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>

Output

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.