README.md

1449. Form Largest Integer With Digits That Add up to Target

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.
• The integer does not have 0 digits.

Since the answer may be very large, return it as a 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 is impossible to paint any integer with total cost equal to target. 

Constraints:

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

Solutions (Rust)

1. Solution

implSolution{pubfnlargest_number(cost:Vec<i32>,target:i32) -> String{let target = target asusize;let cost = cost.iter().map(|&x| x asusize).collect::<Vec<_>>();letmut dp = vec![[-1;10]; target + 1]; dp[0] = [0;10];for i in0..=target {if dp[i][0] == -1{continue;}for j in0..9{letmut count = dp[i]; count[9 - j] += 1; count[0] += 1;if i + cost[j] <= target && dp[i + cost[j]] < count { dp[i + cost[j]] = count;}}}if dp[target][0] == -1{return"0".to_string();}(0..9).rev().map(|i| vec![std::char::from_u32(49 + i asu32).unwrap(); dp[target][9 - i]asusize]).flatten().collect()}}