Leetcode 2600: K Items With the Maximum Sum
You are given a bag with items, where each item has a number written on it, which could be 1, 0, or -1. You are also given the counts of items with 1s, 0s, and -1s, as well as a positive integer k. Your task is to pick exactly k items from the bag, maximizing the sum of the numbers written on the selected items.
Problem
Approach
Steps
Complexity
Input: The input consists of four non-negative integers: numOnes, numZeros, numNegOnes, and an integer k.
Example: For numOnes = 5, numZeros = 3, numNegOnes = 2, k = 4.
Constraints:
• 0 <= numOnes, numZeros, numNegOnes <= 50
• 0 <= k <= numOnes + numZeros + numNegOnes
Output: Return the maximum possible sum of the k selected items.
Example: For numOnes = 5, numZeros = 3, numNegOnes = 2, k = 4, the output is 5.
Constraints:
• The sum will be a non-negative integer.
Goal: Maximize the sum of selected items by picking the most valuable items, prioritizing 1s first, then 0s, and finally -1s.
Steps:
• 1. First, pick as many 1s as possible from the bag until reaching k items.
• 2. If k items are still not picked, choose 0s next (as they do not affect the sum).
• 3. If k items are still left, select -1s (which decrease the sum).
Goal: The constraints ensure that the problem is solvable within the given limits.
Steps:
• 0 <= numOnes, numZeros, numNegOnes <= 50
• 0 <= k <= numOnes + numZeros + numNegOnes
Assumptions:
• The array size is manageable, as the maximum sum of numOnes, numZeros, and numNegOnes is 150.
• Input: For numOnes = 5, numZeros = 3, numNegOnes = 2, k = 4
• Explanation: Select 4 items with 1s to maximize the sum, which gives a total sum of 5.
• Input: For numOnes = 4, numZeros = 2, numNegOnes = 1, k = 3
• Explanation: Select 3 items with 1s to maximize the sum, which gives a total sum of 3.
Approach: The approach is to select items with the highest possible values (1s first), followed by zeros, and finally, -1s, ensuring that exactly k items are selected.
Observations:
• Selecting 1s first is optimal as they contribute the most to the sum.
• Zeros do not affect the sum, so we can use them as fillers if needed.
• Negatives should be avoided if possible, as they reduce the sum.
• We need to carefully prioritize the selection to maximize the sum.
Steps:
• 1. Start by selecting as many 1s as possible.
• 2. If additional items are needed, select 0s (since they do not impact the sum).
• 3. If still more items are needed, select -1s, which will reduce the sum.
Empty Inputs:
• If k is 0, the sum will always be 0, regardless of the contents of the bag.
Large Inputs:
• For large inputs, the solution should ensure that the selection logic is efficient to handle up to 150 items.
Special Values:
• If k is exactly the sum of available 1s and 0s, no negative numbers will be selected.
Constraints:
• Handle cases where there are no 1s or no -1s efficiently.
int kItemsWithMaximumSum(int one, int zero, int neg, int k) {
int sum = 0;
sum += min(one, k);
k -= min(one, k);
if(k > 0) {
k -= min(zero, k);
}
if(k > 0) {
sum -= min(neg, k);
k -= min(neg, k);
}
return sum;
}
1 : Function Definition
int kItemsWithMaximumSum(int one, int zero, int neg, int k) {
Defines the function 'kItemsWithMaximumSum' that takes four arguments: 'one', 'zero', 'neg' representing counts of 1s, 0s, and negative numbers, and 'k' representing the number of items to select.
2 : Variable Initialization
int sum = 0;
Initializes the variable 'sum' to 0, which will accumulate the maximum sum by selecting the appropriate number of items.
3 : Add Positive Items
sum += min(one, k);
Adds the minimum of 'one' and 'k' to 'sum', representing the positive numbers that can be selected.
4 : Update Remaining Selection
k -= min(one, k);
Decreases 'k' by the number of positive items selected, to track the remaining items to select.
5 : Check for Zero Items
if(k > 0) {
Checks if there are still items left to select (if 'k' is greater than 0).
6 : Add Zero Items
k -= min(zero, k);
Subtracts the minimum of 'zero' and 'k' from 'k', selecting zeros if any remaining items are left to choose.
7 : Check for Negative Items
if(k > 0) {
Checks if there are still items to be selected after choosing positive and zero items.
8 : Subtract Negative Items
sum -= min(neg, k);
Subtracts the minimum of 'neg' and 'k' from 'sum', selecting negative values if necessary.
9 : Update Remaining Negative Selection
k -= min(neg, k);
Decreases 'k' by the number of negative items selected.
10 : Return Result
return sum;
Returns the accumulated 'sum' as the result, which represents the maximum sum achievable by selecting 'k' items.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: The time complexity is constant as the number of operations is minimal for each input set.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant as no extra space is needed beyond the input values.
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