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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