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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

Solutions

• Java
• C++
• Python
• Go

• 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;
      }
  };
  

• 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
  
  

• 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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