# Leetcode - Build an Array With Stack Operations Solution

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.

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] <= n`

`1 <= n <= 100`

`target`

is strictly increasing.

## Solution in Python

```
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
stack = []
target = [0]+target
for i in range(1,len(target)):
diff = target[i]-target[i-1]-1
stack.extend(["Push", "Pop"]*diff)
stack.append("Push")
return stack
```