Check if Queue Elements are pairwise consecutive
Given a Queue of integers. The task is to check if consecutive elements in the queue are pairwise consecutive.
Examples:
Input : 1 2 5 6 9 10 Output : Yes Input : 2 3 9 11 8 7 Output : No
Approach:
Using two stacks :
- Transfer all elements of the queue to one auxiliary stack aux.
- Now, transfer the elements from this stack to another auxiliary stack aux2.
- Start popping two elements from aux2 and then check the difference between these two elements. If their difference is 1, continue for other pairs till one element is left in the stack.
- Also, after popping, push them to queue simultaneously to maintain the original queue.
- Finally, if there is any pair with a difference not equal to 1, return false, else return true.
Below is the implementation of the above approach:
C++
// C++ program to check if successive // pair of numbers in the queue are // consecutive or not #include <bits/stdc++.h> using namespace std; // Function to check if elements are // pairwise consecutive in queue bool pairWiseConsecutive(queue< int > q) { // Transfer elements of q to aux. stack< int > aux; while (!q.empty()) { aux.push(q.front()); q.pop(); } // Again transfer the // elements of aux to aux2 stack< int > aux2; while (!aux.empty()) { aux2.push(aux.top()); aux.pop(); } // Traverse aux2 and see if // elements are pairwise // consecutive or not. We also // need to make sure that original // content is retained. bool result = true ; while (aux2.size() > 1) { // Fetch current top two // elements of aux2 and check // if they are consecutive. int x = aux2.top(); aux2.pop(); int y = aux2.top(); aux2.pop(); if ( abs (x - y) != 1) result = false ; // Push the elements to queue q.push(x); q.push(y); } if (aux2.size() == 1) q.push(aux2.top()); return result; } // Driver program int main() { // Pushing elements into the queue queue< int > q; q.push(4); q.push(5); q.push(-2); q.push(-3); q.push(11); q.push(10); q.push(5); q.push(6); if (pairWiseConsecutive(q)) cout << "Yes" << endl; else cout << "No" << endl; // Printing the original queue while (!q.empty()) { cout << q.front() << " " ; q.pop(); } cout << endl; return 0; } |
Java
// Java program to check if successive // pair of numbers in the queue are // consecutive or not import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class GFG { // Function to check if elements are // pairwise consecutive in queue static boolean pairWiseConsecutive(Queue<Integer> q) { // Transfer elements of q to aux. Stack<Integer> aux = new Stack<>(); while (!q.isEmpty()) { aux.push(q.peek()); q.poll(); } // Again transfer the // elements of aux to aux2 Stack<Integer> aux2 = new Stack<>(); while (!aux.empty()) { aux2.push(aux.peek()); aux.pop(); } // Traverse aux2 and see if // elements are pairwise // consecutive or not. We also // need to make sure that original // content is retained. boolean result = true ; while (aux2.size() > 1 ) { // Fetch current top two // elements of aux2 and check // if they are consecutive. int x = aux2.peek(); aux2.pop(); int y = aux2.peek(); aux2.pop(); if (Math.abs(x - y) != 1 ) result = false ; // Push the elements to queue q.add(x); q.add(y); } if (aux2.size() == 1 ) q.add(aux2.peek()); return result; } // Driver program static public void main(String[] args) { // Pushing elements into the queue Queue<Integer> q= new LinkedList<Integer>(); q.add( 4 ); q.add( 5 ); q.add(- 2 ); q.add(- 3 ); q.add( 11 ); q.add( 10 ); q.add( 5 ); q.add( 6 ); if (pairWiseConsecutive(q)) System.out.println( "Yes" ); else System.out.println( "No" ); // Printing the original queue while (!q.isEmpty()) { System.out.print(q.peek() + " " ); q.remove(); } System.out.println(); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program to check # if successive pair of numbers # in the queue are consecutive or not import queue # Function to check if elements are # pairwise consecutive in queue def pairWiseConsecutive(q): # Transfer elements of # q to aux. aux = [] while (q.qsize() ! = 0 ): aux.append(q.queue[ 0 ]) q.get() # Again transfer the # elements of aux to aux2 aux2 = [] while ( len (aux) ! = 0 ): aux2.append(aux[ len (aux) - 1 ]) aux.pop() # Traverse aux2 and see if # elements are pairwise # consecutive or not. We also # need to make sure that original # content is retained. result = bool ( True ) while ( len (aux2) > 1 ): # Fetch current top two # elements of aux2 and check # if they are consecutive. x = aux2[ len (aux2) - 1 ] aux2.pop() y = aux2[ len (aux2) - 1 ] aux2.pop() if ( abs (x - y) ! = 1 ): result = bool ( False ) # Push the elements # to queue q.put(x) q.put(y) if ( len (aux2) = = 1 ): q.put(aux2[ len (aux2) - 1 ]) return result # Driver code # Pushing elements # into the queue q = queue.Queue() q.put( 4 ) q.put( 5 ) q.put( - 2 ) q.put( - 3 ) q.put( 11 ) q.put( 10 ) q.put( 5 ) q.put( 6 ) if (pairWiseConsecutive(q)): print ( "Yes" ) else : print ( "No" ) # Printing the original queue while ( not q.empty()): print (q.queue[ 0 ] , end = " " ) q.get() # This code is contributed by divyeshrabadiya07 |
C#
// C# program to check if successive // pair of numbers in the queue are // consecutive or not using System; using System.Collections.Generic; class GFG { // Function to check if elements are // pairwise consecutive in queue static Boolean pairWiseConsecutive(Queue< int > q) { // Transfer elements of q to aux. Stack< int > aux = new Stack< int >(); while (q.Count != 0) { aux.Push(q.Peek()); q.Dequeue(); } // Again transfer the // elements of aux to aux2 Stack< int > aux2 = new Stack< int >(); while (aux.Count != 0) { aux2.Push(aux.Peek()); aux.Pop(); } // Traverse aux2 and see if // elements are pairwise // consecutive or not. We also // need to make sure that original // content is retained. Boolean result = true ; while (aux2.Count > 1) { // Fetch current top two // elements of aux2 and check // if they are consecutive. int x = aux2.Peek(); aux2.Pop(); int y = aux2.Peek(); aux2.Pop(); if (Math.Abs(x - y) != 1) result = false ; // Push the elements to queue q.Enqueue(x); q.Enqueue(y); } if (aux2.Count == 1) q.Enqueue(aux2.Peek()); return result; } // Driver code static public void Main(String[] args) { // Pushing elements into the queue Queue< int > q = new Queue< int >(); q.Enqueue(4); q.Enqueue(5); q.Enqueue(-2); q.Enqueue(-3); q.Enqueue(11); q.Enqueue(10); q.Enqueue(5); q.Enqueue(6); if (pairWiseConsecutive(q)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); // Printing the original queue while (q.Count != 0) { Console.Write(q.Peek() + " " ); q.Dequeue(); } Console.WriteLine(); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to check if successive // pair of numbers in the queue are // consecutive or not // Function to check if elements are // pairwise consecutive in queue function pairWiseConsecutive(q) { // Transfer elements of q to aux. var aux = []; while (q.length!=0) { aux.push(q[0]); q.shift(); } // Again transfer the // elements of aux to aux2 var aux2 = []; while (aux.length!=0) { aux2.push(aux[aux.length-1]); aux.pop(); } // Traverse aux2 and see if // elements are pairwise // consecutive or not. We also // need to make sure that original // content is retained. var result = true ; while (aux2.length > 1) { // Fetch current top two // elements of aux2 and check // if they are consecutive. var x = aux2[aux2.length-1]; aux2.pop(); var y = aux2[aux2.length-1]; aux2.pop(); if (Math.abs(x - y) != 1) result = false ; // Push the elements to queue q.push(x); q.push(y); } if (aux2.length == 1) q.push(aux2[aux2.length-1]); return result; } // Driver program // Pushing elements into the queue var q = []; q.push(4); q.push(5); q.push(-2); q.push(-3); q.push(11); q.push(10); q.push(5); q.push(6); if (pairWiseConsecutive(q)) document.write( "Yes<br>" ); else document.write( "No<br>" ); // Printing the original queue while (q.length!=0) { document.write( q[0] + " " ); q.shift(); } document.write( "<br>" ); </script> |
Output
Yes 4 5 -2 -3 11 10 5 6
Complexity Analysis:
- Time Complexity: O(n), where n is the size of the queue.
- Auxiliary Space: O(n), where n is the size of the stack.