2460. Apply Operations to an Array

Problem

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

Solution (Java)

class Solution {
    public int[] moveZeroes(int[] nums) {
        if(nums.length == 0 || nums == null) return nums;
        int j = 0;
        for (int i : nums){
            if(i!=0) nums[j++] = i;
        }
        while (j< nums.length){
            nums[j++] = 0;
        }
        return nums;
    }
    public int[] applyOperations(int[] nums) {
        ArrayList<Integer> list = new ArrayList<>();
        for (int i = 0; i < nums.length - 1; i++) {
            if(nums[i] == nums[i+1]){
                nums[i] = nums[i]*2;
                nums[i+1] = 0;
            }
        }
        return moveZeroes(nums);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).