A faster bundler can join up to k chunks in one step. Each step picks between 2 and k chunks and joins them into one; the cost of a step is the total size of the chunks it joins.
Write cheapestBundle(sizes, k) that returns the smallest total cost to end with a single chunk. With zero or one chunk, the cost is 0.
Input: sizes = [1,2,3,4,5], k = 3 Output: 21
Join 1 + 2 + 3 = 6 (cost 6), then 4 + 5 + 6 = 15 (cost 15): 21.
Input: sizes = [1,2,3,4], k = 3 Output: 13
Join just 1 + 2 = 3 first (cost 3), then 3 + 3 + 4 = 10: 13. Joining three at first would cost 6 + 10 = 16.
0 <= sizes.length <= 20001 <= sizes[i] <= 10002 <= k <= 20Run tries your code on the examples. Submit runs every test, including hidden edge cases, and solves the problem when they all pass.
Ctrl + Enter runs, Ctrl + Shift + Enter submits (⌘ on a Mac).
I'm here to help with #602 Bundle Chunks k at a Time. I give hints, never the solution, so you get to solve it yourself. Ask for a hint, select code to ask about it, or press Ask Tuto why on a failing test.
Your first 3 answers need no account. for free to keep asking.