = len ( str ) print (smallestStr( str , n)) |
C#
// C# implementation of the approach using System; class GFG { static int MAX = 26; // Function to return the lexicographically // smallest string after swapping all the // occurrences of any two characters static String smallestStr( char []str, int n) { int i, j = 0; // To store the first index of // every character of str int []chk = new int [MAX]; for (i = 0; i < MAX; i++) chk[i] = -1; // Store the first occurring // index every character for (i = 0; i < n; i++) { // If current character is appearing // for the first time in str if (chk[str[i] - 'a' ] == -1) chk[str[i] - 'a' ] = i; } // Starting from the leftmost character for (i = 0; i < n; i++) { Boolean flag = false ; // For every character smaller than str[i] for (j = 0; j < str[i] - 'a' ; j++) { // If there is a character in str which is // smaller than str[i] and appears after it if (chk[j] > chk[str[i] - 'a' ]) { flag = true ; break ; } } // If the required character pair is found if (flag) break ; } // If swapping is possible if (i < n-1) { // Characters to be swapped char ch1 = str[i]; char ch2 = ( char ) (j + 'a' ); // For every character for (i = 0; i < n; i++) { // Replace every ch1 with ch2 // and every ch2 with ch1 if (str[i] == ch1) str[i] = ch2; else if (str[i] == ch2) str[i] = ch1; } } return String.Join( "" , str); } // Driver code public static void Main(String[] args) { String str = "ccad" ; int n = str.Length; Console.WriteLine(smallestStr( str.ToCharArray(), n)); } } |
Javascript
<script> // JavaScript Implementation of the above approach var MAX = 26; // utility function to replace a char at particular string position String.prototype.replaceAt = function (index, replacement) { return this .substring(0, index) + replacement + this .substring(index + replacement.length); } function smallestStr(str, n) { let i, j; // To store the first index of // every character of str const chk=[]; for (i = 0; i < MAX; i++) chk[i] = -1; // Store the first occurring // index every character for (i = 0; i < n; i++) { // If current character is appearing // for the first time in str if (chk[str[i].charCodeAt(0) - 'a' .charCodeAt(0)] == -1) chk[str[i].charCodeAt(0) - 'a' .charCodeAt(0)] = i; } // Starting from the leftmost character for (i = 0; i < n; i++) { let flag = false ; // For every character smaller than str[i] for (j = 0; j < str[i].charCodeAt(0) - 'a' .charCodeAt(0); j++) { // If there is a character in str which is // smaller than str[i] and appears after it if (chk[j] > chk[str[i].charCodeAt(0) - 'a' .charCodeAt(0)]) { flag = true ; break ; } } // If the required character pair is found if (flag) break ; } // If swapping is possible if (i < n-1) { // Characters to be swapped let ch1 = str[i]; let ch2 = String.fromCharCode(j + 'a' .charCodeAt(0)); // For every character for (i = 0; i < n; i++) { // Replace every ch1 with ch2 // and every ch2 with ch1 if (str[i] == ch1) str=str.replaceAt(i,ch2); else if (str[i] == ch2) str=str.replaceAt(i,ch1); } } return str; } // Driver Code let str = "ccad" ; let n = str.length; document.write(smallestStr(str, n)); // This code is contributed by Ishan Khandelwal </script> |
aacd
Time Complexity: O(n * 26) ⇒ O(n), where n is the length of the given string.
Auxiliary Space: O(26) ⇒ O(1), no extra space is required, so it is a constant.
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