LeetCode 2763: Sum of Imbalance Numbers of All Subarrays Solution

Master LeetCode problem 2763 (Sum of Imbalance Numbers of All Subarrays), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2763. Sum of Imbalance Numbers of All Subarrays

Hard

Array Hash Table Ordered Set

Problem Explanation

Explanation

To solve this problem, we can iterate over all subarrays of the input array and calculate the imbalance number for each subarray. The imbalance number of a subarray is the count of pairs of elements in the sorted version of that subarray where the difference between the two elements is greater than 1.

We can calculate the imbalance number for a subarray efficiently by sorting the subarray and then counting the number of indices where the difference between adjacent elements is greater than 1.

The overall approach involves iterating over all subarrays, sorting each subarray, and then calculating the imbalance number for that subarray. Finally, we sum up the imbalance numbers of all subarrays to get the total sum of imbalance numbers.

Time complexity: O(n^2 log n) where n is the length of the input array.
Space complexity: O(n) for storing the sorted subarrays.

Solution Code

class Solution {
    public int sumOfImbalanceNumbers(int[] nums) {
        int n = nums.length;
        int ans = 0;
        
        for (int start = 0; start < n; start++) {
            for (int end = start; end < n; end++) {
                int[] subarray = Arrays.copyOfRange(nums, start, end + 1);
                Arrays.sort(subarray);
                
                int imbalanceCount = 0;
                for (int i = 0; i < subarray.length - 1; i++) {
                    if (subarray[i + 1] - subarray[i] > 1) {
                        imbalanceCount++;
                    }
                }
                
                ans += imbalanceCount;
            }
        }
        
        return ans;
    }
}

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