# Build an Array With Stack Operations

Given an array `target` and an integer `n`. In each iteration, you will read a number from  `list = {1,2,3..., n}`.

Build the `target` array using the following operations:

* **Push**: Read a new element from the beginning `list`, and push it in the array.
* **Pop**: delete the last element of the array.
* If the target array is already built, stop reading more elements.

You are guaranteed that the target array is strictly increasing, only containing numbers between 1 to `n` inclusive.

Return the operations to build the target array.

You are guaranteed that the answer is unique.

**Example 1:**

```
Input: target = [1,3], n = 3
Output: ["Push","Push","Pop","Push"]
Explanation: 
Read number 1 and automatically push in the array -> [1]
Read number 2 and automatically push in the array then Pop it -> [1]
Read number 3 and automatically push in the array -> [1,3]
```

**Example 2:**

```
Input: target = [1,2,3], n = 3
Output: ["Push","Push","Push"]
```

**Example 3:**

```
Input: target = [1,2], n = 4
Output: ["Push","Push"]
Explanation: You only need to read the first 2 numbers and stop.
```

**Example 4:**

```
Input: target = [2,3,4], n = 4
Output: ["Push","Pop","Push","Push","Push"]
```

**Constraints:**

* `1 <= target.length <= 100`
* `1 <= target[i] <= 100`
* `1 <= n <= 100`
* `target` is strictly increasing.

```java
class Solution {
    public List<String> buildArray(int[] target, int n) {
        List<String> ans = new ArrayList<>();
        int index = 0;
        for (int i = 1; i <= n; i++) {
            if (index == target.length)
                break;
            else if (i == target[index]) {
                index++;
                ans.add("Push");
            } else {
                ans.add("Push");
                ans.add("Pop");
            }
        }
        return ans;
    }
}
```
