Leetcode 2670: Find the Distinct Difference Array
You are given a 0-indexed integer array
nums of length n. Your task is to compute the ‘distinct difference array’ of nums. The distinct difference array, diff, is defined such that diff[i] equals the difference between the number of distinct elements in the prefix nums[0,...,i] and the number of distinct elements in the suffix nums[i+1,...,n-1]. Specifically, for each index i, compute diff[i] = distinct elements in the prefix - distinct elements in the suffix.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` of length `n`.
Example: Input: nums = [4,5,6,7]
Constraints:
• 1 <= n == nums.length <= 50
• 1 <= nums[i] <= 50
Output: The output should be an array `diff` where each element represents the distinct difference for the corresponding index in `nums`.
Example: Output: [-3, -1, 1, 3]
Constraints:
• The output array should have the same length as the input array `nums`.
Goal: The goal is to calculate the number of distinct elements in the prefix and suffix for each index, then compute the difference.
Steps:
• Step 1: Use two maps to store the count of distinct elements in the prefix and suffix of the array.
• Step 2: Iterate through the array, updating the prefix and suffix distinct counts as you go, and compute the difference at each index.
• Step 3: Store the result in the `diff` array and return it.
Goal: The solution must be able to handle inputs within the specified constraints efficiently.
Steps:
• Ensure the solution can compute the result in a time-efficient manner due to the constraint on array length.
Assumptions:
• The input array `nums` has unique values within the range [1,50].
• Input: Input: nums = [4,5,6,7]
• Explanation: For index i = 0, the prefix has 1 distinct element (4) and the suffix has 3 distinct elements (5,6,7), so diff[0] = 1 - 3 = -2. For index i = 1, the prefix has 2 distinct elements (4,5) and the suffix has 2 distinct elements (6,7), so diff[1] = 2 - 2 = 0, and so on.
• Input: Input: nums = [3, 2, 3, 4, 2]
• Explanation: For index i = 0, the prefix has 1 distinct element (3) and the suffix has 3 distinct elements (2,4), so diff[0] = 1 - 3 = -2. For index i = 1, the prefix has 2 distinct elements (3,2) and the suffix has 3 distinct elements (4,2), so diff[1] = 2 - 3 = -1, and so on.
Approach: The solution calculates the number of distinct elements in both the prefix and suffix for each index in the array, then computes the difference between these two counts.
Observations:
• We need to keep track of distinct elements in both the prefix and suffix for each index.
• The task can be broken down into updating counters efficiently as we iterate over the array.
• Using maps to store the distinct element counts for both the prefix and suffix should allow us to compute the result efficiently.
Steps:
• Step 1: Initialize two maps, one for the prefix (`l`) and one for the suffix (`r`).
• Step 2: Populate the suffix map by iterating through the array once.
• Step 3: Iterate over the array, updating the prefix and suffix maps and calculating the distinct difference for each index.
Empty Inputs:
• There will always be at least one element in the array, as `n >= 1`.
Large Inputs:
• The solution should work efficiently even with the maximum input size of `n = 50`.
Special Values:
• If the array contains only one unique element, the result will be zero for all indices.
Constraints:
• Ensure the solution can handle arrays with repeated elements properly.
vector<int> distinctDifferenceArray(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n, 0);
map<int, int> l, r;
for(int i = 0; i < n; i++) {
r[nums[i]]++;
}
for(int i = 0; i < n; i++) {
r[nums[i]]--;
if(r[nums[i]] == 0) r.erase(nums[i]);
l[nums[i]]++;
ans[i] = l.size() - r.size();
}
return ans;
}
1 : Function Definition
vector<int> distinctDifferenceArray(vector<int>& nums) {
The function starts by taking a vector 'nums' as input and will return a vector 'ans', where each element represents the difference in distinct elements between the left and right parts of the array.
2 : Array Size Calculation
int n = nums.size();
We retrieve the size of the input array 'nums' and store it in the variable 'n'.
3 : Array Initialization
vector<int> ans(n, 0);
We initialize a vector 'ans' of size 'n', filled with zeros, to store the results.
4 : Map Initialization
map<int, int> l, r;
We declare two maps, 'l' and 'r', to count the frequency of distinct elements in the left and right subarrays, respectively.
5 : Right Subarray Frequency Initialization
for(int i = 0; i < n; i++) {
We begin a loop to populate the map 'r' with the frequency of each element in the 'nums' array.
6 : Right Subarray Frequency Update
r[nums[i]]++;
For each element in 'nums', we increment its count in the map 'r', representing the frequency in the right subarray.
7 : Main Loop Start
for(int i = 0; i < n; i++) {
We start a loop over the array 'nums' to compute the difference in distinct elements between the left and right subarrays.
8 : Right Subarray Frequency Decrease
r[nums[i]]--;
We decrement the frequency of the current element in 'r' as it is being moved to the left subarray.
9 : Right Subarray Frequency Clean-up
if(r[nums[i]] == 0) r.erase(nums[i]);
If the frequency of the current element in 'r' becomes zero, we remove it from the map to maintain only the distinct elements.
10 : Left Subarray Frequency Update
l[nums[i]]++;
We increment the frequency of the current element in 'l', adding it to the left subarray.
11 : Answer Update
ans[i] = l.size() - r.size();
We calculate the difference in the number of distinct elements between the left and right subarrays by comparing the sizes of maps 'l' and 'r', and store the result in 'ans[i]'.
12 : Return Result
return ans;
We return the vector 'ans', which contains the distinct difference values for each index in the input array.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) since we iterate through the array twice, which is efficient given the problem's constraints.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the prefix and suffix maps.
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