Remove Min Sum Ii in Java

Given an input array of integers and a number k, pick exactly k numbers, some from the beginning of the array and some from the end of it, in such a way that their sum is minimized. Return this minimum sum.

Example

Input: nums = [-2, 5, 2, -1, 3, -10, 9, -2], k = 4
Output: -5
Explanation: Pick the first and the last three numbers
             (-2) + (-10) + 9 + (-2) = -5

Understanding the Problem

The core challenge of this problem is to find a way to select exactly k elements from the array such that their sum is minimized. The elements can be picked from either the beginning or the end of the array, but not from the middle. This problem is significant in scenarios where we need to optimize resource allocation or minimize costs under certain constraints.

Approach

To solve this problem, we need to consider the following steps:

Calculate the sum of the first k elements from the beginning of the array.
Iteratively replace elements from the beginning with elements from the end, updating the sum and keeping track of the minimum sum encountered.

Let's break down the approach:

Initialize the sum with the first k elements.
Iterate from 0 to k and in each iteration, replace one element from the beginning with one element from the end, updating the sum and checking if it's the minimum sum.

Algorithm

Here is a step-by-step breakdown of the algorithm:

Initialize currentSum with the sum of the first k elements.
Set minSum to currentSum.
Iterate from 0 to k:
- Subtract the element at the current index from the beginning and add the element at the corresponding index from the end to currentSum.
- Update minSum if currentSum is smaller.

Code Implementation

public class RemoveMinSumII {
    public static int minSum(int[] nums, int k) {
        int n = nums.length;
        int currentSum = 0;
        
        // Calculate the sum of the first k elements
        for (int i = 0; i < k; i++) {
            currentSum += nums[i];
        }
        
        int minSum = currentSum;
        
        // Replace elements from the beginning with elements from the end
        for (int i = 0; i < k; i++) {
            currentSum = currentSum - nums[k - 1 - i] + nums[n - 1 - i];
            minSum = Math.min(minSum, currentSum);
        }
        
        return minSum;
    }

    public static void main(String[] args) {
        int[] nums = {-2, 5, 2, -1, 3, -10, 9, -2};
        int k = 4;
        System.out.println(minSum(nums, k)); // Output: -5
    }
}

Complexity Analysis

The time complexity of this approach is O(n) because we iterate through the array a constant number of times. The space complexity is O(1) as we only use a few extra variables.

Edge Cases

Consider the following edge cases:

k is 0: The sum should be 0.
k is equal to the length of the array: The sum should be the sum of all elements.
All elements are positive or negative: The algorithm should still find the minimum sum.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases with small arrays.
Cases where k is 0 or equal to the length of the array.
Arrays with all positive or all negative numbers.
Arrays with a mix of positive and negative numbers.

Thinking and Problem-Solving Tips

When approaching such problems, consider breaking down the problem into smaller parts and solving each part step-by-step. Practice similar problems to improve your problem-solving skills and understand different algorithms and their applications.

Conclusion

In this blog post, we discussed how to solve the "Remove Min Sum II" problem using a Java solution. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources:

LeetCode - Practice coding problems and challenges.
GeeksforGeeks - Tutorials and explanations on various algorithms and data structures.
Coursera - Online courses on algorithms and data structures.

Remove Min Sum II - Time Complexity: O(n) - Java Solution /homework