K Items With the Maximum Sum

As a Systems Engineer at Tata Consultancy Services, I deliver exceptional software products for mobile and web platforms, using agile methodologies and robust quality maintenance. I am experienced in performance testing, automation testing, API testing, and manual testing, with various tools and technologies such as Jmeter, Azure LoadTest, Selenium, Java, OOPS, Maven, TestNG, and Postman.
I have successfully developed and executed detailed test plans, test cases, and scripts for Android and web applications, ensuring high-quality standards and user satisfaction. I have also demonstrated my proficiency in manual REST API testing with Postman, as well as in end-to-end performance and automation testing using Jmeter and selenium with Java, TestNG and Maven. Additionally, I have utilized Azure DevOps for bug tracking and issue management.
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:
numOnesitems with1s written on them.numZeroesitems with0s written on them.numNegOnesitems 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.
LeetCode Problem - 2600
class Solution {
public int kItemsWithMaximumSum(int numOnes, int numZeros, int numNegOnes, int k) {
int result = 0; // Initialize result to store the sum of selected items.
// Process 'k' items to maximize the sum.
while (k > 0) {
if (numOnes > 0) {
// If there are '1's available, add 1 to the result and decrease their count.
result++;
numOnes--;
} else if (numZeros > 0) {
// If no '1's are left but there are '0's, simply decrement their count.
// Adding 0 does not change the result.
numZeros--;
} else {
// If no '1's or '0's are left, use '-1's.
// Subtract 1 from the result and decrease the count of '-1's.
result--;
numNegOnes--;
}
k--; // Decrease the count of items to process.
}
return result; // Return the final maximum sum.
}
}




