Sum of Distances - Prefix Sum [JS]

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Description 

Solution: Prefix Sum Per Value

  1. Collect the indices of each group by value.
  2. For each group of indices, count the running prefix sum from the left and right and add the sum of the differences of indices to the answer.
  • From left: ans[indices[j]] += indices[j] * (j + 1) - leftSum
  • From right: ans[indices[j]] += rightSum - indices[j] * (indices.length - j)

The idea is we use prefix sum to calculate the sum of all differences in indices, this way we don't have to loop through and compare each individual pair of indices.

Time Complexity: O(n) 367ms
Space Complexity: O(n) 87.5MB

var distance = function(nums) {
  let n = nums.length, map = {};
  for (let i = 0; i < n; i++) {
    if (!map[nums[i]]) map[nums[i]] = [];
    map[nums[i]].push(i);
  }
  let ans = Array(n).fill(0);
  for (let num in map) {
    let indices = map[num], m = indices.length, leftSum = 0;
    for (let j = 0; j < m; j++) {
      leftSum += indices[j];
      ans[indices[j]] += indices[j] * (j + 1) - leftSum;
    }
    let rightSum = 0;
    for (let j = m - 1; j >= 0; j--) {
      rightSum += indices[j];
      ans[indices[j]] += rightSum - indices[j] * (m - j);
    }
  }
  return ans;
};
Location: Auckland, New Zealand

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