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 minimize the sum of selected numbers from the beginning and the end of the array. This problem is significant in scenarios where we need to minimize costs or losses represented by the array elements. A common pitfall is to assume that we need to pick a contiguous subarray, but we can pick elements from both ends.
To solve this problem, we need to consider all possible combinations of picking elements from the start and the 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.
The naive solution involves iterating through all possible combinations of elements from the start and the end of the array. This approach has a time complexity of O(n^2), which is not efficient for large arrays.
An optimized solution involves using a sliding window technique. We can precompute the prefix sums for the first k elements and the suffix sums for the last k elements. Then, we iterate through all possible values of k to find the minimum sum.
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 values of k and find the minimum sum of the prefix and suffix sums.
// Function to find the minimum sum by picking elements from the start and end
function minSum(nums) {
const n = nums.length;
let minSum = 0;
// Compute prefix sums
let prefixSums = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
prefixSums[i + 1] = prefixSums[i] + nums[i];
}
// Compute suffix sums
let suffixSums = new Array(n + 1).fill(0);
for (let i = n - 1; i >= 0; i--) {
suffixSums[n - i] = suffixSums[n - i - 1] + nums[i];
}
// Find the minimum sum by combining prefix and suffix sums
for (let k = 0; k <= n; k++) {
minSum = Math.min(minSum, prefixSums[k] + suffixSums[n - k]);
}
return minSum;
}
// Example usage
const nums = [-2, 5, 2, -1, 3, -10, 9, -2];
console.log(minSum(nums)); // Output: -5
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 also O(n) due to the storage of prefix and suffix sums.
Potential edge cases include:
Examples:
Input: nums = [] Output: 0 Input: nums = [1, 2, 3, 4] Output: 0 Input: nums = [-1, -2, -3, -4] Output: -10
To test the solution comprehensively, we should include a variety of test cases:
When approaching such problems, it's important to:
In this blog post, we discussed how to solve the problem of minimizing the sum of selected numbers from the beginning and end of an array. We explored both naive and optimized solutions, provided a detailed algorithm, and analyzed the complexity. Understanding and solving such problems is crucial for developing strong problem-solving skills.
For further reading and practice, consider the following resources: