###### Validate Stack Sequences | February Leetcoding challenge 2021 | Day 26 Manas Sinha
Developer | Designer

## Validate Stack Sequences

`PROBLEM  STATEMENT:`
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
```strong>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.

## Explanation

The simplest way to do it is to just have a stack and see if the given operations are feasible.We traverse through the `pushed` array and also we have a pointer `i` that moves over the popped array. If the stack is empty or the top element of the stack is not equal to `popped[i]` we push elements from `pushed` array to the stack. Else we pop the element from the stack and increment `i` until the stack is empty or the top element is not equal to `popped[i].` At the end if stack is empty that means the given operations are feasible.
`ALGORITHM: `
• Maintain a stack.
• Traverse the pushed array from left to right and have a variable i initialised to zero.
• Push the current element to the stack.
• While stack is not empty and st.top() == popped[i], pop the top element of stack and increment i.
• After traversing the entire pushed array return true if the stack is empty else return false.
• Time complexity : O(n)
• Space complexity : O(n) ## C++

``````class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
stack<int> st;

int j=0;
for(int i=0;i<pushed.size();++i){
st.push(pushed[i]);
while(!st.empty() && st.top() == popped[j]){
st.pop();
++j;
}
}
return st.empty();
}
};``````

## PYTHON

``````class Solution:
def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
st = []
i=0
for num in pushed:
st.append(num)
while len(st) and st[-1] == popped[i]:
st.pop()
i += 1
return not len(st)``````