LeetCode in Kotlin
3791. Number of Balanced Integers in a Range
Hard
You are given two integers low and high.
An integer is called balanced if it satisfies both of the following conditions:
- It contains at least two digits.
- The sum of digits at even positions is equal to the sum of digits at odd positions (the leftmost digit has position 1).
Return an integer representing the number of balanced integers in the range [low, high] (both inclusive).
Example 1:
Input: low = 1, high = 100
Output: 9
Explanation:
The 9 balanced numbers between 1 and 100 are 11, 22, 33, 44, 55, 66, 77, 88, and 99.
Example 2:
Input: low = 120, high = 129
Output: 1
Explanation:
Only 121 is balanced because the sum of digits at even and odd positions are both 2.
Example 3:
Input: low = 1234, high = 1234
Output: 0
Explanation:
1234 is not balanced because the sum of digits at odd positions (1 + 3 = 4) does not equal the sum at even positions (2 + 4 = 6).
Constraints:
1 <= low <= high <= 1015
Solution
class Solution {
private fun solve(number: String, idx: Int, diff: Int, tight: Int, dp: Array<Array<LongArray>>): Long {
if (idx == number.length) {
return (if (diff == 0) 1 else 0).toLong()
}
if (dp[idx][diff + dp[0].size / 2][tight] != -1L) {
return dp[idx][diff + dp[0].size / 2][tight]
}
var ans: Long = 0
val ub = if (tight == 1) number[idx].code - '0'.code else 9
for (i in 0..ub) {
val newTight = if (tight == 1 && i == ub) 1 else 0
if (idx % 2 == 0) {
ans += solve(number, idx + 1, diff - i, newTight, dp)
} else {
ans += solve(number, idx + 1, diff + i, newTight, dp)
}
}
dp[idx][diff + dp[0].size / 2][tight] = ans
return dp[idx][diff + dp[0].size / 2][tight]
}
private fun solve(num: Long): Long {
if (num == 0L) {
return 1
}
val number = num.toString()
val len = number.length
val dp: Array<Array<LongArray>> =
if (len % 2 ==
1
) {
Array<Array<LongArray>>(len) { Array<LongArray>((len + 1) * 9 + 1) { LongArray(2) } }
} else {
Array<Array<LongArray>>(
len,
) { Array<LongArray>(len * 9 + 1) { LongArray(2) } }
}
for (tbl in dp) {
for (row in tbl) {
row.fill(-1)
}
}
return solve(number, 0, 0, 1, dp)
}
fun countBalanced(low: Long, high: Long): Long = solve(high) - solve(low - 1)
}