Leetcode Problem of the Day. Data Structures and Algorithms (DSA) #dsa #coding #leetcode #interview

Problem Summary

You are given an integer array called nums of length n.
The cost of any array is defined as the value of its first element. For example, the cost of 1,2,3 is 1, and the cost of 3,4,1 is 3.

You need to divide nums into exactly three contiguous, non empty, and disjoint subarrays that together cover all elements of the original array.
Return the minimum possible sum of costs of these three subarrays.

Key Details

You must create exactly three contiguous parts subarrays.

The cost of each subarray is the first element of that segment.

The goal is to minimize the sum of the three costs.

Core Insights

Observation
The first subarray always begins at index 0, so its cost is nums[0]. For the second and third subarray, you have freedom to choose where to split, but both must start somewhere after index 0.

To minimize total cost, you want the first elements of the other two subarrays to be as small as possible among all elements starting at index 1 onward, because those will define the costs of the second and third segments.

Therefore, the minimum cost is:

nums[0] + smallest value among nums[1 onward] + second smallest value among nums[1 onward]\

Solution Approach
Greedy Selection of Two Minimum Values

Let firstCost be nums[0].

Scan the rest of the array (from index 1 to end) to find the smallest and the second smallest element.

The answer is the sum firstCost + smallest + second smallest.

Time Complexity

Time complexity is O(n) because you just scan the array once to find two minimums.
Space complexity is O(1) as no extra substantial memory is used.

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