Remove Min Sum Ii in C++

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 numbers from either the beginning or the end of the array such that their sum is minimized. This problem is significant in scenarios where we need to optimize resource allocation or minimize costs under certain constraints.

Potential pitfalls include misunderstanding the requirement to pick numbers from both ends of the array and not considering all possible combinations of picks from the start and end.

Approach

To solve this problem, we need to consider all possible ways to pick k numbers from the start and end of the array. A naive solution would involve checking all combinations, but this is not optimal. Instead, we can use a sliding window approach to efficiently find the minimum sum.

Naive Solution

The naive solution involves generating all possible combinations of picking k numbers from the start and end of the array. This approach is not optimal due to its high time complexity.

Optimized Solution

We can optimize the solution by using prefix sums and a sliding window technique. The idea is to precompute the sums of the first i elements and the last j elements for all possible values of i and j such that i + j = k. Then, we can find the minimum sum by iterating through these precomputed sums.

Algorithm

1. Compute the prefix sums for the first k elements.

2. Compute the suffix sums for the last k elements.

3. Iterate through all possible splits of k into i and j where i + j = k and find the minimum sum.

Code Implementation


#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int minSum(vector<int>& nums, int k) {
    int n = nums.size();
    vector<int> prefixSum(k + 1, 0);
    vector<int> suffixSum(k + 1, 0);

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

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

    int minSum = INT_MAX;

    // Find the minimum sum by considering all splits of k
    for (int i = 0; i <= k; ++i) {
        minSum = min(minSum, prefixSum[i] + suffixSum[k - i]);
    }

    return minSum;
}

int main() {
    vector<int> nums = {-2, 5, 2, -1, 3, -10, 9, -2};
    int k = 4;
    cout << "Minimum sum: " << minSum(nums, k) << endl;
    return 0;
}

Complexity Analysis

The time complexity of the optimized solution is O(n) because we compute the prefix and suffix sums in linear time. The space complexity is O(k) for storing the prefix and suffix sums.

Edge Cases

Potential edge cases include:

When k is 0, the sum should be 0.
When k is equal to the length of the array, the sum should be the sum of all elements.
Arrays with all positive or all negative numbers.

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 mixed positive and negative numbers.
Arrays with all positive or all negative numbers.

Thinking and Problem-Solving Tips

When approaching such problems, it's essential to break down the problem into smaller parts and think about how to optimize each part. Practice solving similar problems and study different algorithms to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to solve the problem of minimizing the sum of k numbers picked from the start and end of an array. We explored a naive solution and an optimized solution using prefix and suffix sums. Understanding and solving such problems is crucial for developing strong problem-solving skills.

Additional Resources

For further reading and practice, consider the following resources:

Remove Min Sum II - Time Complexity: O(n) - C++