# 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:gtrinsExplanation:

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++

`// 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;` `}` |

## Java

`// 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` |

## Python3

`# 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` |

## C#

`// 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` |

## Javascript

`<script>` `// Javascript 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.` `function` `findSmallest(s)` `{` ` ` `let len = s.length;` ` ` `// Store last occurrence of every character` ` ` `let loccur = ` `new` `Array(26);` ` ` `// Set -1 as default for every character.` ` ` `loccur.fill(-1);` ` ` `for` `(let i = len - 1; i >= 0; --i)` ` ` `{` ` ` ` ` `// Character index to fill` ` ` `// in the last occurrence array` ` ` `let chI = s[i].charCodeAt() -` ` ` `'a'` `.charCodeAt();` ` ` `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;` ` ` `}` ` ` `}` ` ` `let sorted_s = s;` ` ` `sorted_s.sort();` ` ` `for` `(let i = 0; i < len; ++i)` ` ` `{` ` ` `if` `(s[i] != sorted_s[i])` ` ` `{` ` ` ` ` `// Character to replace` ` ` `let chI = sorted_s[i].charCodeAt() -` ` ` `'a'` `.charCodeAt();` ` ` `// Find the last occurrence` ` ` `// of this character.` ` ` `let last_occ = loccur[chI];` ` ` `// Swap this with the last occurrence` ` ` `let temp = s[i];` ` ` `s[i] = s[last_occ];` ` ` `s[last_occ] = temp;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `return` `s.join(` `""` `);` `}` `// Driver code` `let s = ` `"geeks"` `;` `document.write(findSmallest(s.split(` `''` `)));` `// This code is contributed by vaibhavrabadiya3` `</script>` |

**Output:**

eegks

**Time Complexity:** O(N)**Auxiliary Space: **O(1)

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