The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i - j|.

Return an array intervals of length n where intervals[i] is the sum of intervals between arr[i] and each element in arr with the same value as arr[i].

Note: |x| is the absolute value of x.

Example 1:

Input: arr = [2,1,3,1,2,3,3]
Output: [4,2,7,2,4,4,5]
Explanation:
- Index 0: Another 2 is found at index 4. |0 - 4| = 4
- Index 1: Another 1 is found at index 3. |1 - 3| = 2
- Index 2: Two more 3s are found at indices 5 and 6. |2 - 5| + |2 - 6| = 7
- Index 3: Another 1 is found at index 1. |3 - 1| = 2
- Index 4: Another 2 is found at index 0. |4 - 0| = 4
- Index 5: Two more 3s are found at indices 2 and 6. |5 - 2| + |5 - 6| = 4
- Index 6: Two more 3s are found at indices 2 and 5. |6 - 2| + |6 - 5| = 5

Example 2:

Input: arr = [10,5,10,10]
Output: [5,0,3,4]
Explanation:
- Index 0: Two more 10s are found at indices 2 and 3. |0 - 2| + |0 - 3| = 5
- Index 1: There is only one 5 in the array, so its sum of intervals to identical elements is 0.
- Index 2: Two more 10s are found at indices 0 and 3. |2 - 0| + |2 - 3| = 3
- Index 3: Two more 10s are found at indices 0 and 2. |3 - 0| + |3 - 2| = 4

Constraints:

• n == arr.length
• 1 <= n <= 105
• 1 <= arr[i] <= 105

Solution: Math / Hashtable + Prefix Sum

For each arr[i], suppose it occurs in the array of total c times, among which k of them are in front of it and c – k – 1 of them are after it. Then the total sum intervals:
(i – j₁) + (i – j₂) + … + (i – j_k) + (j_k+1-i) + (j_k+2-i) + … + (j_c-i)
<=> k * i – sum(j₁~j_k) + sum(j_k+1~j_c) – (c – k – 1) * i

Use a hashtable to store the indies of each unique number in the array and compute the prefix sum for fast range sum query.

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<long long> getDistances(vector<int>& arr) {
    const int n = arr.size();
    unordered_map<int, vector<long long>> m;
    vector<int> pos(n);
    for (int i = 0; i < n; ++i) {
      m[arr[i]].push_back(i);
      pos[i] = m[arr[i]].size() - 1;
    }
    for (auto& [k, idx] : m)
      partial_sum(begin(idx), end(idx), begin(idx));      
    vector<long long> ans(n);
    for (int i = 0; i < n; ++i) {
      const auto& sums = m[arr[i]];
      const long long k = pos[i];
      const long long c = sums.size();      
      if (k > 0) ans[i] += k * i - sums[k - 1];
      if (k + 1 < c) ans[i] += (sums[c - 1] - sums[k]) - (c - k - 1) * i;
    }
    return ans;
  }
};

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