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2763. Sum of Imbalance Numbers of All Subarrays

Description

The imbalance number of a 0-indexed integer array arr of length n is defined as the number of indices in sarr = sorted(arr) such that:

  • 0 <= i < n - 1, and
  • sarr[i+1] - sarr[i] > 1

Here, sorted(arr) is the function that returns the sorted version of arr.

Given a 0-indexed integer array nums, return the sum of imbalance numbers of all its subarrays.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [2,3,1,4]
Output: 3
Explanation: There are 3 subarrays with non-zero imbalance numbers:
- Subarray [3, 1] with an imbalance number of 1.
- Subarray [3, 1, 4] with an imbalance number of 1.
- Subarray [1, 4] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.

Example 2:

Input: nums = [1,3,3,3,5]
Output: 8
Explanation: There are 7 subarrays with non-zero imbalance numbers:
- Subarray [1, 3] with an imbalance number of 1.
- Subarray [1, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3] with an imbalance number of 1.
- Subarray [1, 3, 3, 3, 5] with an imbalance number of 2. 
- Subarray [3, 3, 3, 5] with an imbalance number of 1. 
- Subarray [3, 3, 5] with an imbalance number of 1.
- Subarray [3, 5] with an imbalance number of 1.
The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 8.

 

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= nums.length

Solutions

Solution 1: Enumeration + Ordered Set

We can first enumerate the left endpoint \(i\) of the subarray. For each \(i\), we enumerate the right endpoint \(j\) of the subarray from small to large, and maintain all the elements in the current subarray with an ordered list. We also use a variable \(cnt\) to maintain the unbalanced number of the current subarray.

For each number \(nums[j]\), we find the first element \(nums[k]\) in the ordered list that is greater than or equal to \(nums[j]\), and the last element \(nums[h]\) that is less than \(nums[j]\):

  • If \(nums[k]\) exists, and the difference between \(nums[k]\) and \(nums[j]\) is more than \(1\), the unbalanced number increases by \(1\);
  • If \(nums[h]\) exists, and the difference between \(nums[j]\) and \(nums[h]\) is more than \(1\), the unbalanced number increases by \(1\);
  • If both \(nums[k]\) and \(nums[h]\) exist, then inserting the element \(nums[j]\) between \(nums[h]\) and \(nums[k]\) will reduce the unbalanced number by \(1\).

Then, we add the unbalanced number of the current subarray to the answer, and continue the iteration until we finish iterating over all subarrays.

The time complexity is \(O(n^2 \times \log n)\) and the space complexity is \(O(n)\), where \(n\) is the length of the array \(nums\).

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class Solution:
    def sumImbalanceNumbers(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        for i in range(n):
            sl = SortedList()
            cnt = 0
            for j in range(i, n):
                k = sl.bisect_left(nums[j])
                h = k - 1
                if h >= 0 and nums[j] - sl[h] > 1:
                    cnt += 1
                if k < len(sl) and sl[k] - nums[j] > 1:
                    cnt += 1
                if h >= 0 and k < len(sl) and sl[k] - sl[h] > 1:
                    cnt -= 1
                sl.add(nums[j])
                ans += cnt
        return ans

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class Solution {
    public int sumImbalanceNumbers(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            TreeMap<Integer, Integer> tm = new TreeMap<>();
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                Integer k = tm.ceilingKey(nums[j]);
                if (k != null && k - nums[j] > 1) {
                    ++cnt;
                }
                Integer h = tm.floorKey(nums[j]);
                if (h != null && nums[j] - h > 1) {
                    ++cnt;
                }
                if (h != null && k != null && k - h > 1) {
                    --cnt;
                }
                tm.merge(nums[j], 1, Integer::sum);
                ans += cnt;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int sumImbalanceNumbers(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            multiset<int> s;
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                auto it = s.lower_bound(nums[j]);
                if (it != s.end() && *it - nums[j] > 1) {
                    ++cnt;
                }
                if (it != s.begin() && nums[j] - *prev(it) > 1) {
                    ++cnt;
                }
                if (it != s.end() && it != s.begin() && *it - *prev(it) > 1) {
                    --cnt;
                }
                s.insert(nums[j]);
                ans += cnt;
            }
        }
        return ans;
    }
};

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