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