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The key idea is to use a stack to simulate the push and pop operations. We iterate over each element in the pushed array, pushing them onto the stack. After each push, we check the top of the stack to see if it matches the next element in the popped array. If it does, we pop the stack. By the end of both arrays, if all operations are valid, the stack should be empty.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(n), for the stack storage used during the process.
1import java.util.Stack;
2
3public class Solution {
4 public boolean validateStackSequences(int[] pushed, int[] popped) {
5 Stack<Integer> stack = new Stack<>();
6 int j = 0;
7 for (int x : pushed) {
8 stack.push(x);
9 while (!stack.isEmpty() && stack.peek() == popped[j]) {
10 stack.pop();
11 j++;
12 }
13 }
14 return stack.isEmpty();
15 }
16}In this Java solution, we use a Stack to perform the push and pop operations as described above, iterating over the pushed array and matching with the popped.
In this approach, you can use two pointers to bypass the use of an explicit stack. You iterate over the pushed array with one pointer while another pointer operates over the popped array. Match elements and mimic stack operations conceptually.
Time Complexity: O(n), where n is the length of the arrays.
Space Complexity: O(1), as no extra space is used aside from input arrays.
1Java technique employs the same in-place array modifications. The iteration logic tracks index swapping between pushed and popped arrays simulating a stack effectively without additional memory allocation.