1103. Distribute Candies to People LeetCode Solution

In this guide, you will get 1103. Distribute Candies to People LeetCode Solution with the best time and space complexity. The solution to Distribute Candies to People problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Distribute Candies to People solution in C++
  4. Distribute Candies to People solution in Java
  5. Distribute Candies to People solution in Python
  6. Additional Resources
1103. Distribute Candies to People LeetCode Solution image

Problem Statement of Distribute Candies to People

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].

See also  306. Additive NumberLeetCode Solution

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

Complexity Analysis

  • Time Complexity:
  • Space Complexity:

1103. Distribute Candies to People LeetCode Solution in C++

class Solution {
 public:
  vector<int> distributeCandies(int candies, long n) {
    vector<int> ans(n);
    int rows = (-n + sqrt(n * n + 8 * n * n * candies)) / (2 * n * n);
    int accumN = rows * (rows - 1) * n / 2;

    for (int i = 0; i < n; ++i)
      ans[i] = accumN + rows * (i + 1);

    int givenCandies = (n * n * rows * rows + n * rows) / 2;
    candies -= givenCandies;

    for (int i = 0, lastGiven = rows * n + 1; candies > 0; ++i, ++lastGiven) {
      int actualGiven = min(lastGiven, candies);
      candies -= actualGiven;
      ans[i] += actualGiven;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1103. Distribute Candies to People LeetCode Solution in Java

class Solution {
  public int[] distributeCandies(int candies, int num_people) {
    int[] ans = new int[num_people];
    long c = (long) candies;
    long n = (long) num_people;
    int rows = (int) (-n + Math.sqrt(n * n + 8 * n * n * c)) / (int) (2 * n * n);
    int accumN = rows * (rows - 1) * num_people / 2;

    for (int i = 0; i < n; ++i)
      ans[i] = accumN + rows * (i + 1);

    int givenCandies = (num_people * num_people * rows * rows + num_people * rows) / 2;
    candies -= givenCandies;

    for (int i = 0, lastGiven = rows * num_people + 1; candies > 0; ++i, ++lastGiven) {
      int actualGiven = Math.min(lastGiven, candies);
      candies -= actualGiven;
      ans[i] += actualGiven;
    }

    return ans;
  }
}
// code provided by PROGIEZ

1103. Distribute Candies to People LeetCode Solution in Python

class Solution:
  def distributeCandies(self, candies: int, n: int) -> list[int]:
    ans = [0] * n
    rows = int((-n + (n**2 + 8 * n**2 * candies)**0.5) / (2 * n**2))
    accumN = rows * (rows - 1) * n // 2

    for i in range(n):
      ans[i] = accumN + rows * (i + 1)

    givenCandies = (n**2 * rows**2 + n * rows) // 2
    candies -= givenCandies
    lastGiven = rows * n
    i = 0

    while candies > 0:
      lastGiven += 1
      actualGiven = min(lastGiven, candies)
      candies -= actualGiven
      ans[i] += actualGiven
      i += 1

    return ans
# code by PROGIEZ

Additional Resources

See also  823. Binary Trees With Factors LeetCode Solution

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