Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
Alice has n candies, where the ith candy is of type candyType[i]. Alice noticed that she started to gain weight, so she visited a doctor.
The doctor advised Alice to only eat n / 2 of the candies she has (n is always even). Alice likes her candies very much, and she wants to eat the maximum number of different types of candies while still following the doctor's advice.
Given the integer array candyType of length n, return the maximum number of different types of candies she can eat if she only eats n / 2 of them.
Example 1:
Input: candyType = [1,1,2,2,3,3] Output: 3 Explanation: Alice can only eat 6 / 2 = 3 candies. Since there are only 3 types, she can eat one of each type.
Example 2:
Input: candyType = [1,1,2,3] Output: 2 Explanation: Alice can only eat 4 / 2 = 2 candies. Whether she eats types [1,2], [1,3], or [2,3], she still can only eat 2 different types.
Example 3:
Input: candyType = [6,6,6,6] Output: 1 Explanation: Alice can only eat 4 / 2 = 2 candies. Even though she can eat 2 candies, she only has 1 type.
Constraints:
n == candyType.length2 <= n <= 104n is even.-105 <= candyType[i] <= 105Problem summary: Alice has n candies, where the ith candy is of type candyType[i]. Alice noticed that she started to gain weight, so she visited a doctor. The doctor advised Alice to only eat n / 2 of the candies she has (n is always even). Alice likes her candies very much, and she wants to eat the maximum number of different types of candies while still following the doctor's advice. Given the integer array candyType of length n, return the maximum number of different types of candies she can eat if she only eats n / 2 of them.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map
[1,1,2,2,3,3]
[1,1,2,3]
[6,6,6,6]
minimum-number-of-operations-to-satisfy-conditions)check-if-grid-satisfies-conditions)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #575: Distribute Candies
class Solution {
public int distributeCandies(int[] candyType) {
Set<Integer> s = new HashSet<>();
for (int c : candyType) {
s.add(c);
}
return Math.min(candyType.length >> 1, s.size());
}
}
// Accepted solution for LeetCode #575: Distribute Candies
func distributeCandies(candyType []int) int {
s := hashset.New()
for _, c := range candyType {
s.Add(c)
}
return min(len(candyType)>>1, s.Size())
}
# Accepted solution for LeetCode #575: Distribute Candies
class Solution:
def distributeCandies(self, candyType: List[int]) -> int:
return min(len(candyType) >> 1, len(set(candyType)))
// Accepted solution for LeetCode #575: Distribute Candies
struct Solution;
use std::collections::HashSet;
impl Solution {
fn distribute_candies(candies: Vec<i32>) -> i32 {
let n = candies.len();
let mut hs: HashSet<i32> = HashSet::new();
for val in candies {
hs.insert(val);
}
usize::min(n / 2, hs.len()) as i32
}
}
#[test]
fn test() {
let candies = vec![1, 1, 2, 2, 3, 3];
assert_eq!(Solution::distribute_candies(candies), 3);
let candies = vec![1, 1, 2, 3];
assert_eq!(Solution::distribute_candies(candies), 2);
}
// Accepted solution for LeetCode #575: Distribute Candies
function distributeCandies(candyType: number[]): number {
const s = new Set(candyType);
return Math.min(s.size, candyType.length >> 1);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.