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

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  • Last Updated : 01 Jun, 2021

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 an 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++




// 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;
}

Java




// Java implementation of the approach
 
class GFG
{
 
// Function to print the required string
static void printString(char[] 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
    boolean used = false;
 
    for (int i = 0; i < n; i++)
    {
        if (str[i] == '2' && !used)
        {
            used = true;
 
            // Print all the 1s if any 2 is encountered
            for (int j = 0; j < ones; j++)
                System.out.print("1");
        }
 
        // If str[i] = 0 or str[i] = 2
        if (str[i] != '1')
            System.out.print(str[i]);
 
    }
 
    // If 1s are not printed yet
    if (!used)
        for (int j = 0; j < ones; j++)
            System.out.print("1");
}
 
// Driver code
public static void main(String[] args)
{
    String str = "100210";
    int n = str.length();
    printString(str.toCharArray(), n);
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 implementation of the approach
 
# Function to print the 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
 
            # Print all 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

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to print the required string
static void printString(char[] 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 = true;
 
            // Print all the 1s if any 2 is encountered
            for (int j = 0; j < ones; j++)
                Console.Write("1");
        }
 
        // If str[i] = 0 or str[i] = 2
        if (str[i] != '1')
            Console.Write(str[i]);
 
    }
 
    // If 1s are not printed yet
    if (!used)
        for (int j = 0; j < ones; j++)
            Console.Write("1");
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "100210";
    int n = str.Length;
    printString(str.ToCharArray(), n);
}
}
 
// This code has been contributed by 29AjayKumar

PHP




<?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
?>

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Function to print the required string
    function printString(str, n)
    {
        // count number of 1s
        let ones = 0;
        for (let i = 0; i < n; i++)
            if (str[i] == '1')
                ones++;
 
        // To check if the all the 1s
        // have been used or not
        let used = false;
 
        for (let i = 0; i < n; i++)
        {
            if (str[i] == '2' && !used)
            {
                used = true;
 
                // Print all the 1s if any 2 is encountered
                for (let j = 0; j < ones; j++)
                    document.write("1");
            }
 
            // If str[i] = 0 or str[i] = 2
            if (str[i] != '1')
                document.write(str[i]);
 
        }
 
        // If 1s are not printed yet
        if (!used)
            for (let j = 0; j < ones; j++)
                document.write("1");
    }
     
    let str = "100210";
    let n = str.length;
    printString(str.split(''), n);
     
</script>

Output: 

001120

 


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