Given a stack of integers, write a function pairWiseSorted() that checks whether numbers in the stack are pairwise sorted or not.
The pairs must be increasing, and if the stack has an odd number of elements, the element at the top is left out of a pair. The function should retain the original stack content.
Only following standard operations are allowed on the stack.
- push(X): Enter a element X on top of stack.
- pop(): Removes top element of the stack.
- empty(): To check if stack is empty.
Examples:
Input: 4, 5, 6, 7, 8, 9 Output: Yes Input: 4, 9, 2, 1, 10, 8 Output: No
Approach: The idea is to use another stack.
- Create an auxiliary stack aux.
- Transfer contents of given stack to aux.
- Traverse aux. While traversing fetch top two elements and check if they are sorted or not.
- After checking put these elements back to original stack.
Below is the implementation of above approach:
C++
// C++ program to check if successive// pair of numbers in the stack are// sorted or not#include <bits/stdc++.h>using namespace std;// Function to check if elements are// pairwise sorted in stackbool pairWiseConsecutive(stack<int> s){ // Transfer elements of s to aux. stack<int> aux; while (!s.empty()) { aux.push(s.top()); s.pop(); } // Traverse aux and see if // elements are pairwise // sorted or not. We also // need to make sure that original // content is retained. bool result = true; while (!aux.empty()) { // Fetch current top two // elements of aux and check // if they are sorted. int x = aux.top(); aux.pop(); int y = aux.top(); aux.pop(); if (x > y) result = false; // Push the elements to original // stack. s.push(x); s.push(y); } if (aux.size() == 1) s.push(aux.top()); return result;}// Driver programint main(){ stack<int> s; s.push(4); s.push(5); s.push(-3); s.push(-2); s.push(10); s.push(11); s.push(5); s.push(6); // s.push(20); if (pairWiseConsecutive(s)) cout << "Yes" << endl; else cout << "No" << endl; cout << "Stack content (from top)" " after function call\n"; while (s.empty() == false) { cout << s.top() << " "; s.pop(); } return 0;} |
Java
// Java program to check if successive // pair of numbers in the stack are // sorted or not import java.util.Stack;class GFG {// Function to check if elements are // pairwise sorted in stack static boolean pairWiseConsecutive(Stack<Integer> s) { // Transfer elements of s to aux. Stack<Integer> aux = new Stack<>(); while (!s.empty()) { aux.push(s.peek()); s.pop(); } // Traverse aux and see if // elements are pairwise // sorted or not. We also // need to make sure that original // content is retained. boolean result = true; while (!aux.empty()) { // Fetch current top two // elements of aux and check // if they are sorted. int x = aux.peek(); aux.pop(); int y = aux.peek(); aux.pop(); if (x > y) { result = false; } // Push the elements to original // stack. s.push(x); s.push(y); } if (aux.size() == 1) { s.push(aux.peek()); } return result; }// Driver program public static void main(String[] args) { Stack<Integer> s = new Stack<>(); s.push(4); s.push(5); s.push(-3); s.push(-2); s.push(10); s.push(11); s.push(5); s.push(6); // s.push(20); if (pairWiseConsecutive(s)) { System.out.println("Yes"); } else { System.out.println("No"); } System.out.println("Stack content (from top)" + " after function call"); while (s.empty() == false) { System.out.print(s.peek() + " "); s.pop(); } }} |
Python 3
# Python program to check if successive# pair of numbers in the stack are# sorted or not# using deque as stackfrom collections import deque# Function to check if elements are# pairwise sorted in stackdef pairWiseConsecutive(s): # Transfer elements of s to aux. aux = deque() while len(s) > 0: aux.append(s.pop()) # Traverse aux and see if # elements are pairwise # sorted or not. We also # need to make sure that original # content is retained. result = True while len(aux) != 0: # Fetch current top two # elements of aux and check # if they are sorted. x = aux.pop() y = aux.pop() if x > y: result = False # Push the elements to original # stack. s.append(x) s.append(y) if len(aux) == 1: s.append(aux.pop()) return result# Driver Codeif __name__ == "__main__": s = deque() s.append(4) s.append(5) s.append(-3) s.append(-2) s.append(10) s.append(11) s.append(5) s.append(6) if pairWiseConsecutive(s): print("Yes") else: print("No") print("Stack content (from top) after function call") while len(s) > 0: print(s.pop(), end=" ")# This code is contributed by# sanjeev2552 |
C#
// C# program to check if successive // pair of numbers in the stack are using System;using System.Collections.Generic;public class GFG{// Function to check if elements are // pairwise sorted in stack static bool pairWiseConsecutive(Stack<int> s) { // Transfer elements of s to aux. Stack<int> aux = new Stack<int>(); while (!(s.Count==0)) { aux.Push(s.Peek()); s.Pop(); } // Traverse aux and see if // elements are pairwise // sorted or not. We also // need to make sure that original // content is retained. bool result = true; while (!(aux.Count==0)) { // Fetch current top two // elements of aux and check // if they are sorted. int x = aux.Peek(); aux.Pop(); int y = aux.Peek(); aux.Pop(); if (x > y) { result = false; } // Push the elements to original // stack. s.Push(x); s.Push(y); } if (aux.Count == 1) { s.Push(aux.Peek()); } return result; }// Driver program public static void Main() { Stack<int> s = new Stack<int>(); s.Push(4); s.Push(5); s.Push(-3); s.Push(-2); s.Push(10); s.Push(11); s.Push(5); s.Push(6); // s.push(20); if (pairWiseConsecutive(s)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } Console.WriteLine("Stack content (from top)" + " after function call"); while (!(s.Count == 0)) { Console.Write(s.Peek() + " "); s.Pop(); } }} |
Javascript
<script> // JavaScript program to check if successive// pair of numbers in the stack are// sorted or not// Function to check if elements are// pairwise sorted in stackfunction pairWiseConsecutive(s){ // Transfer elements of s to aux. var aux = []; while (s.length!=0) { aux.push(s[s.length-1]); s.pop(); } // Traverse aux and see if // elements are pairwise // sorted or not. We also // need to make sure that original // content is retained. var result = true; while (aux.length!=0) { // Fetch current top two // elements of aux and check // if they are sorted. var x = aux[aux.length-1]; aux.pop(); var y = aux[aux.length-1]; aux.pop(); if (x > y) result = false; // Push the elements to original // stack. s.push(x); s.push(y); } if (aux.length == 1) s.push(aux[aux.length-1]); return result;}// Driver programvar s = [];s.push(4);s.push(5);s.push(-3);s.push(-2);s.push(10);s.push(11);s.push(5);s.push(6);// s.push(20);if (pairWiseConsecutive(s)) document.write( "Yes" + "<br>");else document.write( "No" + "<br>");document.write( "Stack content (from top)"+ " after function call<br>");while (s.length!=0) { document.write( s[s.length-1] + " "); s.pop();}</script> |
Output
Yes Stack content (from top) after function call 6 5 11 10 -2 -3 5 4
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
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