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946. Validate Stack Sequences

Description

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.

Solutions

Solution 1: Stack Simulation

We iterate through the $\textit{pushed}$ array. For the current element $x$ being iterated, we push it into the stack $\textit{stk}$. Then, we check if the top element of the stack is equal to the next element to be popped in the $\textit{popped}$ array. If they are equal, we pop the top element from the stack and increment the index $i$ of the next element to be popped in the $\textit{popped}$ array. Finally, if all elements can be popped in the order specified by the $\textit{popped}$ array, return $\textit{true}$; otherwise, return $\textit{false}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the $\textit{pushed}$ array.

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class Solution:
    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
        stk = []
        i = 0
        for x in pushed:
            stk.append(x)
            while stk and stk[-1] == popped[i]:
                stk.pop()
                i += 1
        return i == len(popped)

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class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Deque<Integer> stk = new ArrayDeque<>();
        int i = 0;
        for (int x : pushed) {
            stk.push(x);
            while (!stk.isEmpty() && stk.peek() == popped[i]) {
                stk.pop();
                ++i;
            }
        }
        return i == popped.length;
    }
}

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class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        stack<int> stk;
        int i = 0;
        for (int x : pushed) {
            stk.push(x);
            while (stk.size() && stk.top() == popped[i]) {
                stk.pop();
                ++i;
            }
        }
        return i == popped.size();
    }
};

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func validateStackSequences(pushed []int, popped []int) bool {
    stk := []int{}
    i := 0
    for _, x := range pushed {
        stk = append(stk, x)
        for len(stk) > 0 && stk[len(stk)-1] == popped[i] {
            stk = stk[:len(stk)-1]
            i++
        }
    }
    return i == len(popped)
}

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function validateStackSequences(pushed: number[], popped: number[]): boolean {
    const stk: number[] = [];
    let i = 0;
    for (const x of pushed) {
        stk.push(x);
        while (stk.length && stk.at(-1)! === popped[i]) {
            stk.pop();
            i++;
        }
    }
    return i === popped.length;
}

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impl Solution {
    pub fn validate_stack_sequences(pushed: Vec<i32>, popped: Vec<i32>) -> bool {
        let mut stk: Vec<i32> = Vec::new();
        let mut i = 0;
        for &x in &pushed {
            stk.push(x