LeetCode in Kotlin
3352. Count K-Reducible Numbers Less Than N
Hard
You are given a binary string s representing a number n in its binary form.
You are also given an integer k.
An integer x is called k-reducible if performing the following operation at most k times reduces it to 1:
- Replace
xwith the count of set bits in its binary representation.
For example, the binary representation of 6 is "110". Applying the operation once reduces it to 2 (since "110" has two set bits). Applying the operation again to 2 (binary "10") reduces it to 1 (since "10" has one set bit).
Return an integer denoting the number of positive integers less than n that are k-reducible.
Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: s = “111”, k = 1
Output: 3
Explanation:
n = 7. The 1-reducible integers less than 7 are 1, 2, and 4.
Example 2:
Input: s = “1000”, k = 2
Output: 6
Explanation:
n = 8. The 2-reducible integers less than 8 are 1, 2, 3, 4, 5, and 6.
Example 3:
Input: s = “1”, k = 3
Output: 0
Explanation:
There are no positive integers less than n = 1, so the answer is 0.
Constraints:
1 <= s.length <= 800shas no leading zeros.sconsists only of the characters'0'and'1'.1 <= k <= 5
Solution
class Solution {
fun countKReducibleNumbers(s: String, k: Int): Int {
val n = s.length
val reducible = IntArray(n + 1)
for (i in 2..<reducible.size) {
reducible[i] = 1 + reducible[Integer.bitCount(i)]
}
val dp = LongArray(n + 1)
var curr = 0
for (i in 0..<n) {
for (j in i - 1 downTo 0) {
dp[j + 1] += dp[j]
dp[j + 1] %= MOD.toLong()
}
if (s[i] == '1') {
dp[curr]++
dp[curr] %= MOD.toLong()
curr++
}
}
var result: Long = 0
for (i in 1..s.length) {
if (reducible[i] < k) {
result += dp[i]
result %= MOD.toLong()
}
}
return (result % MOD).toInt()
}
companion object {
private val MOD = (1e9 + 7).toInt()
}
}