Class Solution


  • public class Solution
    extends Object
    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
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • applyOperations

        public int[] applyOperations​(int[] nums)