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Given two integer arrays pushed and popped each with distinct values, return true if this could have been the result of a sequence of push and pop operations on an initially empty stack, or false otherwise.

 

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.

 

Constraints:

  • 1 <= pushed.length <= 1000
  • 0 <= pushed[i] <= 1000
  • All the elements of pushed are unique.
  • popped.length == pushed.length
  • popped is a permutation of pushed.

Companies:
Amazon, Microsoft, tiktok

Related Topics:
Array, Stack, Simulation

Solution 1.

// OJ: https://leetcode.com/problems/validate-stack-sequences/
// Author: github.com/lzl124631x
// 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();
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/validate-stack-sequences/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) extra space.
class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        int N = pushed.size(), i = 0, j = 0;
        for (int n : pushed) {
            pushed[i++] = n;
            while (i && pushed[i - 1] == popped[j]) --i, ++j;
        }
        return i == 0;
    }
};