Question

2763. Sum of Imbalance Numbers of All Subarrays

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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 $nums$.

from sortedcontainers import SortedList


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

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;
  }
}

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;
  }
};