Remove Min Sum in Java

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

Note: You are allowed to pick 0 numbers, in that case return 0. Or you can pick all numbers.

Example

Input: nums = [-2, 5, 2, -1, 3, -10, 9, -2]
Output: -5
Explanation: You can pick the first and the last 3 numbers
             (-2) + (-10) + 9 + (-2) = -5
             Or you can remove the first 6 and the last one
             (-2) + 5 + 2 + (-1) + 3 + (-10) + (-2) = -5

Understanding the Problem

The core challenge of this problem is to find a way to minimize the sum of selected elements from the array, where the elements can only be picked from the beginning or the end of the array. This problem is significant in scenarios where we need to optimize resource usage or minimize costs.

Potential pitfalls include misunderstanding the requirement to pick elements only from the beginning or the end, and not considering the possibility of picking zero elements.

Approach

To solve this problem, we need to consider all possible ways of picking elements from the beginning and the end of the array. A naive solution would involve checking all combinations, but this would be inefficient. Instead, we can use a more optimized approach:

Calculate prefix sums for the beginning of the array.
Calculate suffix sums for the end of the array.
Combine these sums to find the minimum possible sum.

Naive Solution

The naive solution involves iterating through all possible combinations of elements from the beginning and the end of the array. This approach is not optimal due to its high time complexity.

Optimized Solution

The optimized solution involves using prefix and suffix sums to efficiently calculate the minimum sum. This approach reduces the time complexity to O(n).

Algorithm

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

Initialize two arrays, prefixSum and suffixSum, to store the cumulative sums from the beginning and the end of the array, respectively.
Calculate the prefix sums by iterating from the start to the end of the array.
Calculate the suffix sums by iterating from the end to the start of the array.
Iterate through all possible combinations of prefix and suffix sums to find the minimum sum.

Code Implementation

public class RemoveMinSum {
    public static int minSum(int[] nums) {
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        int[] suffixSum = new int[n + 1];

        // Calculate prefix sums
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        // Calculate suffix sums
        for (int i = n - 1; i >= 0; i--) {
            suffixSum[n - i] = suffixSum[n - i - 1] + nums[i];
        }

        // Find the minimum sum
        int minSum = Integer.MAX_VALUE;
        for (int i = 0; i <= n; i++) {
            minSum = Math.min(minSum, prefixSum[i] + suffixSum[n - i]);
        }

        return minSum;
    }

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

Complexity Analysis

The time complexity of the optimized solution is O(n) because we iterate through the array a constant number of times to calculate the prefix and suffix sums. The space complexity is also O(n) due to the additional arrays used to store the prefix and suffix sums.

Edge Cases

Potential edge cases include:

An empty array, which should return 0.
An array with all positive or all negative numbers.
An array with a single element.

Each of these cases should be tested to ensure the algorithm handles them correctly.

Testing

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

Simple cases with small arrays.
Arrays with a mix of positive and negative numbers.
Edge cases as mentioned above.

Using a testing framework like JUnit can help automate and manage these tests effectively.

Thinking and Problem-Solving Tips

When approaching such problems, it's important to:

Break down the problem into smaller, manageable parts.
Consider both naive and optimized solutions.
Think about edge cases and how to handle them.
Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to solve the problem of minimizing the sum of selected elements from an array. We explored both naive and optimized solutions, provided a detailed algorithm, and implemented the solution in Java. Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources:

LeetCode - A platform for practicing coding problems.
GeeksforGeeks - A comprehensive resource for learning algorithms and data structures.
Coursera - Online courses on algorithms and problem-solving.

Remove Min Sum - Time Complexity: O(n) - Java