LeetCode in Kotlin
2963. Count the Number of Good Partitions
Hard
You are given a 0-indexed array nums consisting of positive integers.
A partition of an array into one or more contiguous subarrays is called good if no two subarrays contain the same number.
Return the total number of good partitions of nums.
Since the answer may be large, return it modulo 109 + 7.
Example 1:
Input: nums = [1,2,3,4]
Output: 8
Explanation: The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).
Example 2:
Input: nums = [1,1,1,1]
Output: 1
Explanation: The only possible good partition is: ([1,1,1,1]).
Example 3:
Input: nums = [1,2,1,3]
Output: 2
Explanation: The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
import kotlin.math.max
class Solution {
fun numberOfGoodPartitions(nums: IntArray): Int {
val mp: MutableMap<Int, Int> = HashMap()
val n = nums.size
for (i in 0 until n) {
mp[nums[i]] = i
}
var i = 0
var j = 0
var cnt = 0
while (i < n) {
j = max(j, mp[nums[i]]!!)
if (i == j) {
cnt++
}
i++
}
var res = 1
for (k in 1 until cnt) {
res *= 2
val mod = 1000000007
res %= mod
}
return res
}
}