# [Leet Code] Build an Array With Stack Operations

Leetcode: https://leetcode.com/problems/build-an-array-with-stack-operations/

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 = 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:

`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