1449.form largest integer with digits that add up to target

using OffsetArrays: fill

title: 1449. Form Largest Integer With Digits That Add up to Target id: problem1449 author: Indigo date: 2021-06-12 difficulty: Hard categories: String, Dynamic Programming link: https://leetcode.com/problems/form-largest-integer-with-digits-that-add-up-to-target/description/ hidden: true –-

Given an array of integers cost and an integer target. Return the maximum integer you can paint under the following rules:

• The cost of painting a digit (i+1) is given by cost[i] (0 indexed).
• The total cost used must be equal to target.
• Integer does not have digits 0.

Since the answer may be too large, return it as string.

If there is no way to paint any integer given the condition, return "0".

Example 1:

Input: cost = [4,3,2,5,6,7,2,5,5], target = 9
Output: "7772"
Explanation: The cost to paint the digit '7' is 2, and the digit '2' is 3. Then cost("7772") = 2*3+ 3*1 = 9. You could also paint "977", but "7772" is the largest number.
**Digit    cost**
  1  ->   4
  2  ->   3
  3  ->   2
  4  ->   5
  5  ->   6
  6  ->   7
  7  ->   2
  8  ->   5
  9  ->   5

Example 2:

Input: cost = [7,6,5,5,5,6,8,7,8], target = 12
Output: "85"
Explanation: The cost to paint the digit '8' is 7, and the digit '5' is 5. Then cost("85") = 7 + 5 = 12.

Example 3:

Input: cost = [2,4,6,2,4,6,4,4,4], target = 5
Output: "0"
Explanation: It's not possible to paint any integer with total cost equal to target.

Example 4:

Input: cost = [6,10,15,40,40,40,40,40,40], target = 47
Output: "32211"

Constraints:

• cost.length == 9
• 1 <= cost[i] <= 5000
• 1 <= target <= 5000

# @lc code=start
using LeetCode

using OffsetArrays
function largest_number(cost::Vector{Int}, target::Int)::BigInt
    dp = OffsetArray(fill(typemin(Int), target + 1), -1)
    dp[0] = 0
    for c in cost, j in c:target
        dp[j] = max(dp[j], dp[j - c] + 1)
    end
    res = big(0)
    dp[end] < 0 && return res
    for i in 9:-1:1
        c = cost[i]
        while target >= c && dp[target] == dp[target - c] + 1
            res = res * 10 + i
            target -= c
        end
    end
    return res
end
# @lc code=end

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