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 :



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                        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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