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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.
1function validateStackSequences(pushed, popped) {
2 const stack = [];
3 let j = 0;
4 for (const x of pushed) {
5 stack.push(x);
6 while (stack.length && stack[stack.length - 1] === popped[j]) {
7 stack.pop();
8 j++;
9 }
10 }
11 return stack.length === 0;
12}This JavaScript solution employs an array as a stack to check the sequence by pushing items from pushed and popping them when they match the popped sequence. JavaScript arrays efficiently perform as stacks with push and pop operations.
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.
1In this Python code, the same two-pointer approach is leveraged by directly working on indexes of the pushed array similar to using a stack. It checks and manages elements consistency using these pointers.