Sort the given stack elements based on their modulo with K

Given a stack of integers and an integer K, the task is to sort the elements of the given stack using another stack in the increasing order of their modulo with K. If two numbers have the same remainder then the smaller number should come first.
Examples

Input: stack = {10, 3, 2, 6, 12}, K = 4 
Output: 12 2 6 10 3 
{12, 2, 6, 10, 3} is the required sorted order as the modulo 
of these elements with K = 4 is {2, 3, 2, 2, 0}

Input: stack = {3, 4, 5, 10, 11, 1}, K = 3 
Output: 3 1 4 10 5 11 

Approach: An approach to sort the elements of the stack using another temporary stack has been discussed in this article, the same approach can be used here to sort the elements based on their modulo with K, the only difference is that when the elements being compared give the same modulo value then they will be compared based on their values.

Below is the implementation of the above approach:  

C++

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;

// Function to sort the stack using
// another stack based on the
// values of elements modulo k
void sortStack(stack<int>& input, int k)
{
    stack<int> tmpStack;

    while (!input.empty()) {

        // Pop out the first element
        int tmp = input.top();
        input.pop();

        // While temporary stack is not empty
        while (!tmpStack.empty()) {
            int tmpStackMod = tmpStack.top() % k;
            int tmpMod = tmp % k;

            // The top of the stack modulo k is
            // greater than (temp & k) or if they
            // are equal then compare the values
            if ((tmpStackMod > tmpMod)
                || (tmpStackMod == tmpMod
                    && tmpStack.top() > tmp)) {

                // Pop from temporary stack and push
                // it to the input stack
                input.push(tmpStack.top());
                tmpStack.pop();
            }
            else
                break;
        }

        // Push temp in temporary of stack
        tmpStack.push(tmp);
    }

    // Push all the elements in the original
    // stack to get the ascending order
    while (!tmpStack.empty()) {
        input.push(tmpStack.top());
        tmpStack.pop();
    }

    // Print the sorted elements
    while (!input.empty()) {
        cout << input.top() << " ";
        input.pop();
    }
}

// Driver code
int main()
{
    stack<int> input;
    input.push(10);
    input.push(3);
    input.push(2);
    input.push(6);
    input.push(12);

    int k = 4;

    sortStack(input, k);

    return 0;
}

Java

// Java implementation of the
// above approach
import java.io.*;
import java.util.*;

class GFG{

// Function to sort the stack using
// another stack based on the
// values of elements modulo k
static void sortStack(Stack<Integer> input, int k)
{
    Stack<Integer> tmpStack = new Stack<Integer>();

    while (!input.isEmpty())
    {

        // Pop out the first element
        int tmp = input.peek();
        input.pop();

        // While temporary stack is not empty
        while (!tmpStack.isEmpty())
        {
            int tmpStackMod = tmpStack.peek() % k;
            int tmpMod = tmp % k;

            // The top of the stack modulo k is
            // greater than (temp & k) or if they
            // are equal then compare the values
            if ((tmpStackMod > tmpMod) ||
                (tmpStackMod == tmpMod &&
                 tmpStack.peek() > tmp))
            {

                // Pop from temporary stack and push
                // it to the input stack
                input.push(tmpStack.peek());
                tmpStack.pop();
            }
            else
                break;
        }

        // Push temp in temporary of stack
        tmpStack.push(tmp);
    }

    // Push all the elements in the original
    // stack to get the ascending order
    while (!tmpStack.isEmpty())
    {
        input.push(tmpStack.peek());
        tmpStack.pop();
    }

    // Print the sorted elements
    while (!input.empty())
    {
        System.out.print(input.peek() + " ");
        input.pop();
    }
}

// Driver code
public static void main(String args[])
{
    Stack<Integer> input = new Stack<Integer>();
    input.push(10);
    input.push(3);
    input.push(2);
    input.push(6);
    input.push(12);

    int k = 4;

    sortStack(input, k);
}
}

// This code is contributed by adityapande88

Python3

# Python3 implementation of the
# above approach

# Function to sort the stack using
# another stack based on the
# values of elements modulo k
def sortStack(input1, k):

    tmpStack = []

    while (len(input1) != 0):

        # Pop out the first element
        tmp = input1[-1]
        input1.pop()

        # While temporary stack is
        # not empty
        while (len(tmpStack) != 0):
            tmpStackMod = tmpStack[-1] % k
            tmpMod = tmp % k

            # The top of the stack modulo
            # k is greater than (temp & k)
            # or if they are equal then
            # compare the values
            if ((tmpStackMod > tmpMod) or
                (tmpStackMod == tmpMod and
                 tmpStack[-1] > tmp)):

                # Pop from temporary stack
                # and push it to the input
                # stack
                input1.append(tmpStack[-1])
                tmpStack.pop()
            else:
                break

        # Push temp in temporary of stack
        tmpStack.append(tmp)

    # Push all the elements in
    # the original stack to get
    # the ascending order
    while (len(tmpStack) != 0):
        input1.append(tmpStack[-1])
        tmpStack.pop()

    # Print the sorted elements
    while (len(input1) != 0):
        print(input1[-1], end = " ")
        input1.pop()

# Driver code
if __name__ == "__main__":

    input1 = []
    input1.append(10)
    input1.append(3)
    input1.append(2)
    input1.append(6)
    input1.append(12)
    k = 4
    sortStack(input1, k)

# This code is contributed by Chitranayal

C#

// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{

// Function to sort the stack using
// another stack based on the
// values of elements modulo k
static void sortStack(Stack input,
                      int k)
{
  Stack tmpStack = new Stack();

  while(input.Count != 0)
  {
    // Pop out the first element
    int tmp = (int)input.Peek();
    input.Pop();

    // While temporary stack is not empty
    while (tmpStack.Count != 0)
    {
      int tmpStackMod = (int)tmpStack.Peek() % k;
      int tmpMod = tmp % k;

      // The top of the stack modulo k is
      // greater than (temp & k) or if they
      // are equal then compare the values
      if ((tmpStackMod > tmpMod) ||
          (tmpStackMod == tmpMod &&
          (int)tmpStack.Peek() > tmp))
      {
        // Pop from temporary stack and push
        // it to the input stack
        input.Push((int)tmpStack.Peek());
        tmpStack.Pop();
      }
      else
        break;
    }

    // Push temp in temporary of stack
    tmpStack.Push(tmp);
  }

  // Push all the elements in the original
  // stack to get the ascending order
  while (tmpStack.Count != 0)
  {
    input.Push((int)tmpStack.Peek());
    tmpStack.Pop();
  }

  // Print the sorted elements
  while (input.Count != 0)
  {
    Console.Write((int)input.Peek() + " ");
    input.Pop();
  }
}

// Driver Code
public static void Main(string[] args)
{
  Stack input = new Stack();
  input.Push(10);
  input.Push(3);
  input.Push(2);
  input.Push(6);
  input.Push(12);
  int k = 4;
  sortStack(input, k);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// JavaScript implementation of the
// above approach

// Function to sort the stack using
// another stack based on the
// values of elements modulo k
function sortStack(input,k)
{
    let tmpStack = [];

    while (input.length!=0)
    {

        // Pop out the first element
           let tmp = input.pop();


        // While temporary stack is not empty
        while (tmpStack.length!=0)
        {
            let tmpStackMod = tmpStack[tmpStack.length-1] % k;
            let tmpMod = tmp % k;

            // The top of the stack modulo k is
            // greater than (temp & k) or if they
            // are equal then compare the values
            if ((tmpStackMod > tmpMod) ||
                (tmpStackMod == tmpMod &&
                 tmpStack[tmpStack.length-1] > tmp))
            {

                // Pop from temporary stack and push
                // it to the input stack
                input.push(tmpStack[tmpStack.length-1]);
                tmpStack.pop();
            }
            else
                break;
        }

        // Push temp in temporary of stack
        tmpStack.push(tmp);
    }

    // Push all the elements in the original
    // stack to get the ascending order
    while (tmpStack.length!=0)
    {
        input.push(tmpStack[tmpStack.length-1]);
        tmpStack.pop();
    }

    // Print the sorted elements
    while (input.length!=0)
    {
        document.write(input.pop() + " ");

    }
}

// Driver code
let input =[];
input.push(10);
input.push(3);
input.push(2);
input.push(6);
input.push(12);

let k = 4;

sortStack(input, k);


// This code is contributed by rag2127

</script>

Output:

12 2 6 10 3

Time Complexity: O(n²) where n is the total number of integers in the given stack.
Auxiliary Space: O(n)