2121. Intervals Between Identical Elements

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

 

Note: This question is the same as 2615: Sum of Distances.

Solutions

Solution 1

Python3JavaC++Go

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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