Find the Distinct Difference Array
As a Systems Engineer at Tata Consultancy Services, I deliver exceptional software products for mobile and web platforms, using agile methodologies and robust quality maintenance. I am experienced in performance testing, automation testing, API testing, and manual testing, with various tools and technologies such as Jmeter, Azure LoadTest, Selenium, Java, OOPS, Maven, TestNG, and Postman.
I have successfully developed and executed detailed test plans, test cases, and scripts for Android and web applications, ensuring high-quality standards and user satisfaction. I have also demonstrated my proficiency in manual REST API testing with Postman, as well as in end-to-end performance and automation testing using Jmeter and selenium with Java, TestNG and Maven. Additionally, I have utilized Azure DevOps for bug tracking and issue management.
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.
LeetCode Problem - 2670
class Solution {
// Method to calculate distinct difference array
public int[] distinctDifferenceArray(int[] nums) {
int[] answer = new int[nums.length]; // Initialize array to store results
// Iterate through each element in the input array
for(int i=0; i<nums.length; i++){
// Calculate distinct elements in the prefix (from start to current index)
int prefix = findDistinctElement(Arrays.copyOfRange(nums, 0, i+1));
// Calculate distinct elements in the suffix (from current index to end)
int suffix = findDistinctElement(Arrays.copyOfRange(nums, i+1, nums.length));
// Store the difference of distinct elements between prefix and suffix
answer[i] = prefix - suffix;
}
return answer; // Return the resulting array
}
// Method to find number of distinct elements in an array
public int findDistinctElement(int[] arr){
List<Integer> al = new ArrayList<>(); // Create a list to store distinct elements
int count = 0; // Initialize count of distinct elements
// Iterate through the array
for(int i=0; i<arr.length; i++){
if(!al.contains(arr[i])){ // Check if element is not already in list
al.add(arr[i]); // Add element to list
count++; // Increment count of distinct elements
}
}
return count; // Return total count of distinct elements
}
}
