Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Build confidence with an intuition-first walkthrough focused on math fundamentals.
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:
Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
7 4
10 3
distribute-money-to-maximum-children)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1103: Distribute Candies to People
class Solution {
public int[] distributeCandies(int candies, int num_people) {
int[] ans = new int[num_people];
for (int i = 0; candies > 0; ++i) {
ans[i % num_people] += Math.min(candies, i + 1);
candies -= Math.min(candies, i + 1);
}
return ans;
}
}
// Accepted solution for LeetCode #1103: Distribute Candies to People
func distributeCandies(candies int, num_people int) []int {
ans := make([]int, num_people)
for i := 0; candies > 0; i++ {
ans[i%num_people] += min(candies, i+1)
candies -= min(candies, i+1)
}
return ans
}
# Accepted solution for LeetCode #1103: Distribute Candies to People
class Solution:
def distributeCandies(self, candies: int, num_people: int) -> List[int]:
ans = [0] * num_people
i = 0
while candies:
ans[i % num_people] += min(candies, i + 1)
candies -= min(candies, i + 1)
i += 1
return ans
// Accepted solution for LeetCode #1103: Distribute Candies to People
struct Solution;
impl Solution {
fn distribute_candies(mut candies: i32, num_people: i32) -> Vec<i32> {
let mut i = 0;
let n = num_people as usize;
let mut res: Vec<i32> = vec![0; n];
let mut gift = 0;
while candies > 0 {
gift = i32::min(gift + 1, candies);
res[i] += gift;
candies -= gift;
i = (i + 1) % n;
}
res
}
}
#[test]
fn test() {
let candies = 7;
let num_people = 4;
let res = vec![1, 2, 3, 1];
assert_eq!(Solution::distribute_candies(candies, num_people), res);
let candies = 10;
let num_people = 3;
let res = vec![5, 2, 3];
assert_eq!(Solution::distribute_candies(candies, num_people), res);
}
// Accepted solution for LeetCode #1103: Distribute Candies to People
function distributeCandies(candies: number, num_people: number): number[] {
const ans: number[] = Array(num_people).fill(0);
for (let i = 0; candies > 0; ++i) {
ans[i % num_people] += Math.min(candies, i + 1);
candies -= Math.min(candies, i + 1);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.
Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.