Total Appeal of A String - Last Index [JS]

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Description 

Solution: Last Index

Keep track of the last index of each character in s, on the fly.
To get the number of subarrays each character appears in, the formula is (n - i) * (i + 1)
(i + 1) -> number of starting subarrays ending with arr[i]
(n - i) -> number of different ending subarrays after arr[i]
e.g: arr = [1,2,3,4]
Let's look for the number of times arr[1] appears in a subarray:
(i + 1) = [[1,2],[2]]
(n - i) = [[2],[2,3],[2,3,4]]
(n - i) * (i + 1) = all the combinations (6) ->[[1,2],[1,2,3],[1,2,3,4],[2],[2,3],[2,3,4]]

However, we are only counting the distinct characters.
To solve this, keep track of the last index of each character, then exclude counting the subarrays which overlap with the last occurance.
The formula turns into: (n - i) * (i - lastIdx)

e.g: arr = [1,2,3,2,4]
Let's calculate the 'appeal' of arr[3].
We only want to count these subarrays (don't include the earlier 2) -> [[3,2],[3,2,4],[2],[2,4]]
The last occurance of 2 was at index 1.
(n - i) * (i - lastIdx) = (5 - 3) * (3 - 1) = 2 * 2 = 4

Time Complexity: O(n) 74ms
Space Complexity: O(1) 44.5MB

var appealSum = function(s) {
  let ans = 0, n = s.length;
  let lastIndex = Array(26).fill(-1);
  for (let i = 0; i < n; i++) {
    let charcode = s.charCodeAt(i) - 97;
    let lastIdx = lastIndex[charcode];
    ans += (n - i) * (i - lastIdx);
    lastIndex[charcode] = i;
  }  
  return ans;
};

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