2551 Apply Operations To An Array

Apply Operations to an Array

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.

 

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:

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

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

        for i, k in enumerate(nums):
            if i == len(nums) -1 :
                continue
            if nums[i] == nums[i+1]:
                nums[i] = nums[i] * 2
                nums[i+1] = 0
        
        for k in nums:
            if k != 0:
                arr.append(k)

        for i in range(len(nums) - len(arr)):
            arr.append(0)
        return arr