# Stack Sequence - Google Top Interview Questions

### Problem Statement :

```Given a list of distinct integers pushes, and another list of integers pops, return whether this is a valid sequence of stack push and pop actions.

Constraints

n ≤ 100,000 where n is the length of pushes

m ≤ 100,000 where m is the length of pops

Example 1

Input

pushes = [0, 1, 4, 6, 8]

pops = [1, 0, 8, 6, 4]

Output

True

Explanation

We can first push [0, 1], then pop both off. Then push [4, 6, 8] and then pop them all off.

Example 2

Input

pushes = [1, 2, 3, 4]

pops = [4, 1, 2, 3]

Output

False

Explanation

This is not valid since 3 was pushed after 1 but is popped earlier.```

### Solution :

```                        ```Solution in C++ :

bool solve(vector<int>& pushes, vector<int>& pops) {
stack<int> st;
int i = 0, j = 0;
while (i < pops.size()) {
if (st.empty())
st.push(pushes[j++]);
else {
if (st.top() == pops[i]) {
st.pop();
i++;
} else
st.push(pushes[j++]);
}
}
return st.size() == 0;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[] pushes, int[] pops) {
Stack<Integer> stack = new Stack();
int p = 0;
for (int i = 0; i < pushes.length; i++) {
stack.push(pushes[i]);
while (!stack.isEmpty() && stack.peek() == pops[p]) {
p++;
stack.pop();
}
}
return stack.isEmpty();
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, pushes, pops):
stack = []
i = 0
l = 0
for n in pushes:
stack.append(n)
l += 1
while l and stack[-1] == pops[i]:
stack.pop()
i += 1
l -= 1
return not l```
```

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