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发布于 2024-06-17 01:03:09 字数 5012 浏览 0 评论 0 收藏 0

2121. Intervals Between Identical Elements

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Description

You are given a 0-indexed array of n integers arr.

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

Solutions

Solution 1

class Solution:
  def getDistances(self, arr: List[int]) -> List[int]:
    d = defaultdict(list)
    n = len(arr)
    for i, v in enumerate(arr):
      d[v].append(i)
    ans = [0] * n
    for v in d.values():
      m = len(v)
      val = sum(v) - v[0] * m
      for i, p in enumerate(v):
        delta = v[i] - v[i - 1] if i >= 1 else 0
        val += i * delta - (m - i) * delta
        ans[p] = val
    return ans
class Solution {
  public long[] getDistances(int[] arr) {
    Map<Integer, List<Integer>> d = new HashMap<>();
    int n = arr.length;
    for (int i = 0; i < n; ++i) {
      d.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
    }
    long[] ans = new long[n];
    for (List<Integer> v : d.values()) {
      int m = v.size();
      long val = 0;
      for (int e : v) {
        val += e;
      }
      val -= (m * v.get(0));
      for (int i = 0; i < v.size(); ++i) {
        int delta = i >= 1 ? v.get(i) - v.get(i - 1) : 0;
        val += i * delta - (m - i) * delta;
        ans[v.get(i)] = val;
      }
    }
    return ans;
  }
}
class Solution {
public:
  vector<long long> getDistances(vector<int>& arr) {
    unordered_map<int, vector<int>> d;
    int n = arr.size();
    for (int i = 0; i < n; ++i) d[arr[i]].push_back(i);
    vector<long long> ans(n);
    for (auto& item : d) {
      auto& v = item.second;
      int m = v.size();
      long long val = 0;
      for (int e : v) val += e;
      val -= m * v[0];
      for (int i = 0; i < v.size(); ++i) {
        int delta = i >= 1 ? v[i] - v[i - 1] : 0;
        val += i * delta - (m - i) * delta;
        ans[v[i]] = val;
      }
    }
    return ans;
  }
};
func getDistances(arr []int) []int64 {
  d := make(map[int][]int)
  n := len(arr)
  for i, v := range arr {
    d[v] = append(d[v], i)
  }
  ans := make([]int64, n)
  for _, v := range d {
    m := len(v)
    val := 0
    for _, e := range v {
      val += e
    }
    val -= m * v[0]
    for i, p := range v {
      delta := 0
      if i >= 1 {
        delta = v[i] - v[i-1]
      }
      val += i*delta - (m-i)*delta
      ans[p] = int64(val)
    }
  }
  return ans
}

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