LeetCode 2600. K Items With the Maximum Sum Solution in Java, C++, Python & More | Explanation + Code

Description

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 <= 50
• 0 <= k <= numOnes + numZeros + numNegOnes

Solutions

Solution 1: Greedy

According to the problem description, we should take as many items marked as 1 as possible, then take items marked as 0, and finally take items marked as -1.

Thus:

• If the number of items marked as 1 in the bag is greater than or equal to k, we take k items, and the sum of the numbers is k.
• If the number of items marked as 1 is less than k, we take numOnes items, resulting in a sum of numOnes. If the number of items marked as 0 is greater than or equal to k - numOnes, we take k - numOnes more items, keeping the sum at numOnes.
• Otherwise, we take k - numOnes - numZeros items from those marked as -1, resulting in a sum of numOnes - (k - numOnes - numZeros).

The time complexity is O(1), and the space complexity is O(1).

PythonJavaC++GoTypeScriptRustC#

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)(code-box)

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); } }(code-box)

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); } };(code-box)

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) }(code-box)

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); }(code-box)

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) } }(code-box)

public 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); } }(code-box)