# Validate Stack Sequences

Updated: Mar 25, 2021

Given two sequences pushed and popped **with distinct values**, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.

**Example 1:**

**Input****:**** **pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
**Output****:**** **true
**Explanation****:**** **We might do the following sequence: push(1), push(2), push(3), push(4), pop() -> 4, push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1

**Example 2:**

**Input****:**** **pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
**Output****:**** **false
**Explanation****:**** **1 cannot be popped before 2.

**Constraints:**

0 <= pushed.length == popped.length <= 1000

0 <= pushed[i], popped[i] < 1000

pushed is a permutation of popped.

pushed and popped have distinct values.

**Solution:**

**Approach 1: **

**Time Complexity:** O(N) **Space Complexity: **O(N)

```
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
Stack<Integer> stack = new Stack<>();
int i=0;
for(int x:pushed)
{
stack.push(x);
while(!stack.isEmpty() && stack.peek()==popped[i])
{
stack.pop();
i++;
}
}
return stack.isEmpty();
}
}
```

**Approach 2: **

**Time Complexity: **O(N) **Space Complexity: **O(1)

```
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
int i=0,j=0;
for(int x:pushed)
{
pushed[i++]=x;
while(i>0 && pushed[i-1]==popped[j])
{
--i;
j++;
}
}
return i==0;
}
}
```