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# Sort the given stack elements based on their modulo with K

• Difficulty Level : Easy
• Last Updated : 29 Jun, 2022

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 ``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 input, ``int` `k)``{``    ``Stack tmpStack = ``new` `Stack();` `    ``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 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 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

 ``

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)

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