Breaking down a large dataset into smaller, manageable chunks is a fundamental skill in algorithm design. In this problem, we explore how to minimize the "cost" of partitioning an array by identifying the most influential elements. This challenge provides a great introduction to greedy logic and optimization.
Problem Summary
You're given:
- An array of integers called
numswith a length of . - A definition of "cost" which is simply the value of the first element of a subarray.
Your goal:
- Split the array into exactly 3 contiguous subarrays.
- Find the minimum possible total cost, which is the sum of the first elements of these three subarrays.
Intuition
To solve this efficiently, we need to look at the rules of the split. Since we must divide the array into 3 contiguous subarrays, the very first element of the original array (nums[0]) will always be the first element of the first subarray. Therefore, nums[0] is a mandatory part of our total cost.
To minimize the total sum, we need to pick two other elements from the remaining part of the array (from index 1 to ) to serve as the starting points for the second and third subarrays. Because the subarrays must be contiguous, any two elements we pick will naturally define the boundaries of the split. To get the minimum sum, we simply need to find the two smallest values in the rest of the array.
Walkthrough: Understanding the Examples
Example 1: `nums = [1, 2, 3, 12]`
- The first element is
1. This is our first cost. - We look at the remaining elements:
[2, 3, 12]. - The two smallest values here are
2and3. - Total cost: .
Example 3: `nums = [10, 3, 1, 1]`
- The first element is
10. This is our first cost. - We look at the remaining elements:
[3, 1, 1]. - The two smallest values here are
1and1. - Total cost: .
Code Blocks
C++
class Solution {
public:
int minimumCost(vector<int>& nums) {
// The maximum possible value based on constraints
int min1 = 51;
int min2 = 51;
// Start from index 1 because nums[0] is always included
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] < min1) {
min2 = min1;
min1 = nums[i];
} else if (nums[i] < min2) {
min2 = nums[i];
}
}
return nums[0] + min1 + min2;
}
};
Python
class Solution:
def minimumCost(self, nums: list[int]) -> int:
# nums[0] is always the start of the first subarray
first_val = nums[0]
# Get the rest of the array
remaining = nums[1:]
# Sort to easily find the two smallest elements
remaining.sort()
# Sum the first element and the two smallest from the rest
return first_val + remaining[0] + remaining[1]
JavaScript
/**
* @param {number[]} nums
* @return {number}
*/
var minimumCost = function(nums) {
const firstVal = nums[0];
// Extract everything after the first element
let remaining = nums.slice(1);
// Sort numerically in ascending order
remaining.sort((a, b) => a - b);
// The result is the sum of the first element and the two smallest remaining
return firstVal + remaining[0] + remaining[1];
};
Key Takeaways
- Fixed Constraints: Recognizing that the first element is "locked in" simplifies the problem from a complex partitioning task to a simple search task.
- Greedy Selection: By choosing the two smallest available numbers, we guarantee the lowest possible sum, which is a classic greedy strategy.
- Efficiency: This solution runs in time if we use a single pass to find the two minimums, or if we use sorting. Given the small constraints (), both are extremely fast.
Final Thoughts
This problem is a fantastic exercise in "cutting through the noise." At first glance, it might look like a Dynamic Programming problem involving array partitions. However, by observing that the cost is only tied to the first element of each split, it becomes a simple search for the smallest values. In real-world software engineering, this is similar to optimizing load balancers: you often want to identify the "entry points" that incur the least latency or cost.
Top comments (1)
Well explained OM!