Formatted question description: https://leetcode.ca/all/946.html

# 946. Validate Stack Sequences (Medium)

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.


Note:

1. 0 <= pushed.length == popped.length <= 1000
2. 0 <= pushed[i], popped[i] < 1000
3. pushed is a permutation of popped.
4. pushed and popped have distinct values.

Related Topics:
Stack

## Solution 1.

// OJ: https://leetcode.com/problems/validate-stack-sequences/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
stack<int> s;
int p = 0;
for (int i : pushed) {
s.push(i);
while (s.size() && s.top() == popped[p]) {
s.pop();
++p;
}
}
return s.empty();
}
};


Java

• class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
Stack<Integer> stack = new Stack<Integer>();
int length = pushed.length;
int pushIndex = 0, popIndex = 0;
while (pushIndex < length && popIndex < length) {
stack.push(pushed[pushIndex]);
pushIndex++;
while (!stack.isEmpty() && stack.peek() == popped[popIndex]) {
stack.pop();
popIndex++;
}
}
return stack.isEmpty();
}
}

• // OJ: https://leetcode.com/problems/validate-stack-sequences/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
stack<int> s;
int p = 0;
for (int i : pushed) {
s.push(i);
while (s.size() && s.top() == popped[p]) {
s.pop();
++p;
}
}
return s.empty();
}
};

• class Solution(object):
def validateStackSequences(self, pushed, popped):
"""
:type pushed: List[int]
:type popped: List[int]
:rtype: bool
"""
stack = []
N = len(pushed)
pi = 0
for i in range(N):
if stack and popped[i] == stack[-1]:
stack.pop()
else:
while pi < N and pushed[pi] != popped[i]:
stack.append(pushed[pi])
pi += 1
pi += 1
return not stack