LeetCode 2600 - K Items With the Maximum Sum

The problem asks us to determine the maximum possible sum when picking exactly k items from a bag containing items labeled 1, 0, or -1.

leetcodeeasymathgreedy

LeetCode Problem 2600

Difficulty: 🟢 Easy
Topics: Math, Greedy

Solution

Problem Understanding

The problem asks us to determine the maximum possible sum when picking exactly k items from a bag containing items labeled 1, 0, or -1. The bag's composition is given by three non-negative integers: numOnes for the number of 1s, numZeros for the number of 0s, and numNegOnes for the number of -1s. The integer k specifies how many items we must pick.

The expected output is a single integer representing the largest sum obtainable by selecting exactly k items. Positive numbers contribute positively to the sum, zeros contribute nothing, and negative numbers reduce the sum.

The constraints guarantee that all input numbers are non-negative and the sum of all items is at least k, ensuring it is always possible to pick k items. Edge cases include scenarios where k is less than the number of 1s, where k exceeds the sum of 1s and 0s, or where some counts are zero.

This problem is straightforward in its requirement but demands careful handling of the item counts to maximize the sum efficiently.

Approaches

The brute-force approach would involve generating all possible subsets of size k from the bag, calculating the sum for each subset, and then taking the maximum. While this guarantees the correct answer, it is inefficient because the number of combinations grows combinatorially with the total number of items, which can reach up to 150. Calculating all subsets would be infeasible.

The key insight for an optimal solution is that the sum is maximized by picking as many 1s as possible first, then 0s if needed, and only picking -1s if k exceeds the sum of 1s and 0s. This works because 1 > 0 > -1, so the greedy choice of taking the largest available number at each step guarantees the maximum sum. With this approach, we can compute the sum using simple arithmetic without iterating over subsets.

Approach Time Complexity Space Complexity Notes
Brute Force O(C(n, k)) O(k) Generates all subsets of size k and computes sums. Too slow for n up to 150.
Optimal (Greedy) O(1) O(1) Picks items greedily based on value, using simple arithmetic.

Algorithm Walkthrough

  1. Start by picking items labeled 1. Compute how many 1s we can take: ones_taken = min(numOnes, k). Add ones_taken to the sum.
  2. Update k to reflect the remaining items to pick: k -= ones_taken.
  3. Next, pick items labeled 0. The number of zeros we can take is zeros_taken = min(numZeros, k). Adding zeros does not change the sum, so the sum remains the same.
  4. Update k again: k -= zeros_taken.
  5. If there are any remaining items to pick (k > 0), these must be -1s. The sum decreases by k, since each -1 reduces the sum by 1.
  6. Return the final sum.

Why it works: The greedy invariant is that at each step, we pick the available number with the largest value. Since 1 > 0 > -1, this ensures that every pick contributes as much as possible to the sum. The algorithm handles all edge cases because it always respects the available counts of each number.

Python Solution

class Solution:
    def kItemsWithMaximumSum(self, numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int:
        ones_taken = min(numOnes, k)
        k -= ones_taken
        
        zeros_taken = min(numZeros, k)
        k -= zeros_taken
        
        # Remaining items must be -1s
        neg_ones_taken = k
        
        return ones_taken - neg_ones_taken

This Python implementation follows the algorithm directly. First, it maximizes the number of 1s, then accounts for 0s, and finally reduces the sum by any -1s that must be picked to reach k. Each step is a simple arithmetic operation, which makes the solution concise and efficient.

Go Solution

func kItemsWithMaximumSum(numOnes int, numZeros int, numNegOnes int, k int) int {
    onesTaken := min(numOnes, k)
    k -= onesTaken
    
    zerosTaken := min(numZeros, k)
    k -= zerosTaken
    
    // Remaining items are -1
    negOnesTaken := k
    
    return onesTaken - negOnesTaken
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

The Go version mirrors the Python logic, using a helper function min to handle the minimum calculation. The key difference in Go is explicit integer arithmetic and the need for a small helper function instead of using built-in min.

Worked Examples

Example 1

Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 2

Steps:

Step Action k Sum
1 Take min(3, 2) ones k = 2 - 2 = 0 sum = 2
2 No zeros needed k = 0 sum = 2
3 No -1s needed k = 0 sum = 2

Output: 2

Example 2

Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 4

Steps:

Step Action k Sum
1 Take min(3, 4) ones k = 4 - 3 = 1 sum = 3
2 Take min(2, 1) zeros k = 1 - 1 = 0 sum = 3
3 No -1s needed k = 0 sum = 3

Output: 3

Complexity Analysis

Measure Complexity Explanation
Time O(1) All operations are arithmetic and comparisons; no loops over input.
Space O(1) Only a few integer variables are used; no additional data structures.

The algorithm is optimal because it directly calculates the maximum sum using the counts of each item type without enumerating subsets.

Test Cases

# Provided examples
assert Solution().kItemsWithMaximumSum(3, 2, 0, 2) == 2  # Take 2 ones
assert Solution().kItemsWithMaximumSum(3, 2, 0, 4) == 3  # Take 3 ones + 1 zero

# Edge and boundary cases
assert Solution().kItemsWithMaximumSum(0, 0, 0, 0) == 0  # Empty bag, k=0
assert Solution().kItemsWithMaximumSum(5, 0, 0, 5) == 5  # All ones, pick all
assert Solution().kItemsWithMaximumSum(0, 5, 0, 3) == 0  # Only zeros, sum is 0
assert Solution().kItemsWithMaximumSum(0, 0, 5, 3) == -3 # Only -1s, sum negative
assert Solution().kItemsWithMaximumSum(2, 2, 2, 5) == 2  # Mix: 2 ones + 2 zeros + 1 -1
Test Why
k = 2 with 3 ones Tests picking fewer than all ones
k = 4 with 3 ones + 2 zeros Tests need to include zeros
k = 0, empty bag Edge case with zero items
All ones Maximum sum scenario
Only zeros Sum remains 0
Only -1s Sum negative, edge case
Mix Tests combination logic with all types

Edge Cases

The first important edge case is when k is zero, meaning no items are picked. The algorithm correctly returns 0 because no additions or subtractions are performed.

The second edge case is when k is larger than the sum of 1s and 0s, forcing the selection of -1s. This tests whether the subtraction logic is correctly applied, and the algorithm handles this by reducing the sum by the number of -1s picked.

The third edge case is when one or more counts are zero, such as no 1s or no 0s. The algorithm uses min for each type, ensuring it never tries to pick more items than are available, so it handles these scenarios naturally without errors.

These edge cases are critical because they could trip up naive implementations that assume all item types are always present or that k is always smaller than the count of positive items.