Description
https://leetcode.com/problems/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.
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.
Explanation
Track the numbers added and compare with the target array to decide whether “push” or “pop” operation to use.
Python Solution
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
results = []
numbers = []
for i in range(1, n + 1):
if numbers == target:
return results
numbers.append(i)
if numbers[len(numbers) - 1] == target[len(numbers) - 1]:
results.append("Push")
else:
results.append("Push")
results.append("Pop")
numbers.pop()
return results
- Time Complexity: O(N).
- Space Complexity: O(N).