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Java-based LeetCode algorithm problem solutions, regularly updated
2460\. Apply Operations to an Array
Easy
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`