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
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 just from one side.
To solve this problem, we need to consider the following steps:
A naive solution would involve checking all possible combinations, but this would be inefficient. Instead, we can use a sliding window approach to optimize the solution.
Here is a step-by-step breakdown of the optimized algorithm:
/**
* Function to find the minimum sum of k elements taken from the start and end of the array.
* @param {number[]} nums - The input array of integers.
* @param {number} k - The number of elements to pick.
* @return {number} - The minimum sum of k elements.
*/
function minSum(nums, k) {
// Calculate the initial sum of the first k elements
let currentSum = 0;
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
// Initialize the minimum sum with the initial sum
let minSum = currentSum;
// Iterate through the array to find the minimum sum
for (let i = 0; i < k; i++) {
// Adjust the sum by removing an element from the start and adding an element from the end
currentSum = currentSum - nums[k - 1 - i] + nums[nums.length - 1 - i];
// Update the minimum sum if the current sum is smaller
minSum = Math.min(minSum, currentSum);
}
return minSum;
}
// Example usage
const nums = [-2, 5, 2, -1, 3, -10, 9, -2];
const k = 4;
console.log(minSum(nums, k)); // Output: -5
The time complexity of this approach is O(k) because we iterate through the array a fixed number of times, which is proportional to k. The space complexity is O(1) as we only use a few extra variables.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
// Test case 1: General case
console.log(minSum([-2, 5, 2, -1, 3, -10, 9, -2], 4)); // Output: -5
// Test case 2: k is 0
console.log(minSum([1, 2, 3, 4], 0)); // Output: 0
// Test case 3: k is equal to the length of the array
console.log(minSum([1, 2, 3, 4], 4)); // Output: 10
// Test case 4: Array with negative numbers
console.log(minSum([-1, -2, -3, -4], 2)); // Output: -7
// Test case 5: Array with zeros
console.log(minSum([0, 0, 0, 0], 2)); // Output: 0
When approaching such problems, consider the following tips:
In this blog post, we discussed how to solve the problem of finding the minimum sum of k elements taken from the start and end of an array. We explored the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills.
For further reading and practice, consider the following resources: