1441. Build an Array With Stack Operations

Explanation

To solve this problem, we can simulate the stack operations based on the given rules. We iterate through the numbers from 1 to n and check if the current number is in the target array. If it is, we push it onto the stack. If it's not, we push the number onto the stack and immediately pop it to simulate not using it. We continue this process until the stack matches the target array or we reach the end of the numbers from 1 to n.

Algorithm:

Initialize an empty stack to store the elements.
Iterate through the numbers from 1 to n:
- If the current number is in the target array, push it onto the stack.
- If the current number is not in the target array, push it onto the stack and immediately pop it.
- Check if the stack matches the target array. If it does, stop the iterations.
Return the stack operations as the output.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the value of the input parameter n.

Space Complexity:

The space complexity of this algorithm is O(n) since the stack can potentially store all n elements.

View Leetcode Problem

class Solution {
    public List<String> buildArray(int[] target, int n) {
        List<String> result = new ArrayList<>();
        Stack<Integer> stack = new Stack<>();
        
        int index = 0;
        for (int i = 1; i <= n && index < target.length; i++) {
            stack.push(i);
            result.add("Push");
            
            if (stack.peek() != target[index]) {
                stack.pop();
                result.add("Pop");
            } else {
                index++;
            }
        }
        
        return result;
    }
}

Code Editor (Testing phase)

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