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 ways of picking elements from the beginning and the end of the array. A naive solution would involve checking all combinations, but this is not efficient. Instead, we can use a sliding window approach to find the minimum sum efficiently.
The naive solution involves iterating through all possible combinations of elements from the beginning and the end of the array. This approach has a time complexity of O(n^2), which is not optimal for large arrays.
An optimized solution involves using prefix sums and a sliding window technique. We can precompute the prefix sums for the beginning and the end of the array and then use a sliding window to find the minimum sum.
1. Compute the prefix sums for the beginning and the end of the array.
2. Use a sliding window to find the minimum sum by combining elements from the beginning and the end.
def min_sum(nums):
n = len(nums)
if n == 0:
return 0
# Compute prefix sums for the beginning and the end of the array
prefix_sum_start = [0] * (n + 1)
prefix_sum_end = [0] * (n + 1)
for i in range(n):
prefix_sum_start[i + 1] = prefix_sum_start[i] + nums[i]
prefix_sum_end[i + 1] = prefix_sum_end[i] + nums[n - 1 - i]
# Initialize the minimum sum to a large value
min_sum = float('inf')
# Use a sliding window to find the minimum sum
for i in range(n + 1):
current_sum = prefix_sum_start[i] + prefix_sum_end[n - i]
min_sum = min(min_sum, current_sum)
return min_sum
# Example usage
nums = [-2, 5, 2, -1, 3, -10, 9, -2]
print(min_sum(nums)) # Output: -5
The time complexity of the optimized solution is O(n) because we compute the prefix sums in O(n) time and then use a sliding window in O(n) time. The space complexity is also O(n) due to the storage of prefix sums.
1. An empty array should return 0.
2. An array with all positive numbers should return 0 if we pick 0 elements.
3. An array with all negative numbers should return the sum of all elements.
To test the solution comprehensively, we should include a variety of test cases:
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.
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 naive solution and an optimized solution using prefix sums and a sliding window technique. Understanding and solving such problems is crucial for developing strong problem-solving skills.