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

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

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

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

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

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

Rust Code
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impl Solution {
    pub fn validate_stack_sequences(pushed: Vec<i32>, popped: Vec<i32>) -> bool {
        let mut stack = Vec::new();
        let mut i = 0;
        for &num in pushed.iter() {
            stack.push(num);
            while !stack.is_empty() && *stack.last().unwrap() == popped[i] {
                stack.pop();
                i += 1;
            }
        }
        stack.len() == 0
    }
}

JavaScript Code
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/**
 * @param {number[]} pushed
 * @param {number[]} popped
 * @return {boolean}
 */
var validateStackSequences = function (pushed, popped) {
    let stk = [];
    let j = 0;
    for (const v of pushed) {
        stk.push(v);
        while (stk.length && stk[stk.length - 1] == popped[j]) {
            stk.pop();
            ++j;
        }
    }
    return j == pushed.length;
};

C# Code
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public class Solution {
    public bool ValidateStackSequences(int[] pushed, int[] popped) {
        Stack<int> stk = new Stack<int>();
        int j = 0;
        foreach (int x in pushed)
        {
            stk.Push(x);
            while (stk.Count != 0 && stk.Peek() == popped[j]) {
                stk.Pop();
                ++j;
            }
        }
        return stk.Count == 0;
    }
}