1103. Distribute Candies to People

Description

We distribute some number of candies, to a row of n = num_people people in the following way:

We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.

Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.

This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift).

Return an array (of length num_people and sum candies) that represents the final distribution of candies.

Example 1:

Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].

Example 2:

Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation: 
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].

Constraints:

1 <= candies <= 10^9
1 <= num_people <= 1000

My Solution

Source Code

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/**
 * @param {number} candies
 * @param {number} num_people
 * @return {number[]}
 */
let distributeCandies = function(candies, num_people) {
    let arr = new Array(num_people)
    arr.fill(0);
    let sumCandies = 0;
    
    for (let x = 0; x < candies; x++){
        sumCandies += x; 
        if(sumCandies + x+1 >= candies){
            arr[x % num_people] += candies - sumCandies;
            break;
        }
        else
            arr[x % num_people] += x + 1;
            }
    
    return arr;
};

Analysis

My solution is simple. We create an array of num_people and then iterate over the number of candies. The running time will be O(max(candies, num_people)).

One interesting thing I noticed while submitting my solution is that it's a little bit faster to keep a running total of the number of candies given away versus calculating the total using x(x+1)/2. Also, it's significantly faster to do

arr[x % num_people] += x + 1;

rather than

arr[x % num_people] = arr[x % num_people] + x + 1;

I know it's because when you assign through +=, the system only has to access that memory location once versus twice the other way. Normally I always use += but for whatever reason, I forgot this time. I just wanted to point that out here because it was kind of cool seeing just how much of a difference it makes. The first way yields a runtime of 48ms/faster than 97% of other JavaScript solutions versus 60ms/40% faster than other solutions. 12ms does not seem like a huge difference but I imagine when you're dealing with massive datasets and you don't have unlimited resources to process them, that difference will matter.