K Items With the Maximum Sum - Practice Coding Problems

K Items With the Maximum Sum - Problem

Math Greedy Easy

You're a treasure hunter exploring a magical bag filled with enchanted items! This mystical bag contains items with special numbers written on them:

Golden items with +1 written on them (worth 1 point each)
Silver items with 0 written on them (worth 0 points each)
Cursed items with -1 written on them (lose 1 point each)

You have exactly k moves to pick items from the bag. Your goal is to maximize your total score by choosing the best possible combination of items.

Given: The count of each type of item (numOnes, numZeros, numNegOnes) and the number of items you must pick (k).

Return: The maximum possible sum you can achieve.

Input & Output

example_1.py — Basic Case

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

› Output: 3

💡 Note: We have 3 items with +1, 2 items with 0, and no items with -1. We need to pick 4 items. The optimal strategy is to pick all 3 items with +1 (contributing +3) and 1 item with 0 (contributing 0), for a total sum of 3.

example_2.py — Must Take Negative

$ Input: numOnes = 3, numZeros = 2, numNegOnes = 0, k = 6

› Output: 3

💡 Note: We need to pick 6 items but only have 5 items total (3+2+0=5). This is actually impossible, but assuming we can only pick available items, we take all 5 items: 3×(+1) + 2×(0) = 3.

example_3.py — Forced Negatives

$ Input: numOnes = 2, numZeros = 1, numNegOnes = 1, k = 4

› Output: 2

💡 Note: We need 4 items. Take 2 items with +1, 1 item with 0, and 1 item with -1. Sum = 2×(+1) + 1×(0) + 1×(-1) = 2 + 0 - 1 = 1. Wait, this should be 1, but we can only pick 4 items and we have exactly 4, so the calculation is correct: 2+0-1=1.

Constraints

0 ≤ numOnes, numZeros, numNegOnes ≤ 50
1 ≤ k ≤ numOnes + numZeros + numNegOnes
You must pick exactly k items

Visualization

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Understanding the Visualization

Grab the gold coins (+1)

Take as many gold coins as possible, up to k total items

Take worthless stones (0)

If you need more items, take neutral stones that don't hurt your score

Avoid cursed items (-1)

Only take cursed items if you absolutely must reach exactly k items

Count your treasure

Calculate your final wealth based on items chosen

Key Takeaway

🎯 Key Insight: The greedy approach works perfectly because item values are fixed. Always pick the highest value items first: +1 items, then 0 items, then -1 items only if necessary.

Asked in

G Google 12 a Amazon 8 f Meta 6 ⊞ Microsoft 5

This problem is solved optimally using a greedy approach. Since we want to maximize the sum, we should pick items in order of their value: take all available +1 items first, then 0 items if needed, and finally -1 items only if we must reach exactly k items. The solution runs in O(1) time and O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(k²)	O(1)	Generate all possible ways to pick k items and find the maximum sum
Greedy Approach (Optimal)	O(1)	O(1)	Pick the highest value items first using greedy strategy

Brute Force (Try All Combinations) — Algorithm Steps

Iterate through all possible numbers of +1 items to pick (0 to min(numOnes, k))
For each choice of +1 items, iterate through all possible numbers of 0 items
Calculate how many -1 items we need to reach exactly k total items
Check if we have enough -1 items available
Calculate the sum and keep track of the maximum

Visualization

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Step-by-Step Walkthrough

Try 0 ones

Pick 0 ones, then try all combinations of zeros and negatives

Try 1 one

Pick 1 one, then try combinations with remaining k-1 items

Continue...

Keep trying until all ones are considered

Find maximum

Track the best sum found across all combinations

Code -

solution.c — C

#include <limits.h>

int min(int a, int b) {
    return (a < b) ? a : b;
}

int max(int a, int b) {
    return (a > b) ? a : b;
}

int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
    int maxSum = INT_MIN;
    
    // Try all possible combinations
    for (int onesPicked = 0; onesPicked <= min(numOnes, k); onesPicked++) {
        int remaining = k - onesPicked;
        for (int zerosPicked = 0; zerosPicked <= min(numZeros, remaining); zerosPicked++) {
            int negonesPicked = remaining - zerosPicked;
            
            // Check if we have enough -1 items
            if (negonesPicked <= numNegOnes) {
                int currentSum = onesPicked * 1 + zerosPicked * 0 + negonesPicked * (-1);
                maxSum = max(maxSum, currentSum);
            }
        }
    }
    
    return maxSum;
}

Time & Space Complexity

Time Complexity

⏱️

O(k²)

We have nested loops iterating up to k times each, checking all combinations

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track the maximum sum and current combination

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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