Find lexicographically smallest string in at most one swaps
Last Updated :
15 Jan, 2024
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++
// 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
// and set -1 as default for every character.
vector<int> loccur(26,-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;
}
}
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 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 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# 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
<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>
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Optimized Approach in O(N)
As we know lexicographical order evaluates from left to right and decides whether its order is smaller or bigger to the compared string.
So , we find the first element from the left that have smaller element on its right indexes of it and note the current index and last occurrence of smaller element. After knowing their indexes we simply swap them. to track whether swapping need to be done or not at end , we use a flag variable called found because in some cases like 'aaaa' swapping not requires.
now return the array that contains String characters after swapping if found = 1 or without swapping if found = 0 as a 'String'.
implementation of above approach in java
in java Strings are immutable . so instead of swapping , we note the indexes and also elements. at last we replace both elements with opposite index.
C++
#include <cstring>
#include <iostream>
#include <string>
using namespace std;
string findSmallest(string s)
{
int len = s.length();
// Used to track last occurrences
int loccur[26];
memset(loccur, -1, sizeof(loccur));
for (int i = len - 1; i >= 0; i--) {
int ch = s[i] - 'a';
if (loccur[ch] == -1) {
loccur[ch] = i;
}
}
int found = 0, index1 = -1,
index2 = -1; // Tracks the indexes
string c1 = "", c2 = ""; // Tracks the elements
for (int i = 0; i < len; i++) {
if (i > 0 && s[i] == s[i - 1]) {
continue; // Eliminates continuously repeated
// elements
}
int ch = s[i] - 'a';
if (loccur[ch] >= 0 && found == 0) {
int j = 0;
for (j = 0; j < ch; j++) {
// If loop will be executed when smaller
// element on the right of the current
// position is found
if (loccur[j] >= 0
&& loccur[j] > loccur[ch]) {
found = 1;
index1 = i; // Note the current element
// index
c1 = s[i]; // and current element
index2 = loccur[j];
c2 = s[index2]; // Last occurrence of
// the smaller element
break;
}
}
}
}
string res = "";
if (found == 1) { // Perform swapping if found=1
for (int i = 0; i < len; i++) {
if (i == index1) {
res += c2;
}
else if (i == index2) {
res += c1;
}
else {
res += s[i];
}
}
}
else {
res = s;
}
return res; // Result
}
int main()
{
cout << findSmallest("geeks") << endl;
return 0;
}
import java.io.*;
import java.util.*;
class GFG {
static String findSmallest(String s)
{
int len = s.length();
// used to track last occurrences
int[] loccur = new int[26];
Arrays.fill(loccur, -1);
for (int i = len - 1; i >= 0; i--) {
int ch = s.charAt(i) - 'a';
if (loccur[ch] == -1) {
loccur[ch] = i;
}
}
int found = 0, index1 = -1,
index2 = -1; // tracks the indexes
String c1 = "", c2 = ""; // tracks the elements
for (int i = 0; i < len; i++) {
if(i>0 && s.charAt(i)==s.charAt(i-1)){
continue; // eliminates continuously repeated elements
}
int ch = (int)(s.charAt(i) - 'a');
if (loccur[ch] >= 0 && found == 0) {
int j = 0;
for (j = 0; j < ch; j++) {
// if loop will be executed when smaller
// element on right of current position
// is found
if (loccur[j] >= 0 && loccur[j] > loccur[ch]) {
found = 1;
index1 = i; // note the current element index
c1 = "" + s.charAt(i); // and current element
index2 = loccur[j];
c2 = "" + s.charAt(index2); // last occurrence of smaller element
break;
}
}
}
}
String res = "";
if (found == 1) { // perform swapping if found=1
for (int i = 0; i < len; i++) {
if (i == index1) {
res += c2;
}
else if (i == index2) {
res += c1;
}
else {
res += "" + s.charAt(i);
}
}
}
else {
for (int i = 0; i < len; i++) {
res += "" + s.charAt(i);
}
}
return res; // result
}
public static void main(String[] args)
{
System.out.println(findSmallest("geeks"));
}
}
def find_smallest(s):
len_s = len(s)
# Used to track last occurrences
loccur = [-1] * 26
for i in range(len_s - 1, -1, -1):
ch = ord(s[i]) - ord('a')
if loccur[ch] == -1:
loccur[ch] = i
found, index1, index2 = 0, -1, -1 # Tracks the indexes
c1, c2 = "", "" # Tracks the elements
for i in range(len_s):
if i > 0 and s[i] == s[i - 1]:
continue # Eliminates continuously repeated elements
ch = ord(s[i]) - ord('a')
if loccur[ch] >= 0 and found == 0:
for j in range(ch):
# If loop will be executed when a smaller
# element on the right of the current
# position is found
if loccur[j] >= 0 and loccur[j] > loccur[ch]:
found = 1
index1 = i # Note the current element index
c1 = s[i] # and current element
index2 = loccur[j]
c2 = s[index2] # Last occurrence of
# the smaller element
break
res = ""
if found == 1: # Perform swapping if found=1
for i in range(len_s):
if i == index1:
res += c2
elif i == index2:
res += c1
else:
res += s[i]
else:
res = s
return res # Result
# Driver Code
print(find_smallest("geeks"))
# This code is contributed by Dwaipayan Bandyopadhyay
using System;
class Program
{
static string FindSmallest(string s)
{
int len = s.Length;
// Used to track last occurrences
int[] loccur = new int[26];
Array.Fill(loccur, -1);
for (int i = len - 1; i >= 0; i--)
{
int ch = s[i] - 'a';
if (loccur[ch] == -1)
{
loccur[ch] = i;
}
}
int found = 0, index1 = -1, index2 = -1; // Tracks the indexes
char c1 = '\0', c2 = '\0'; // Tracks the elements
for (int i = 0; i < len; i++)
{
if (i > 0 && s[i] == s[i - 1])
{
continue; // Eliminates continuously repeated elements
}
int ch = s[i] - 'a';
if (loccur[ch] >= 0 && found == 0)
{
int j = 0;
for (j = 0; j < ch; j++)
{
// If loop will be executed when a smaller
// element on the right of the current
// position is found
if (loccur[j] >= 0 && loccur[j] > loccur[ch])
{
found = 1;
index1 = i; // Note the current element index
c1 = s[i]; // and current element
index2 = loccur[j];
c2 = s[index2]; // Last occurrence of the smaller element
break;
}
}
}
}
string res = "";
if (found == 1)
{
// Perform swapping if found=1
for (int i = 0; i < len; i++)
{
if (i == index1)
{
res += c2.ToString();
}
else if (i == index2)
{
res += c1.ToString();
}
else
{
res += s[i];
}
}
}
else
{
res = s;
}
return res; // Result
}
static void Main()
{
Console.WriteLine(FindSmallest("geeks"));
}
}
function findSmallest(s) {
let len = s.length;
// used to track last occurrences
let loccur = new Array(26).fill(-1);
for (let i = len - 1; i >= 0; i--) {
let ch = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (loccur[ch] === -1) {
loccur[ch] = i;
}
}
let found = 0,
index1 = -1,
index2 = -1; // tracks the indexes
let c1 = "",
c2 = ""; // tracks the elements
for (let i = 0; i < len; i++) {
if (i > 0 && s.charAt(i) === s.charAt(i - 1)) {
continue; // eliminates continuously repeated elements
}
let ch = s.charCodeAt(i) - 'a'.charCodeAt(0);
if (loccur[ch] >= 0 && found === 0) {
let j = 0;
for (j = 0; j < ch; j++) {
// if loop will be executed when smaller
// element on right of current position
// is found
if (loccur[j] >= 0 && loccur[j] > loccur[ch]) {
found = 1;
index1 = i; // note the current element index
c1 = s.charAt(i); // and current element
index2 = loccur[j];
c2 = s.charAt(index2); // last occurrence of smaller element
break;
}
}
}
}
let res = "";
if (found === 1) {
// perform swapping if found = 1
for (let i = 0; i < len; i++) {
if (i === index1) {
res += c2;
} else if (i === index2) {
res += c1;
} else {
res += s.charAt(i);
}
}
} else {
for (let i = 0; i < len; i++) {
res += s.charAt(i);
}
}
return res; // result
}
console.log(findSmallest("geeks"));
Output : eegks
Time Complexity :O(N).
Auxiliary Space : O(26) ~ O(1).
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