2460

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

Related Topics:
Array, Simulation

Similar Questions:

• Remove Duplicates from Sorted Array (Easy)
• Move Zeroes (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/apply-operations-to-an-array
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> applyOperations(vector<int>& A) {
        int N = A.size();
        for (int i = 0; i < N - 1; ++i) {
            if (A[i] == A[i + 1]) {
                A[i] *= 2;
                A[i + 1] = 0;
            }
        }
        int i = 0, j = 0;
        for (; j < N; ++j) {
            if (A[j]) A[i++] = A[j];
        }
        for (; i < N; ++i) A[i] = 0;
        return A;
    }
};