2025-03-01 Daily Challenge

Today I have done leetcode's March LeetCoding Challenge with cpp.

March LeetCoding Challenge 1

Description

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

Solution

class Solution {
public:
  vector<int> applyOperations(vector<int>& nums) {
    for(int i = 0; i < nums.size() - 1; ++i) {
      if(nums[i] == nums[i + 1]) {
        nums[i] *= 2;
        nums[i + 1] = 0;
      }
    }
    int pos = 0;
    for(int i = 0; i < nums.size(); ++i) {
      if(pos >= nums.size()) {
        nums[i] = 0;
      } else {
        while(pos < nums.size() && !nums[pos]) {
          pos += 1;
        }
        nums[i] = pos == nums.size() ? 0 : nums[pos++];
      }
    }
    return nums;
  }
};

// Accepted
// 36/36 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 87.59 % of cpp submissions (13.1 MB)