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
    

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