2615. Sum of Distances

Description

You are given a 0-indexed integer array nums. There exists an array arr of length nums.length, where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there is no such j, set arr[i] to be 0.

Return the array arr.

 

Example 1:

Input: nums = [1,3,1,1,2]
Output: [5,0,3,4,0]
Explanation: 
When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5. 
When i = 1, arr[1] = 0 because there is no other index with value 3.
When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3. 
When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4. 
When i = 4, arr[4] = 0 because there is no other index with value 2. 

Example 2:

Input: nums = [0,5,3]
Output: [0,0,0]
Explanation: Since each element in nums is distinct, arr[i] = 0 for all i.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 109

 

Note: This question is the same as 2121: Intervals Between Identical Elements.

Solutions

Solution 1

Python3JavaC++Go

class Solution:
    def distance(self, nums: List[int]) -> List[int]:
        d = defaultdict(list)
        for i, x in enumerate(nums):
            d[x].append(i)
        ans = [0] * len(nums)
        for idx in d.values():
            left, right = 0, sum(idx) - len(idx) * idx[0]
            for i in range(len(idx)):
                ans[idx[i]] = left + right
                if i + 1 < len(idx):
                    left += (idx[i + 1] - idx[i]) * (i + 1)
                    right -= (idx[i + 1] - idx[i]) * (len(idx) - i - 1)
        return ans

class Solution {
    public long[] distance(int[] nums) {
        int n = nums.length;
        long[] ans = new long[n];
        Map<Integer, List<Integer>> d = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            d.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
        }
        for (var idx : d.values()) {
            int m = idx.size();
            long left = 0;
            long right = -1L * m * idx.get(0);
            for (int i : idx) {
                right += i;
            }
            for (int i = 0; i < m; ++i) {
                ans[idx.get(i)] = left + right;
                if (i + 1 < m) {
                    left += (idx.get(i + 1) - idx.get(i)) * (i + 1L);
                    right -= (idx.get(i + 1) - idx.get(i)) * (m - i - 1L);
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<long long> distance(vector<int>& nums) {
        int n = nums.size();
        vector<long long> ans(n);
        unordered_map<int, vector<int>> d;
        for (int i = 0; i < n; ++i) {
            d[nums[i]].push_back(i);
        }
        for (auto& [_, idx] : d) {
            int m = idx.size();
            long long left = 0;
            long long right = -1LL * m * idx[0];
            for (int i : idx) {
                right += i;
            }
            for (int i = 0; i < m; ++i) {
                ans[idx[i]] = left + right;
                if (i + 1 < m) {
                    left += (idx[i + 1] - idx[i]) * (i + 1);
                    right -= (idx[i + 1] - idx[i]) * (m - i - 1);
                }
            }
        }
        return ans;
    }
};

func distance(nums []int) []int64 {
    n := len(nums)
    ans := make([]int64, n)
    d := map[int][]int{}
    for i, x := range nums {
        d[x] = append(d[x], i)
    }
    for _, idx := range d {
        m := len(idx)
        left, right := 0, -m*idx[0]
        for _, i := range idx {
            right += i
        }
        for i := range idx {
            ans[idx[i]] = int64(left + right)
            if i+1 < m {
                left += (idx[i+1] - idx[i]) * (i + 1)
                right -= (idx[i+1] - idx[i]) * (m - i - 1)
            }
        }
    }
    return ans
}

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