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.
There is a bag that consists of items, each item has a number 1, 0, or -1 written on it.
You are given four non-negative integers numOnes, numZeros, numNegOnes, and k.
The bag initially contains:
numOnes items with 1s written on them.numZeroes items with 0s written on them.numNegOnes items with -1s written on them.We want to pick exactly k items among the available items. Return the maximum possible sum of numbers written on the items.
Example 1:
Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 2
Output: 2
Explanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 2 items with 1 written on them and get a sum in a total of 2.
It can be proven that 2 is the maximum possible sum.
Example 2:
Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 4
Output: 3
Explanation: We have a bag of items with numbers written on them {1, 1, 1, 0, 0}. We take 3 items with 1 written on them, and 1 item with 0 written on it, and get a sum in a total of 3.
It can be proven that 3 is the maximum possible sum.
Constraints:
0 <= numOnes, numZeros, numNegOnes <= 500 <= k <= numOnes + numZeros + numNegOnesProblem summary: There is a bag that consists of items, each item has a number 1, 0, or -1 written on it. You are given four non-negative integers numOnes, numZeros, numNegOnes, and k. The bag initially contains: numOnes items with 1s written on them. numZeroes items with 0s written on them. numNegOnes items with -1s written on them. We want to pick exactly k items among the available items. Return the maximum possible sum of numbers written on the items.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
3 2 0 2
3 2 0 4
maximum-subarray)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2600: K Items With the Maximum Sum
class Solution {
public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
if (numOnes >= k) {
return k;
}
if (numZeros >= k - numOnes) {
return numOnes;
}
return numOnes - (k - numOnes - numZeros);
}
}
// Accepted solution for LeetCode #2600: K Items With the Maximum Sum
func kItemsWithMaximumSum(numOnes int, numZeros int, numNegOnes int, k int) int {
if numOnes >= k {
return k
}
if numZeros >= k-numOnes {
return numOnes
}
return numOnes - (k - numOnes - numZeros)
}
# Accepted solution for LeetCode #2600: K Items With the Maximum Sum
class Solution:
def kItemsWithMaximumSum(
self, numOnes: int, numZeros: int, numNegOnes: int, k: int
) -> int:
if numOnes >= k:
return k
if numZeros >= k - numOnes:
return numOnes
return numOnes - (k - numOnes - numZeros)
// Accepted solution for LeetCode #2600: K Items With the Maximum Sum
impl Solution {
pub fn k_items_with_maximum_sum(
num_ones: i32,
num_zeros: i32,
num_neg_ones: i32,
k: i32,
) -> i32 {
if num_ones > k {
return k;
}
if num_ones + num_zeros > k {
return num_ones;
}
num_ones - (k - num_ones - num_zeros)
}
}
// Accepted solution for LeetCode #2600: K Items With the Maximum Sum
function kItemsWithMaximumSum(
numOnes: number,
numZeros: number,
numNegOnes: number,
k: number,
): number {
if (numOnes >= k) {
return k;
}
if (numZeros >= k - numOnes) {
return numOnes;
}
return numOnes - (k - numOnes - numZeros);
}
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.