# [Leet Code] Build an Array With Stack Operations

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Problem:

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 = 3Output: ["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 = 3Output: ["Push","Push","Push"]`

Example 3:

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

Example 4:

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

Constraints:

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

Solution:

`class Solution(object):    def buildArray(self, target, n):        """        :type target: List[int]        :type n: int        :rtype: List[str]        """        result_list = []        for i in range(1,n+1):            if i>target[len(target)-1]:                return result_list            result_list.append("Push")            if i not in target:                result_list.append("Pop")        return result_list`