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Leetcode 2913. Subarrays Distinct Element Sum of Squares I

Problem Statement:

Given an array nums of integers, the task is to find the sum of the squares of the distinct elements in each subarray of nums.

Clarifying Questions:

  1. What are the constraints on the length of the array?
    • This will help us understand if our solution needs to handle very large arrays efficiently.
  2. Are negative numbers allowed in the array?
    • This could affect our calculations if we are dealing with absolute values or specific properties of negative numbers.
  3. Is the list of integers guaranteed to be non-empty?
    • Confirming this will help us avoid edge cases related to empty input.
  4. How is a subarray defined in the context of this problem?
    • Subarrays are contiguous segments of the array. This shouldn’t change but confirming can clarify assumptions.

Strategy:

Given the problem statement, we can approach it with the following steps:

  1. Generate All Subarrays: Iterate through each possible starting index of a subarray and for each starting index, generate subarrays ending at each possible end index.
  2. Track Distinct Elements: Use a HashSet to keep track of distinct elements in the subarray.
  3. Sum of Squares: For each subarray, calculate the sum of squares of the elements in the HashSet.
  4. Accumulate Results: Accumulate these sums for all subarrays.

Code:

import java.util.HashSet;

public class Solution {
    public long subarraySumOfSquares(int[] nums) {
        long totalSum = 0;
        
        // Iterate over all possible subarray starting points
        for (int start = 0; start < nums.length; start++) {
            HashSet<Integer> distinctElements = new HashSet<>();
            // Iterate over all possible subarray ending points
            for (int end = start; end < nums.length; end++) {
                distinctElements.add(nums[end]); // Add current element to the set
                totalSum += sumOfSquares(distinctElements); // Calculate sum of squares
            }
        }
        
        return totalSum;
    }
    
    // Helper method to calculate sum of squares of all elements in the set
    private long sumOfSquares(HashSet<Integer> set) {
        long sum = 0;
        for (int num : set) {
            sum += (long) num * num;
        }
        return sum;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] nums = {1, 2, 3};
        System.out.println("Total Sum of Squares: " + solution.subarraySumOfSquares(nums));
    }
}

Time Complexity:

Thus, the overall complexity is approximately O(n^3) in the worst case under the assumption that calculating sum of squares O(n) times for each subarray. This approach may face performance issues for large input sizes, and we might need to look into optimization strategies such as sliding window techniques or dynamic programming for larger constraints.

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