LeetCode in Kotlin
2670. Find the Distinct Difference Array
Easy
You are given a 0-indexed array nums of length n.
The distinct difference array of nums is an array diff of length n such that diff[i] is equal to the number of distinct elements in the suffix nums[i + 1, ..., n - 1] subtracted from the number of distinct elements in the prefix nums[0, ..., i].
Return the distinct difference array of nums.
Note that nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j inclusive. Particularly, if i > j then nums[i, ..., j] denotes an empty subarray.
Example 1:
Input: nums = [1,2,3,4,5]
Output: [-3,-1,1,3,5]
Explanation:
For index i = 0, there is 1 element in the prefix and 4 distinct elements in the suffix. Thus, diff[0] = 1 - 4 = -3.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 3 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 3 - 2 = 1.
For index i = 3, there are 4 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 4 - 1 = 3.
For index i = 4, there are 5 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 5 - 0 = 5.
Example 2:
Input: nums = [3,2,3,4,2]
Output: [-2,-1,0,2,3]
Explanation:
For index i = 0, there is 1 element in the prefix and 3 distinct elements in the suffix. Thus, diff[0] = 1 - 3 = -2.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 2 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 2 - 2 = 0.
For index i = 3, there are 3 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 3 - 1 = 2.
For index i = 4, there are 3 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 3 - 0 = 3.
Constraints:
1 <= n == nums.length <= 501 <= nums[i] <= 50
Solution
class Solution {
fun distinctDifferenceArray(nums: IntArray): IntArray {
val n = nums.size
val prefixSet = HashSet<Int>()
val suffixSet = HashSet<Int>()
val preList = IntArray(n)
val sufList = IntArray(n)
val ans = IntArray(n)
for (i in 0..nums.lastIndex) {
prefixSet.add(nums[i])
suffixSet.add(nums[n - 1 - i])
preList[i] = prefixSet.size
sufList[n - 1 - i] = suffixSet.size
}
for (i in 0..nums.lastIndex) {
if (i == nums.lastIndex) {
ans[i] = preList[i]
} else {
ans[i] = preList[i] - sufList[i + 1]
}
}
return ans
}
}