[LeetCode] 2460. Apply Operations to an Array (Python)

 

 

난이도: Easy

Problem Description

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You are given a 0-indexed array nums of size n consisting of non-negative integers.

 

You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:

• If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.

After performing all the operations, shift all the 0's to the end of the array.

• For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].

Return the resulting array.

 

Note that the operations are applied sequentially, not all at once.

 

Problem Example

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Example 1:

Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].

Example 2:

Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.

 

Constraints

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• 2 <= nums.length <= 2000
• 0 <= nums[i] <= 1000

 

✏️ Solution

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class Solution:
    def applyOperations(self, nums: List[int]) -> List[int]:
        n = len(nums)
        answer = []

        for i in range(n):
            if i+1 < n and nums[i] == nums[i+1]:
                nums[i] *= 2
                nums[i+1] = 0
            
            if nums[i] != 0:
                answer.append(nums[i])
        
        for _ in range(n-len(answer)):
            answer.append(0)

        return answer

 

문제에서 주어진대로 `nums`의 요소들을 확인하여 `nums[i] == nums[i+1]`이면 `nums[i]`에 2를 곱하고 `nums[i+1]`은 0으로 만들어 주었다.

 

이후 현재 `nums[i]`가 0이 아니면 `answer` 리스트에 추가해주었다.

 

이렇게 하면 모든 요소를 확인했을 때, `answer`에는 0이 아닌 요소들만 들어가 있다.

 

예를 들어 `nums`가 [1,0,2,0,0,1]이라고 한다면 `answer`에는 [1,2,1]로 채워진다.

 

이제 nums의 길이 - answer의 길이 만큼 0을 answer에 채워주면 된다.

 

마지막에는 answer를 return 하고 정답을 맞을 수 있었다.

 

⚙️ Runtime & Memory

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문제: 2460. Apply Operations to an Array

 

깃허브: github

algorithmPractice/LeetCode/2551-apply-operations-to-an-array at main · laewonJeong/algorithmPractice