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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given an array happiness of length n, and a positive integer k.
There are n children standing in a queue, where the ith child has happiness value happiness[i]. You want to select k children from these n children in k turns.
In each turn, when you select a child, the happiness value of all the children that have not been selected till now decreases by 1. Note that the happiness value cannot become negative and gets decremented only if it is positive.
Return the maximum sum of the happiness values of the selected children you can achieve by selecting k children.
Example 1:
Input: happiness = [1,2,3], k = 2 Output: 4 Explanation: We can pick 2 children in the following way: - Pick the child with the happiness value == 3. The happiness value of the remaining children becomes [0,1]. - Pick the child with the happiness value == 1. The happiness value of the remaining child becomes [0]. Note that the happiness value cannot become less than 0. The sum of the happiness values of the selected children is 3 + 1 = 4.
Example 2:
Input: happiness = [1,1,1,1], k = 2 Output: 1 Explanation: We can pick 2 children in the following way: - Pick any child with the happiness value == 1. The happiness value of the remaining children becomes [0,0,0]. - Pick the child with the happiness value == 0. The happiness value of the remaining child becomes [0,0]. The sum of the happiness values of the selected children is 1 + 0 = 1.
Example 3:
Input: happiness = [2,3,4,5], k = 1 Output: 5 Explanation: We can pick 1 child in the following way: - Pick the child with the happiness value == 5. The happiness value of the remaining children becomes [1,2,3]. The sum of the happiness values of the selected children is 5.
Constraints:
1 <= n == happiness.length <= 2 * 1051 <= happiness[i] <= 1081 <= k <= nProblem summary: You are given an array happiness of length n, and a positive integer k. There are n children standing in a queue, where the ith child has happiness value happiness[i]. You want to select k children from these n children in k turns. In each turn, when you select a child, the happiness value of all the children that have not been selected till now decreases by 1. Note that the happiness value cannot become negative and gets decremented only if it is positive. Return the maximum sum of the happiness values of the selected children you can achieve by selecting k children.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Greedy
[1,2,3] 2
[1,1,1,1] 2
[2,3,4,5] 1
maximum-candies-allocated-to-k-children)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3075: Maximize Happiness of Selected Children
class Solution {
public long maximumHappinessSum(int[] happiness, int k) {
Arrays.sort(happiness);
long ans = 0;
for (int i = 0, n = happiness.length; i < k; ++i) {
int x = happiness[n - i - 1] - i;
ans += Math.max(x, 0);
}
return ans;
}
}
// Accepted solution for LeetCode #3075: Maximize Happiness of Selected Children
func maximumHappinessSum(happiness []int, k int) (ans int64) {
sort.Ints(happiness)
for i := 0; i < k; i++ {
x := happiness[len(happiness)-i-1] - i
ans += int64(max(x, 0))
}
return
}
# Accepted solution for LeetCode #3075: Maximize Happiness of Selected Children
class Solution:
def maximumHappinessSum(self, happiness: List[int], k: int) -> int:
happiness.sort(reverse=True)
ans = 0
for i, x in enumerate(happiness[:k]):
x -= i
ans += max(x, 0)
return ans
// Accepted solution for LeetCode #3075: Maximize Happiness of Selected Children
impl Solution {
pub fn maximum_happiness_sum(mut happiness: Vec<i32>, k: i32) -> i64 {
happiness.sort_unstable_by(|a, b| b.cmp(a));
let mut ans: i64 = 0;
for i in 0..(k as usize) {
let x = happiness[i] as i64 - i as i64;
if x > 0 {
ans += x;
}
}
ans
}
}
// Accepted solution for LeetCode #3075: Maximize Happiness of Selected Children
function maximumHappinessSum(happiness: number[], k: number): number {
happiness.sort((a, b) => b - a);
let ans = 0;
for (let i = 0; i < k; ++i) {
const x = happiness[i] - i;
ans += Math.max(x, 0);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
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: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.