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Smallest number possible by swapping adjacent even odd pairs

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  • Difficulty Level : Medium
  • Last Updated : 30 Mar, 2022
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Given a numeric string str, the task is to find the smallest integer that can be formed by swapping adjacent digits of distinct parity. Examples:

Input: 836360 Output: 338660 Explanation: 1st Swap: 836360 -> 386360 2nd Swap: 386360 -> 383660 3rd Swap: 383660 -> 338660 Input: 1003 Output: 13

Approach: We can observe that on repeated swaps, we can split the even and odd digits of str into two separate blocks. The order of digits in their respective blocks will be the same as their order of appearance in the string as swapping within a block (same parity) is not allowed. Thus, after separating str into two separate blocks, we need to traverse these two blocks and append the minimum of the two values currently pointed, to the answer. The final string generated after this operation followed by the removal of leading 0’s, if any, is the required answer. Example: The even and odd blocks int the order of their appearance in 836360 are {8, 6, 6, 0} and {3, 3} respectively. Hence the smallest number ans formed is as follows:

  1. ans = ans + min(8, 3) => ans = 3
  2. ans = ans + min(8, 3) => ans = 33
  3. Since, all the odd digits are exhausted, the remaining even digits need to be added one by one.
  4. Hence, the required answer is 338660

Below is the implementation of the above approach: 

C++




// C++ Program to find the
// smallest number possible
// by swapping adjacent digits
// of different parity
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// smallest number possible
string findAns(string s)
{
    int digit;
 
    // Arrays to store odd and
    // even digits in the order
    // of their appearance in
    // the given string
    vector<int> odd;
    vector<int> even;
 
    // Insert the odd and
    // even digits
    for (auto c : s) {
        digit = c - '0';
        if (digit & 1)
            odd.push_back(digit);
        else
            even.push_back(digit);
    }
 
    // pointer to odd digit
    int i = 0;
    // pointer to even digit
    int j = 0;
 
    string ans = "";
 
    while (i < odd.size()
           and j < even.size()) {
 
        if (odd[i] < even[j])
            ans += (char)(odd[i++] + '0');
        else
            ans += (char)(even[j++] + '0');
    }
 
    // In case number of even and
    // odd digits are not equal
 
    // If odd digits are remaining
    while (i < odd.size())
        ans += (char)(odd[i++] + '0');
 
    // If even digits are remaining
    while (j < even.size())
        ans += (char)(even[j++] + '0');
 
    // Removal of leading 0's
    while (ans[0] == '0') {
        ans.erase(ans.begin());
    }
 
    return ans;
}
int main()
{
 
    string s = "894687536";
    cout << findAns(s);
    return 0;
}


Java




// Java program to find the smallest
// number possible by swapping adjacent
// digits of different parity
import java.util.*;
 
class GFG{
 
// Function to return the
// smallest number possible
static String findAns(String s)
{
    int digit;
 
    // Arrays to store odd and even
    // digits in the order their
    // appearance in the given String
    Vector<Integer> odd =new Vector<Integer>();
    Vector<Integer> even = new Vector<Integer>();
 
    // Insert the odd and
    // even digits
    for(char c : s.toCharArray())
    {
       digit = c - '0';
       if (digit % 2 == 1)
           odd.add(digit);
       else
           even.add(digit);
    }
 
    // Pointer to odd digit
    int i = 0;
     
    // Pointer to even digit
    int j = 0;
 
    String ans = "";
 
    while (i < odd.size() && j < even.size())
    {
        if (odd.get(i) < even.get(j))
            ans += (char)(odd.get(i++) + '0');
        else
            ans += (char)(even.get(j++) + '0');
    }
 
    // In case number of even and
    // odd digits are not equal
    // If odd digits are remaining
    while (i < odd.size())
        ans += (char)(odd.get(i++) + '0');
 
    // If even digits are remaining
    while (j < even.size())
        ans += (char)(even.get(j++) + '0');
 
    // Removal of leading 0's
    while (ans.charAt(0) == '0')
    {
        ans = ans.substring(1);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "894687536";
     
    System.out.print(findAns(s));
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 Program to find the
# smallest number possible
# by swapping adjacent digits
# of different parity
 
# Function to return the
# smallest number possible
def findAns(s):
 
    # Arrays to store odd and
    # even digits in the order
    # of their appearance in
    # the given string
    odd = []
    even = []
 
    # Insert the odd and
    # even digits
         
    for c in s:
        digit = int(c)
        if (digit & 1):
            odd.append(digit)
        else:
            even.append(digit)
 
    # pointer to odd digit
    i = 0
     
    # pointer to even digit
    j = 0
 
    ans = ""
 
    while (i < len(odd) and j < len(even)):
         
        if (odd[i] < even[j]):
            ans += str(odd[i])
            i = i + 1
        else:
            ans += str(even[j])
            j = j + 1
 
    # In case number of even and
    # odd digits are not equal
 
    # If odd digits are remaining
    while (i < len(odd)):
        ans += str(odd[i])
        i = i + 1
 
    # If even digits are remaining
    while (j < len(even)):
        ans += str(even[j])
        j = j + 1
 
    # Removal of leading 0's
    while (ans[0] == '0'):
        ans = ans[1:]
 
    return ans
 
# Driver Code
s = "894687536"
print(findAns(s))
 
# This code is contributed by yatin


C#




// C# program to find the smallest
// number possible by swapping adjacent
// digits of different parity
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the
// smallest number possible
static String findAns(String s)
{
    int digit;
 
    // Arrays to store odd and even
    // digits in the order their
    // appearance in the given String
    List<int> odd = new List<int>();
    List<int> even = new List<int>();
 
    // Insert the odd and
    // even digits
    foreach(char c in s.ToCharArray())
    {
        digit = c - '0';
        if (digit % 2 == 1)
            odd.Add(digit);
        else
            even.Add(digit);
    }
 
    // Pointer to odd digit
    int i = 0;
     
    // Pointer to even digit
    int j = 0;
 
    String ans = "";
 
    while (i < odd.Count && j < even.Count)
    {
        if (odd[i] < even[j])
            ans += (char)(odd[i++] + '0');
        else
            ans += (char)(even[j++] + '0');
    }
 
    // In case number of even and
    // odd digits are not equal
    // If odd digits are remaining
    while (i < odd.Count)
        ans += (char)(odd[i++] + '0');
 
    // If even digits are remaining
    while (j < even.Count)
        ans += (char)(even[j++] + '0');
 
    // Removal of leading 0's
    while (ans[0] == '0')
    {
        ans = ans.Substring(1);
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "894687536";
     
    Console.Write(findAns(s));
}
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
// javascript program to find the smallest
// number possible by swapping adjacent
// digits of different parity
 
    // Function to return the
    // smallest number possible
     function findAns( s) {
        var digit;
 
        // Arrays to store odd and even
        // digits in the order their
        // appearance in the given String
        var odd = [];
        var even = [];
 
        // Insert the odd and
        // even digits
        for (var i =0;i<s.length;i++) {
            digit = s[i].charCodeAt(0) -'0'.charCodeAt(0);
            if (digit % 2 == 1)
                odd.push(digit);
            else
                even.push(digit);
        }
 
        // Pointer to odd digit
        var i = 0;
 
        // Pointer to even digit
        var j = 0;
 
        var ans = "";
 
        while (i < odd.length && j < even.length) {
            if (odd[i] < even[j])
                ans +=  String.fromCharCode(odd[i++] + '0'.charCodeAt(0));
            else
                ans += String.fromCharCode(even[j++] + '0'.charCodeAt(0));
        }
 
        // In case number of even and
        // odd digits are not equal
        // If odd digits are remaining
        while (i < odd.length)
            ans +=  String.fromCharCode(odd[i++] + '0'.charCodeAt(0));
 
        // If even digits are remaining
        while (j < even.length)
            ans +=  String.fromCharCode(even[j++] + '0'.charCodeAt(0));
 
        // Removal of leading 0's
        while (ans.charAt(0) == '0') {
            ans = ans.substring(1);
        }
        return ans;
    }
 
    // Driver code
        var s = "894687536";
        document.write(findAns(s));
 
// This code is contributed by gauravrajput1
</script>


Output:

846869753

Time Complexity: O(N), where N is the size of the given string.


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