Stack Permutations (Check if an array is stack permutation of other)
A stack permutation is a permutation of objects in the given input queue which is done by transferring elements from input queue to the output queue with the help of a stack and the built-in push and pop functions.
The well defined rules are:
- Only dequeue from the input queue.
- Use inbuilt push, pop functions in the single stack.
- Stack and input queue must be empty at the end.
- Only enqueue to the output queue.
There are a huge number of permutations possible using a stack for a single input queue.
Given two arrays, both of unique elements. One represents the input queue and the other represents the output queue. Our task is to check if the given output is possible through stack permutation.
Examples:
Input : First array: 1, 2, 3 Second array: 2, 1, 3 Output : Yes Procedure: push 1 from input to stack push 2 from input to stack pop 2 from stack to output pop 1 from stack to output push 3 from input to stack pop 3 from stack to output Input : First array: 1, 2, 3 Second array: 3, 1, 2 Output : Not Possible
The idea to do this is we will try to convert the input queue to output queue using a stack, if we are able to do so then the queue is permutable otherwise not.
Below is the step by step algorithm to do this:
- Continuously pop elements from the input queue and check if it is equal to the top of output queue or not, if it is not equal to the top of output queue then we will push the element to stack.
- Once we find an element in input queue such the top of input queue is equal to top of output queue, we will pop a single element from both input and output queues, and compare the top of stack and top of output queue now. If top of both stack and output queue are equal then pop element from both stack and output queue. If not equal, go to step 1.
- Repeat above two steps until the input queue becomes empty. At the end if both of the input queue and stack are empty then the input queue is permutable otherwise not.
Below is implementation of above idea:
C++
// Given two arrays, check if one array is // stack permutation of other. #include<bits/stdc++.h> using namespace std; // function to check if input queue is // permutable to output queue bool checkStackPermutation( int ip[], int op[], int n) { // Input queue queue< int > input; for ( int i=0;i<n;i++) input.push(ip[i]); // output queue queue< int > output; for ( int i=0;i<n;i++) output.push(op[i]); // stack to be used for permutation stack < int > tempStack; while (!input.empty()) { int ele = input.front(); input.pop(); if (ele == output.front()) { output.pop(); while (!tempStack.empty()) { if (tempStack.top() == output.front()) { tempStack.pop(); output.pop(); } else break ; } } else tempStack.push(ele); } // If after processing, both input queue and // stack are empty then the input queue is // permutable otherwise not. return (input.empty()&&tempStack.empty()); } // Driver program to test above function int main() { // Input Queue int input[] = {1, 2, 3}; // Output Queue int output[] = {2, 1, 3}; int n = 3; if (checkStackPermutation(input, output, n)) cout << "Yes" ; else cout << "Not Possible" ; return 0; } |
Java
// Given two arrays, check if one array is // stack permutation of other. import java.util.LinkedList; import java.util.Queue; import java.util.Stack; class Gfg { // function to check if input queue is // permutable to output queue static boolean checkStackPermutation( int ip[], int op[], int n) { Queue<Integer> input = new LinkedList<>(); // Input queue for ( int i = 0 ; i < n; i++) { input.add(ip[i]); } // Output queue Queue<Integer> output = new LinkedList<>(); for ( int i = 0 ; i < n; i++) { output.add(op[i]); } // stack to be used for permutation Stack<Integer> tempStack = new Stack<>(); while (!input.isEmpty()) { int ele = input.poll(); if (ele == output.peek()) { output.poll(); while (!tempStack.isEmpty()) { if (tempStack.peek() == output.peek()) { tempStack.pop(); output.poll(); } else break ; } } else { tempStack.push(ele); } } // If after processing, both input queue and // stack are empty then the input queue is // permutable otherwise not. return (input.isEmpty() && tempStack.isEmpty()); } // Driver code public static void main(String[] args) { // Input Queue int input[] = { 1 , 2 , 3 }; // Output Queue int output[] = { 2 , 1 , 3 }; int n = 3 ; if (checkStackPermutation(input, output, n)) System.out.println( "Yes" ); else System.out.println( "Not Possible" ); } } // This code is contributed by Vivekkumar Singh |
Python3
# Given two arrays, check if one array is # stack permutation of other. from queue import Queue # function to check if Input queue # is permutable to output queue def checkStackPermutation(ip, op, n): # Input queue Input = Queue() for i in range (n): Input .put(ip[i]) # output queue output = Queue() for i in range (n): output.put(op[i]) # stack to be used for permutation tempStack = [] while ( not Input .empty()): ele = Input .queue[ 0 ] Input .get() if (ele = = output.queue[ 0 ]): output.get() while ( len (tempStack) ! = 0 ): if (tempStack[ - 1 ] = = output.queue[ 0 ]): tempStack.pop() output.get() else : break else : tempStack.append(ele) # If after processing, both Input # queue and stack are empty then # the Input queue is permutable # otherwise not. return ( Input .empty() and len (tempStack) = = 0 ) # Driver Code if __name__ = = '__main__' : # Input Queue Input = [ 1 , 2 , 3 ] # Output Queue output = [ 2 , 1 , 3 ] n = 3 if (checkStackPermutation( Input , output, n)): print ( "Yes" ) else : print ( "Not Possible" ) # This code is contributed by PranchalK |
C#
// Given two arrays, check if one array is // stack permutation of other. using System; using System.Collections.Generic; class GFG { // function to check if input queue is // permutable to output queue static bool checkStackPermutation( int []ip, int []op, int n) { Queue< int > input = new Queue< int >(); // Input queue for ( int i = 0; i < n; i++) { input.Enqueue(ip[i]); } // Output queue Queue< int > output = new Queue< int >(); for ( int i = 0; i < n; i++) { output.Enqueue(op[i]); } // stack to be used for permutation Stack< int > tempStack = new Stack< int >(); while (input.Count != 0) { int ele = input.Dequeue(); if (ele == output.Peek()) { output.Dequeue(); while (tempStack.Count != 0) { if (tempStack.Peek() == output.Peek()) { tempStack.Pop(); output.Dequeue(); } else break ; } } else { tempStack.Push(ele); } } // If after processing, both input queue and // stack are empty then the input queue is // permutable otherwise not. return (input.Count == 0 && tempStack.Count == 0); } // Driver code public static void Main(String[] args) { // Input Queue int []input = { 1, 2, 3 }; // Output Queue int []output = { 2, 1, 3 }; int n = 3; if (checkStackPermutation(input, output, n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "Not Possible" ); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Given two arrays, check if one array is // stack permutation of other. // function to check if input queue is // permutable to output queue function checkStackPermutation(ip, op, n) { let input = []; // Input queue for (let i = 0; i < n; i++) { input.push(ip[i]); } // Output queue let output = []; for (let i = 0; i < n; i++) { output.push(op[i]); } // stack to be used for permutation let tempStack = []; while (input.length != 0) { let ele = input.shift(); if (ele == output[0]) { output.shift(); while (tempStack.length != 0) { if (tempStack[tempStack.length - 1] == output[0]) { tempStack.pop(); output.shift(); } else break ; } } else { tempStack.push(ele); } } // If after processing, both input queue and // stack are empty then the input queue is // permutable otherwise not. return (input.length == 0 && tempStack.length == 0); } // Input Queue let input = [ 1, 2, 3 ]; // Output Queue let output = [ 2, 1, 3 ]; let n = 3; if (checkStackPermutation(input, output, n)) document.write( "Yes" ); else document.write( "Not Possible" ); // This code is contributed by rameshtravel07. </script> |
Output:
Yes
Another Approach: –
Idea – we will start iterating on input array and storing its element one by one in a stack and if top of our stack matches with an element in output array we will pop that element from stack and compare next element of output array with top of our stack if again it matches then again pop until our stack isn’t empty
C++
// Given two arrays, check if one array is // stack permutation of other. #include<bits/stdc++.h> using namespace std; // function to check if input array is // permutable to output array bool checkStackPermutation( int ip[], int op[], int n) { // we will be pushing elements from input array to stack uptill top of our stack // matches with first element of output array stack< int >s; // will maintain a variable j to iterate on output array int j=0; // will iterate one by one in input array for ( int i=0;i<n;i++) { // pushed an element from input array to stack s.push(ip[i]); // if our stack isn't empty and top matches with output array // then we will keep popping out from stack uptill top matches with // output array while (!s.empty() and s.top()==op[j]) { s.pop(); // increasing j so next time we can compare next element in output array j++; } } // if output array was a correct permutation of input array then // by now our stack should be empty if (s.empty()) { return true ; } return false ; } // Driver program to test above function int main() { // Input Array int input[] = {4,5,6,7,8}; // Output Array int output[] = {8,7,6,5,4}; int n = 5; if (checkStackPermutation(input, output, n)) cout << "Yes" ; else cout << "Not Possible" ; return 0; } |
Java
// Java program to check if one array is // stack permutation of other. import java.util.Stack; class Rextester { // function to check if input array is // permutable to output array static Boolean checkStackPermutation( int ip[], int op[], int n) { // we will be pushing elements from input array to // stack uptill top of our stack matches with first // element of output array Stack<Integer> s = new Stack<Integer>(); // will maintain a variable j to iterate on output // array int j = 0 ; // will iterate one by one in input array for ( int i = 0 ; i < n; i++) { // pushed an element from input array to stack s.push(ip[i]); // if our stack isn't empty and top matches with // output array then we will keep popping out // from stack uptill top matches with output // array while (!s.isEmpty() && s.peek() == op[j]) { s.pop(); // increasing j so next time we can compare // next element in output array j++; } } // if output array was a correct permutation of // input array then by now our stack should be empty if (s.isEmpty()) { return true ; } return false ; } // Driver program to test above function public static void main(String args[]) { // Input Array int input[] = { 4 , 5 , 6 , 7 , 8 }; // Output Array int output[] = { 8 , 7 , 6 , 5 , 4 }; int n = 5 ; if (checkStackPermutation(input, output, n)) System.out.println( "Yes" ); else System.out.println( "Not Possible" ); } } // This code is contributed by Lovely Jain |
Python3
# Given two arrays, check if one array is # stack permutation of other. # function to check if input array is # permutable to output array def checkStackPermutation(ip, op, n): # we will be appending elements from input array to stack uptill top of our stack # matches with first element of output array s = [] # will maintain a variable j to iterate on output array j = 0 # will iterate one by one in input array for i in range (n): # appended an element from input array to stack s.append(ip[i]) # if our stack isn't empty and top matches with output array # then we will keep popping out from stack uptill top matches with # output array while ( len (s) > 0 and s[ - 1 ] = = op[j]): s.pop() # increasing j so next time we can compare next element in output array j + = 1 # if output array was a correct permutation of input array then # by now our stack should be empty if ( len (s) = = 0 ): return True return False # Driver program to test above function # Input Array input = [ 4 , 5 , 6 , 7 , 8 ] # Output Array output = [ 8 , 7 , 6 , 5 , 4 ] n = 5 if (checkStackPermutation( input , output, n)): print ( "Yes" ) else : print ( "Not Possible" ) # This code is contributed by shinjanpatra |
Javascript
<script> // Given two arrays, check if one array is // stack permutation of other. // function to check if input array is // permutable to output array function checkStackPermutation(ip, op, n) { // we will be pushing elements from input array to stack uptill top of our stack // matches with first element of output array let s = []; // will maintain a variable j to iterate on output array let j = 0; // will iterate one by one in input array for (let i = 0; i < n; i++) { // pushed an element from input array to stack s.push(ip[i]); // if our stack isn't empty and top matches with output array // then we will keep popping out from stack uptill top matches with // output array while (s.length > 0 && s[s.length - 1] == op[j]) { s.pop(); // increasing j so next time we can compare next element in output array j++; } } // if output array was a correct permutation of input array then // by now our stack should be empty if (s.length == 0) { return true ; } return false ; } // Driver program to test above function // Input Array let input = [4,5,6,7,8]; // Output Array let output = [8,7,6,5,4]; let n = 5; if (checkStackPermutation(input, output, n)) document.write( "Yes" ); else document.write( "Not Possible" ); // This code is contributed by shinjanpatra </script> |
Yes
Time Complexity – O(N)
Space Complexity – O(N)
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