Find lexicographically smallest string in at most one swaps

Given a string str of length N. The task is to find out the lexicographically smallest string when at most only one swap is allowed. That is, two indices 1 <= i, j <= n can be chosen and swapped. This operation can be performed at most one time.

Examples:

Input: str = “string”
Output: gtrins
Explanation:
Choose i=1, j=6, string becomes – gtrins. This is lexicographically smallest strings that can be formed.

Input: str = “zyxw”
Output: wyxz

Approach: The idea is to use sorting and compute the smallest lexicographical string possible for the given string. After computing the sorted string, find the first unmatched character from the given string and replace it with the last occurrence of the unmatched character in the sorted string.



For example, let str = “geeks” and the sorted = “eegks”. First unmatched character is in the first place. This character has to swapped such that this character matches the character with sorted string. Resulting lexicographical smallest string. On replacing “g” with the last occurring “e”, the string becomes eegks which is lexicographically smallest.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the lexicographically
// smallest string that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
string findSmallest(string s)
{
    int len = s.size();
  
    // Store last occurrence of every character
    int loccur[26];
  
    // Set -1 as default for every character.
    memset(loccur, -1, sizeof(loccur));
  
    for (int i = len - 1; i >= 0; --i) {
  
        // Character index to fill
        // in the last occurrence array
        int chI = s[i] - 'a';
        if (loccur[chI] == -1) {
  
            // If this is true then this
            // character is being visited
            // for the first time from the last
            // Thus last occurrence of this
            // character is stored in this index
            loccur[chI] = i;
        }
    }
  
    string sorted_s = s;
    sort(sorted_s.begin(), sorted_s.end());
  
    for (int i = 0; i < len; ++i) {
        if (s[i] != sorted_s[i]) {
  
            // Character to replace
            int chI = sorted_s[i] - 'a';
  
            // Find the last occurrence
            // of this character.
            int last_occ = loccur[chI];
  
            // Swap this with the last occurrence
            swap(s[i], s[last_occ]);
            break;
        }
    }
  
    return s;
}
  
// Driver code
int main()
{
    string s = "geeks";
    cout << findSmallest(s);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
import java.util.*;
  
class GFG{
   
// Function to return the lexicographically
// smallest String that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
static String findSmallest(char []s)
{
    int len = s.length;
   
    // Store last occurrence of every character
    int []loccur = new int[26];
   
    // Set -1 as default for every character.
    Arrays.fill(loccur, -1);
   
    for (int i = len - 1; i >= 0; --i) {
   
        // Character index to fill
        // in the last occurrence array
        int chI = s[i] - 'a';
        if (loccur[chI] == -1) {
   
            // If this is true then this
            // character is being visited
            // for the first time from the last
            // Thus last occurrence of this
            // character is stored in this index
            loccur[chI] = i;
        }
    }
   
    char []sorted_s = s;
    Arrays.sort(sorted_s);
   
    for (int i = 0; i < len; ++i) {
        if (s[i] != sorted_s[i]) {
   
            // Character to replace
            int chI = sorted_s[i] - 'a';
   
            // Find the last occurrence
            // of this character.
            int last_occ = loccur[chI];
   
            // Swap this with the last occurrence
            char temp = s[last_occ];
            s[last_occ] = s[i];
            s[i] = temp;
            break;
        }
    }
   
    return String.valueOf(s);
}
   
// Driver code
public static void main(String[] args)
{
    String s = "geeks";
    System.out.print(findSmallest(s.toCharArray()));
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the above approach 
  
# Function to return the lexicographically 
# smallest string that can be formed by 
# swapping at most one character. 
# The characters might not necessarily 
# be adjacent. 
def findSmallest(s) : 
  
    length = len(s); 
  
    # Store last occurrence of every character 
    # Set -1 as default for every character.
    loccur = [-1]*26
  
  
    for i in range(length - 1, -1, -1) : 
  
        # Character index to fill 
        # in the last occurrence array 
        chI = ord(s[i]) - ord('a'); 
        if (loccur[chI] == -1) :
  
            # If this is true then this 
            # character is being visited 
            # for the first time from the last 
            # Thus last occurrence of this 
            # character is stored in this index 
            loccur[chI] = i; 
  
    sorted_s = s; 
    sorted_s.sort();
  
    for i in range(length) :
        if (s[i] != sorted_s[i]) :
  
            # Character to replace 
            chI = ord(sorted_s[i]) - ord('a'); 
  
            # Find the last occurrence 
            # of this character. 
            last_occ = loccur[chI]; 
  
            # Swap this with the last occurrence 
            # swap(s[i], s[last_occ]); 
            s[i],s[last_occ] = s[last_occ],s[i]
            break
  
    return "".join(s); 
  
# Driver code 
if __name__ == "__main__"
  
    s = "geeks"
      
    print(findSmallest(list(s))); 
  
# This code is contributed by Yash_R

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach
using System;
  
class GFG{
    
// Function to return the lexicographically
// smallest String that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
static String findSmallest(char []s)
{
    int len = s.Length;
    
    // Store last occurrence of every character
    int []loccur = new int[26];
    
    // Set -1 as default for every character.
    for (int i = 0; i < 26; i++)
        loccur[i] = -1;
    
    for (int i = len - 1; i >= 0; --i) {
    
        // char index to fill
        // in the last occurrence array
        int chI = s[i] - 'a';
        if (loccur[chI] == -1) {
    
            // If this is true then this
            // character is being visited
            // for the first time from the last
            // Thus last occurrence of this
            // character is stored in this index
            loccur[chI] = i;
        }
    }
    
    char []sorted_s = s;
    Array.Sort(sorted_s);
    
    for (int i = 0; i < len; ++i) {
        if (s[i] != sorted_s[i]) {
    
            // char to replace
            int chI = sorted_s[i] - 'a';
    
            // Find the last occurrence
            // of this character.
            int last_occ = loccur[chI];
    
            // Swap this with the last occurrence
            char temp = s[last_occ];
            s[last_occ] = s[i];
            s[i] = temp;
            break;
        }
    }
    
    return String.Join("", s);
}
    
// Driver code
public static void Main(String[] args)
{
    String s = "geeks";
    Console.Write(findSmallest(s.ToCharArray()));
}
}
  
// This code is contributed by sapnasingh4991

chevron_right


Output:

eegks

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.