2670

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 <= 50
• 1 <= nums[i] <= 50

Related Topics:
Array, Hash Table

Similar Questions:

• Left and Right Sum Differences (Easy)

Solution 1. Left-to-right State Transition

// OJ: https://leetcode.com/problems/find-the-distinct-difference-array
// Author: github.com/lzl124631x
// Time: O(N + C) where C is the range of numbers
// Space: O(C)
class Solution {
public:
    vector<int> distinctDifferenceArray(vector<int>& A) {
        int right[51] = {}, left[51] = {}, rightUnique = 0, leftUnique = 0;
        for (int i : A) right[i]++;
        for (int i = 1; i <= 50; ++i) rightUnique += right[i] > 0;
        vector<int> ans(A.size());
        for (int i = 0; i < A.size(); ++i) {
            rightUnique -= --right[A[i]] == 0;
            leftUnique += ++left[A[i]] == 1;
            ans[i] = leftUnique - rightUnique;
        }
        return ans;
    }
};