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:

0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed is a permutation of popped.
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

946 - Validate Stack Sequences

946. Validate Stack Sequences (Medium)

Solution 1.

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