Generate lexicographically smallest string of 0, 1 and 2 with adjacent swaps allowed

Given a string str containing only characters 0, 1 and 2, you can swap any two adjacent (consecutive) characters 0 and 1 or any two adjacent (consecutive) characters 1 and 2. The task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times.

Examples:

Input: str = “100210”
Output: 001120
We can swap 0 and 1 OR we can swap 1 and 2. Swapping 0 and 2 is not allowed. All the swaps can happen for adjacent only.

Input: str = “2021”
Output: 1202
Note that 0 and 2 cannot be swapped



Approach: You can print all the 1s together as 1 can be swapped with either of the other characters while 0 and 2 can not be swapped, so all the 0s and 2s will follow the same order as the original string.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the required string
void printString(string str, int n)
{
    // count number of 1s
    int ones = 0;
    for (int i = 0; i < n; i++)
        if (str[i] == '1')
            ones++;
  
    // To check if the all the 1s
    // have been used or not
    bool used = false;
  
    for (int i = 0; i < n; i++) {
        if (str[i] == '2' && !used) {
            used = 1;
  
            // Print all the 1s if any 2 is encountered
            for (int j = 0; j < ones; j++)
                cout << "1";
        }
  
        // If str[i] = 0 or str[i] = 2
        if (str[i] != '1')
            cout << str[i];
    }
  
    // If 1s are not printed yet
    if (!used)
        for (int j = 0; j < ones; j++)
            cout << "1";
}
  
// Driver code
int main()
{
    string str = "100210";
    int n = str.length();
    printString(str, n);
    return 0;
}

chevron_right


Python3

# Python3 implementation of the approach

# Function to prthe required string
def printString(Str1, n):

# count number of 1s
ones = 0
for i in range(n):
if (Str1[i] == ‘1’):
ones += 1

# To check if the all the 1s
# have been used or not
used = False

for i in range(n):
if (Str1[i] == ‘2’ and used == False):
used = 1

# Prall the 1s if any 2 is encountered
for j in range(ones):
print(“1”, end = “”)

# If Str1[i] = 0 or Str1[i] = 2
if (Str1[i] != ‘1’):
print(Str1[i], end = “”)

# If 1s are not printed yet
if (used == False):
for j in range(ones):
print(“1”, end = “”)

# Driver code
Str1 = “100210”
n = len(Str1)
printString(Str1, n)

# This code is contributed
# by Mohit Kumar

PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Function to print the required string 
function printString($str, $n
    // count number of 1s 
    $ones = 0; 
    for ($i = 0; $i < $n; $i++) 
        if ($str[$i] == '1'
            $ones++; 
  
    // To check if the all the 1s 
    // have been used or not 
    $used = false; 
  
    for ($i = 0; $i < $n; $i++)
    
        if ($str[$i] == '2' && !$used)
        
            $used = 1; 
  
            // Print all the 1s if any 2
            // is encountered 
            for ($j = 0; $j < $ones; $j++) 
                echo "1"
        
  
        // If str[i] = 0 or str[i] = 2 
        if ($str[$i] != '1'
            echo $str[$i]; 
    
  
    // If 1s are not printed yet 
    if (!$used
        for ($j = 0; $j < $ones; $j++) 
            echo "1"
  
// Driver code 
$str = "100210"
$n = strlen($str);
printString($str, $n); 
  
// This code is contributed by Ryuga
?>

chevron_right


Output:

001120


My Personal Notes arrow_drop_up

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.



Improved By : Ryuga, mohit kumar 29