2670. Find the Distinct Difference Array

Description

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 <= 50
• 1 <= nums[i] <= 50

Solutions

• class Solution {
public int[] distinctDifferenceArray(int[] nums) {
int n = nums.length;
int[] suf = new int[n + 1];
Set<Integer> s = new HashSet<>();
for (int i = n - 1; i >= 0; --i) {
s.add(nums[i]);
suf[i] = s.size();
}
s.clear();
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
s.add(nums[i]);
ans[i] = s.size() - suf[i + 1];
}
return ans;
}
}

• class Solution {
public:
vector<int> distinctDifferenceArray(vector<int>& nums) {
int n = nums.size();
vector<int> suf(n + 1);
unordered_set<int> s;
for (int i = n - 1; i >= 0; --i) {
s.insert(nums[i]);
suf[i] = s.size();
}
s.clear();
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
s.insert(nums[i]);
ans[i] = s.size() - suf[i + 1];
}
return ans;
}
};

• class Solution:
def distinctDifferenceArray(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [0] * n
for i in range(n):
a = len(set(nums[: i + 1]))
b = len(set(nums[i + 1 :]))
ans[i] = a - b
return ans


• func distinctDifferenceArray(nums []int) []int {
n := len(nums)
suf := make([]int, n+1)
s := map[int]bool{}
for i := n - 1; i >= 0; i-- {
s[nums[i]] = true
suf[i] = len(s)
}
ans := make([]int, n)
s = map[int]bool{}
for i, x := range nums {
s[x] = true
ans[i] = len(s) - suf[i+1]
}
return ans
}

• function distinctDifferenceArray(nums: number[]): number[] {
const n = nums.length;
const suf: number[] = new Array(n + 1).fill(0);
const s: Set<number> = new Set();
for (let i = n - 1; i >= 0; --i) {
s.add(nums[i]);
suf[i] = s.size;
}
s.clear();
const ans: number[] = new Array(n);
for (let i = 0; i < n; ++i) {
s.add(nums[i]);
ans[i] = s.size - suf[i + 1];
}
return ans;
}


• use std::collections::HashSet;

impl Solution {
pub fn distinct_difference_array(nums: Vec<i32>) -> Vec<i32> {
let mut ans: Vec<i32> = Vec::new();

for i in 0..nums.len() {
let mut j = 0;
let mut hash1 = HashSet::new();
while j <= i {
hash1.insert(nums[j]);
j += 1;
}

let mut k = i + 1;
let mut hash2 = HashSet::new();
while k < nums.len() {
hash2.insert(nums[k]);
k += 1;
}

ans.push((hash1.len() - hash2.len()) as i32);
}

ans
}
}