946. Validate Stack Sequences

MediumStackArraySimulation
Leetcode Link

Problem Description

In this problem, we are given two integer arrays named pushed and popped respectively. Each array contains distinct values. The aim is to determine whether the sequence of integers in the popped array could be the result of a certain sequence of push and pop operations on an initially empty stack by only using the integers given in the pushed array, or not. To clarify, we want to know if we can push the numbers from the pushed array onto the stack in the order they are given, and then pop them off to match the order they appear in the popped array. We are to return true if this is possible, or false otherwise.

Intuition

The intuition behind the solution is to simulate the push and pop operations of a stack using the given pushed and popped arrays.

  1. We traverse the pushed array and push each element onto the stack.
  2. After each push, we check if the stack's top element is equal to the current element at index j of the popped array.
  3. If it is, we pop that element from the stack and increase the j index to move to the next element in the popped array.
  4. We continue this process until either the stack is empty or the stack's top element is not equal to the popped[j].
  5. The stack simulates the push and pop operations, and the index j keeps track of the number of elements correctly popped from the stack.
  6. If, after all push operations, the j index equals the length of the pushed array, it means all elements were popped in the popped order, so we return true.
  7. If j is not equal to the length of pushed after the simulation, it means the popped sequence is not possible with the given push and pop operations, thus we return false.

The key insight is to recognize that the stack allows us to reorder elements as they are pushed and popped, but the popped array creates a constraint on the order in which elements must be popped. The solution algorithm effectively tries to match these constraints by simulating the stack operations.

Learn more about Stack patterns.

Solution Approach

The solution uses a simple stack data structure and a loop to simulate the stack operations. Below is a step-by-step explanation of the solution:

  1. Initialize an empty list stk that will act as our stack, and a variable j to keep track of the current index in the popped array.

    1j, stk = 0, []
  2. Begin a loop to iterate over each value v in the pushed array.

    1for v in pushed:
  3. Inside the loop, push the current element v onto the stack (stk).

    1stk.append(v)
  4. After the push operation, enter a while loop that will run as long as the stack is not empty and the top element of the stack is equal to the element at the current index j of the popped array. This checks whether we can pop the top element to match the next element to be popped according to the popped sequence.

    1while stk and stk[-1] == popped[j]:
  5. If the top element on the stack is the same as popped[j], pop it from the stack and increment j to check against the next element in the popped array. This simulates the pop operation and progresses the index as matching elements are found and popped.

    1stk.pop()
    2j += 1
  6. After the loop has iterated through all elements in pushed, check if the index j is now the same as the length of the pushed array. If it is, it means all pushed elements have been successfully matched with the popped elements in the right order, so the function returns true.

    1return j == len(pushed)
  7. If j is not equal to len(pushed), it means not all elements could be matched, so the sequence of operations is not possible, and the function returns false.

In this solution, the stack data structure is essential since it allows elements to be accessed and removed in a last-in, first-out manner, which is exactly what we need to simulate the push and pop operations. The while loop within the for loop ensures that as soon as an element is pushed onto the stack, it is checked to see if it can be immediately popped off to follow the popped sequence. This continues the checking and popping of elements without pausing the push operations, effectively interleaving the necessary operations to achieve the end goal.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider the following pushed and popped arrays:

1pushed = [1, 2, 3, 4, 5]
2popped = [4, 5, 3, 2, 1]

Follow the steps outlined in the solution approach:

  1. Initialize an empty stack stk and j index to 0.

  2. Traverse the pushed array:

    • v = 1: Push 1 into stk. stk = [1]. Since stk[-1] (1) is not equal to popped[0] (4), we continue without popping.

    • v = 2: Push 2 into stk. stk = [1, 2]. Since stk[-1] (2) is not equal to popped[0] (4), we continue without popping.

    • v = 3: Push 3 into stk. stk = [1, 2, 3]. Since stk[-1] (3) is not equal to popped[0] (4), we continue without popping.

    • v = 4: Push 4 into stk. stk = [1, 2, 3, 4]. Since stk[-1] (4) is equal to popped[0] (4), we pop from stk and increment j. After popping, stk = [1, 2, 3] and j = 1.

  3. Continue iterating and check the top against popped[j]. The new values of j and popped we compare against are as follows:

    • Now stk[-1] is 3 and popped[j] is 5. They are not the same, so we move to push the next element from pushed.

    • v = 5: Push 5 into stk. stk = [1, 2, 3, 5]. Now, stk[-1] is 5, which matches popped[j] (5), so we pop 5 from stk, and now stk = [1, 2, 3] and increment j to 2.

    • Now stk[-1] is 3, which matches popped[j] (3), so we pop 3 from stk, and now stk = [1, 2] and increment j to 3.

    • Continue this comparison and popping process:

      • Pop 2 from stk because it matches popped[j] (2), and now stk = [1] and increment j to 4.
      • Pop 1 from stk because it matches popped[j] (1), and now stk is empty and increment j to 5.
  4. Since we've finished iterating through pushed, we check if j is now equal to the length of pushed (which is 5):

    • Indeed, j is 5, which is the length of pushed, suggesting that all elements were popped in the popped order.

Hence, for the given pushed and popped arrays, the simulation shows that it is possible to match the push and pop sequence, and the function should return true.

Solution Implementation

1from typing import List
2
3class Solution:
4    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
5        # Initialize index to track the position in the 'popped' sequence
6        pop_index = 0
7      
8        # Initialize an empty list to simulate stack operations
9        stack = []
10      
11        # Iterate through each value in the 'pushed' sequence
12        for value in pushed:
13            # Push the current value onto the stack
14            stack.append(value)
15          
16            # Check if the top of the stack matches the current value in 'popped'
17            # If it does, pop from the stack and advance the index in 'popped'
18            while stack and stack[-1] == popped[pop_index]:
19                stack.pop()
20                pop_index += 1
21      
22        # If the pop_index is equal to the length of 'pushed', all elements were popped
23        # in the correct order, hence we return True. Otherwise, return False.
24        return pop_index == len(pushed)
25
26# Example usage:
27# solution = Solution()
28# result = solution.validateStackSequences([1, 2, 3, 4, 5], [4, 5, 3, 2, 1])  # True
29# result = solution.validateStackSequences([1, 2, 3, 4, 5], [4, 3, 5, 1, 2])  # False
30
1class Solution {
2  
3    // Method to validate stack sequences using provided pushed and popped array sequences
4    public boolean validateStackSequences(int[] pushed, int[] popped) {
5        // Use a Deque as a stack for simulating the push and pop operations
6        Deque<Integer> stack = new ArrayDeque<>();
7      
8        // Index for keeping track of the position in the popped sequence
9        int popIndex = 0;
10      
11        // Iterate over the pushed sequence to simulate the stack operations
12        for (int num : pushed) {
13            // Push the current number onto the stack
14            stack.push(num);
15          
16            // Keep popping from the stack if the top of the stack matches the current
17            // number in the popped sequence
18            while (!stack.isEmpty() && stack.peek() == popped[popIndex]) {
19                stack.pop(); // Pop from stack
20                popIndex++;  // Move to the next index in the popped sequence
21            }
22        }
23      
24        // If all elements were successfully popped in the correct sequence, the popIndex
25        // should be equal to the length of the pushed sequence
26        return popIndex == pushed.length;
27    }
28}
29
1#include <vector>
2#include <stack>
3
4class Solution {
5public:
6    // This function checks whether a given stack push and pop sequence is valid
7    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
8      
9        // Initialize an empty stack
10        stack<int> stack;
11      
12        // The index for the popped sequence
13        int popIndex = 0;
14      
15        // Iterate over each value in the pushed sequence
16        for (int value : pushed) {
17            // Push the current value onto the stack
18            stack.push(value);
19          
20            // While the stack is not empty and the top of the stack is equal to
21            // the next value in the popped sequence
22            while (!stack.empty() && stack.top() == popped[popIndex]) {
23                // Pop the top value off the stack
24                stack.pop();
25                // Increment the pop sequence index
26                popIndex++;
27            }
28        }
29
30        // If the popIndex equals the size of the pushed sequence, 
31        // then all elements were popped in the correct order
32        // Hence, the sequences are valid and the method returns true.
33        // If elements remain in the stack or the popIndex does not reach the end, 
34        // then the sequences are not valid and the method returns false.
35        return popIndex == pushed.size();
36    }
37};
38
1function validateStackSequences(pushed: number[], popped: number[]): boolean {
2    // Initialize an empty array to simulate stack operations
3    const stack: number[] = [];
4    // Index to keep track of the position in the 'popped' sequence
5    let popIndex = 0;
6  
7    // Iterate through each value in the 'pushed' sequence
8    for (const value of pushed) {
9        // Push the current value onto the stack
10        stack.push(value);
11        // Continue popping from the stack if the top element equals 
12        // the next element in the 'popped' sequence
13        while (stack.length && stack[stack.length - 1] === popped[popIndex]) {
14            stack.pop(); // Remove the top element from the stack
15            popIndex++; // Move to the next index in the 'popped' sequence
16        }
17    }
18  
19    // If all elements were successfully popped in the 'popped' sequence order,
20    // then the popIndex should match the length of the 'pushed' array
21    return popIndex === pushed.length;
22}
23

Time and Space Complexity

Time Complexity: The time complexity of the code is O(n), where n is the length of the pushed and popped lists. The for loop runs for each element in pushed, and while each element is pushed onto the stack once, the while loop may not necessarily run for every push operation as it depends on the match with the popped sequence. However, each element will be popped at most once. Therefore, each element from pushed is involved in constant-time push and pop operations on the stack, making the time complexity linear.

Space Complexity: The space complexity of the code is O(n). In the worst case, all the elements from the pushed list could be stacked up in the stk array (which happens when the popped sequence corresponds to reversing the pushed sequence), which would take up a space proportional to the number of elements in pushed.

Learn more about how to find time and space complexity quickly using problem constraints.


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