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
The core challenge of this problem is to find a way to pick numbers from the beginning and the end of the array such that their sum is minimized. This problem is significant in scenarios where we need to minimize the sum of selected elements from a sequence, which can be useful in optimization problems.
Potential pitfalls include misunderstanding the requirement to pick numbers from both ends and not considering the possibility of picking zero numbers.
To solve this problem, we can use a sliding window approach. The idea is to consider all possible ways of picking numbers from the beginning and the end of the array and calculate their sums. We can then find the minimum sum among these possibilities.
1. **Naive Solution**: A naive solution would involve checking all possible combinations of picking numbers from the beginning and the end, which would be inefficient.
2. **Optimized Solution**: We can use a sliding window approach to efficiently calculate the sums and find the minimum sum.
1. Calculate the prefix sums for the array.
2. Calculate the suffix sums for the array.
3. Use a sliding window to find the minimum sum by combining prefix and suffix sums.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Function to find the minimum sum by picking numbers from the beginning and the end
int minSum(vector<int>& nums) {
int n = nums.size();
if (n == 0) return 0;
// Calculate prefix sums
vector<int> prefixSum(n + 1, 0);
for (int i = 0; i < n; ++i) {
prefixSum[i + 1] = prefixSum[i] + nums[i];
}
// Calculate suffix sums
vector<int> suffixSum(n + 1, 0);
for (int i = n - 1; i >= 0; --i) {
suffixSum[n - i] = suffixSum[n - i - 1] + nums[i];
}
// Find the minimum sum
int minSum = INT_MAX;
for (int i = 0; i <= n; ++i) {
minSum = min(minSum, prefixSum[i] + suffixSum[n - i]);
}
return minSum;
}
int main() {
vector<int> nums = {-2, 5, 2, -1, 3, -10, 9, -2};
cout << "Minimum Sum: " << minSum(nums) << endl;
return 0;
}
The time complexity of this approach is O(n) because we are calculating the prefix and suffix sums in linear time. The space complexity is also O(n) due to the additional space used for storing the prefix and suffix sums.
1. Empty array: The function should return 0.
2. Array with all positive numbers: The function should return the sum of the smallest numbers from the beginning and the end.
3. Array with all negative numbers: The function should return the sum of the largest (least negative) numbers from the beginning and the end.
To test the solution comprehensively, we should include a variety of test cases:
When approaching such problems, it's important to break down the problem into smaller parts and think about how to combine the results efficiently. Practice solving similar problems and studying different algorithms to improve problem-solving skills.
In this blog post, we discussed how to solve the problem of minimizing the sum of selected numbers from the beginning and the end of an array. We explored a sliding window approach and provided a detailed explanation of the algorithm and its implementation in C++. Understanding and solving such problems is crucial for developing strong problem-solving skills.
For further reading and practice, consider the following resources: