A stack permutation is a permutation of objects in the given input queue which is done by transferring elements from the input queue to the output queue with the help of a stack and the built-in push and pop functions.
The rules are:
- Only dequeue from the input queue.
- Use inbuilt push, and 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: arr1[] = [ 1, 2, 3 ] , arr2[] = [ 2, 1, 3 ]
Output: YES
Explanation:
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 outputInput: arr1[] = [ 1, 2, 3 ] , arr2[] = [ 3, 1, 2 ]
Output: Not Possible
Stack Permutation Using Stack
The idea is to try to convert the input queue to the output queue using a stack, if we are able to do so then the queue is permutable otherwise not.
Follow the steps mentioned below to implement the approach:
- 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 the implementation of the above approach:
// 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;
} |
// 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 |
# 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 |
// 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 |
<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> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
Optimized Approach
The idea to start iterating on the input array and storing its element one by one in a stack and if the top of our stack matches with an element in the output array we will pop that element from the stack and compare the next element of the output array with the top of our stack if again it matches then again pop until our stack isn’t empty
Below is the implementation of the above approach:
// 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 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 |
# 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 |
// 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 array is
// permutable to output array
static 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 = new Stack< int >();
// 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.Count != 0 && s.Peek() == op[j]) {
// increasing j so next time we can compare
// next element in output array
s.Pop();
j = j + 1;
}
}
// if output array was a correct permutation of
// input array then by now our stack should be empty
if (s.Count == 0) {
return true ;
}
return false ;
}
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 aadityamaharshi21. |
<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)
Auxiliary Space: O(N)
Optimize Approach 2:
The above code already has a linear time complexity, but we can make a few small optimizations to make it more efficient:
Use std::vector instead of a fixed-size array. This will make it easier to pass the arrays to the function and avoid potential buffer overflows.
Reserve memory in the vector to avoid unnecessary allocations. We know the exact size of the arrays, so we can reserve that much memory in the vectors to avoid resizing during the push operation.
Avoid unnecessary comparisons by breaking out of the loop early. If we encounter an element in the input array that is already in the output array, we know that it cannot be a valid stack permutation, so we can return false immediately.
Here’s the optimized code:
#include <iostream> #include <vector> #include <stack> using namespace std;
bool checkStackPermutation( const vector< int >& input, const vector< int >& output) {
stack< int > s;
int j = 0;
for ( int i = 0; i < input.size(); i++) {
s.push(input[i]);
while (!s.empty() && s.top() == output[j]) {
s.pop();
j++;
}
} if (j==output.size())
return true ;
return false ;
} int main() {
vector< int > input = {4, 5, 6, 7, 8};
vector< int > output = {8, 7, 6, 5, 4};
if (input.size() != output.size()) {
cout << "Not Possible" << endl;
return 0;
}
checkStackPermutation(input, output) ? cout << "Yes" << endl : cout << "Not Possible" << endl;
return 0;
} |
import java.util.*;
public class Main {
public static boolean checkStackPermutation(List<Integer> input, List<Integer> output) {
Stack<Integer> s = new Stack<>();
int j = 0 ;
for ( int i = 0 ; i < input.size(); i++) {
s.push(input.get(i));
while (!s.empty() && s.peek() == output.get(j)) {
s.pop();
j++;
}
if (j < output.size() && s.peek() == output.get(j)) {
return false ;
}
}
return true ;
}
public static void main(String[] args) {
List<Integer> input = new ArrayList<>(Arrays.asList( 4 , 5 , 6 , 7 , 8 ));
List<Integer> output = new ArrayList<>(Arrays.asList( 8 , 7 , 6 , 5 , 4 ));
if (input.size() != output.size()) {
System.out.println( "Not Possible" );
return ;
}
if (checkStackPermutation(input, output)) {
System.out.println( "Yes" );
} else {
System.out.println( "Not Possible" );
}
}
} |
from typing import List
def checkStackPermutation(ip: List [ int ], op: List [ int ]) - > bool :
s = []
j = 0
for i in range ( len (ip)):
s.append(ip[i])
while s and s[ - 1 ] = = op[j]:
s.pop()
j + = 1
if j < len (op) and s[ - 1 ] = = op[j]:
return False
return True
input_arr = [ 4 , 5 , 6 , 7 , 8 ]
output_arr = [ 8 , 7 , 6 , 5 , 4 ]
if len (input_arr) ! = len (output_arr):
print ( "Not Possible" )
else :
if checkStackPermutation(input_arr, output_arr):
print ( "Yes" )
else :
print ( "Not Possible" )
|
using System;
using System.Collections.Generic;
public class MainClass {
public static bool CheckStackPermutation(List< int > input, List< int > output) {
Stack< int > s = new Stack< int >();
int j = 0;
for ( int i = 0; i < input.Count; i++) {
s.Push(input[i]);
while (s.Count > 0 && s.Peek() == output[j]) {
s.Pop();
j++;
}
if (j < output.Count && s.Peek() == output[j]) {
return false ;
}
}
return true ;
}
public static void Main( string [] args) {
List< int > input = new List< int >() { 4, 5, 6, 7, 8 };
List< int > output = new List< int >() { 8, 7, 6, 5, 4 };
if (input.Count != output.Count) {
Console.WriteLine( "Not Possible" );
return ;
}
if (CheckStackPermutation(input, output)) {
Console.WriteLine( "Yes" );
} else {
Console.WriteLine( "Not Possible" );
}
}
} |
function checkStackPermutation(ip, op) {
let s = [];
let j = 0;
for (let i = 0; i < ip.length; i++) {
s.push(ip[i]);
while (s.length > 0 && s[s.length - 1] === op[j]) {
s.pop();
j++;
}
if (j < op.length && s[s.length - 1] === op[j]) {
return false ;
}
}
return true ;
} const inputArr = [4, 5, 6, 7, 8]; const outputArr = [8, 7, 6, 5, 4]; if (inputArr.length !== outputArr.length) {
console.log( "Not Possible" );
} else {
if (checkStackPermutation(inputArr, outputArr)) {
console.log( "Yes" );
} else {
console.log( "Not Possible" );
}
} |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)