LeetCode in Kotlin
1449. Form Largest Integer With Digits That Add up to Target
Hard
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 bycost[i](0-indexed). - The total cost used must be equal to
target. - The integer does not have
0digits.
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 == 91 <= cost[i], target <= 5000
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun largestNumber(cost: IntArray, target: Int): String {
var target = target
val dp = Array(10) { IntArray(5001) }
dp[0].fill(-1)
for (i in 1..cost.size) {
for (j in 1..target) {
if (cost[i - 1] > j) {
dp[i][j] = dp[i - 1][j]
} else {
var temp = if (dp[i - 1][j - cost[i - 1]] == -1) -1 else 1 + dp[i - 1][j - cost[i - 1]]
val t = if (dp[i][j - cost[i - 1]] == -1) -1 else 1 + dp[i][j - cost[i - 1]]
temp = if (t != -1 && temp == -1) {
t
} else {
Math.max(t, temp)
}
if (dp[i - 1][j] == -1) {
dp[i][j] = temp
} else if (temp == -1) {
dp[i][j] = dp[i - 1][j]
} else {
dp[i][j] = Math.max(temp, dp[i - 1][j])
}
}
}
}
if (dp[9][target] == -1) {
return "0"
}
var i = 9
val result = StringBuilder()
while (target > 0) {
if (target - cost[i - 1] >= 0 && dp[i][target - cost[i - 1]] + 1 == dp[i][target] ||
(
target - cost[i - 1] >= 0 &&
dp[i - 1][target - cost[i - 1]] + 1 == dp[i][target]
)
) {
result.append(i)
target -= cost[i - 1]
} else {
i--
}
}
return result.toString()
}
}