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 tempory 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 tempory 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 tempory 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 tempory 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 |
12 2 6 10 3
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