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/*
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Suppose N is the length of given array.
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Number of subarrays including element arr[i] is
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i * (N-i) + (N-i) because there are N-i subarrays with arr[i] as first element
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and i * (N-i) subarrays with arr[i] as a not-first element. arr[i] appears in 
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(N-i) subarrays for each preceding element and therefore we have i*(N-i).
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Suppose i * (N-i) + (N-i) is `total`. Ceil(total / 2) is the number of odd-length subarrays and Floor(total / 2) is the number of even-length subarrays. 
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When total is odd, there is one more odd-length subarray because of a single-element subarray.
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For each number, we multiply its value with the total number of subarrays it appears and
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add it to a sum.
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*/
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var sumOddLengthSubarrays = function (arr) {
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  let sum = 0,
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    N = arr.length;
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  for (let i = 0; i < arr.length; i++) {
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    let total = i * (N - i) + (N - i);
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    sum += Math.ceil(total / 2) * arr[i];
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  }
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  return sum;
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  // T.C: O(N)
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  // S.C: O(1)
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};