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.
Move from brute-force thinking to an efficient approach using math strategy.
You are given two integers, n and k.
An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k.
Return the minimum possible sum of a k-avoiding array of length n.
Example 1:
Input: n = 5, k = 4 Output: 18 Explanation: Consider the k-avoiding array [1,2,4,5,6], which has a sum of 18. It can be proven that there is no k-avoiding array with a sum less than 18.
Example 2:
Input: n = 2, k = 6 Output: 3 Explanation: We can construct the array [1,2], which has a sum of 3. It can be proven that there is no k-avoiding array with a sum less than 3.
Constraints:
1 <= n, k <= 50Problem summary: You are given two integers, n and k. An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k. Return the minimum possible sum of a k-avoiding array of length n.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
5 4
2 6
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2829: Determine the Minimum Sum of a k-avoiding Array
class Solution {
public int minimumSum(int n, int k) {
int s = 0, i = 1;
boolean[] vis = new boolean[n + k + 1];
while (n-- > 0) {
while (vis[i]) {
++i;
}
if (k >= i) {
vis[k - i] = true;
}
s += i++;
}
return s;
}
}
// Accepted solution for LeetCode #2829: Determine the Minimum Sum of a k-avoiding Array
func minimumSum(n int, k int) int {
s, i := 0, 1
vis := make([]bool, n+k+1)
for ; n > 0; n-- {
for vis[i] {
i++
}
if k >= i {
vis[k-i] = true
}
s += i
i++
}
return s
}
# Accepted solution for LeetCode #2829: Determine the Minimum Sum of a k-avoiding Array
class Solution:
def minimumSum(self, n: int, k: int) -> int:
s, i = 0, 1
vis = set()
for _ in range(n):
while i in vis:
i += 1
vis.add(k - i)
s += i
i += 1
return s
// Accepted solution for LeetCode #2829: Determine the Minimum Sum of a k-avoiding Array
impl Solution {
pub fn minimum_sum(n: i32, k: i32) -> i32 {
let (mut s, mut i) = (0, 1);
let mut vis = std::collections::HashSet::new();
for _ in 0..n {
while vis.contains(&i) {
i += 1;
}
vis.insert(k - i);
s += i;
i += 1;
}
s
}
}
// Accepted solution for LeetCode #2829: Determine the Minimum Sum of a k-avoiding Array
function minimumSum(n: number, k: number): number {
let s = 0;
let i = 1;
const vis: boolean[] = Array(n + k + 1).fill(false);
while (n--) {
while (vis[i]) {
++i;
}
if (k >= i) {
vis[k - i] = true;
}
s += i++;
}
return s;
}
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: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
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.