Before attempting this problem, you should be comfortable with:
We need to find the index of the largest element, but we can only compare subarray sums, not individual elements. The key observation is that if we split the array into two equal halves and compare their sums, the half containing the largest element will have a greater sum (since all other elements are identical). This naturally leads to binary search: we repeatedly halve the search space based on which half has the larger sum.
left to 0 and length to the total array length.length > 1:left.left to the right subarray.length becomes 1, return left as the index of the largest element.class Solution(object):
def getIndex(self, reader):
left = 0
length = reader.length()
while (length > 1):
length //= 2
cmp = reader.compareSub(left, left + length - 1, left + length,
left + length + length - 1)
if cmp == 0:
return left + length + length
if cmp < 0:
left += length
return leftWhere is the length of the internal array.
When the array length is odd, dividing it into two equal halves leaves one element in the middle. If compareSub returns 0 (both halves have equal sums), the larger integer must be this middle element. Failing to handle this case causes the algorithm to return an incorrect index or loop indefinitely.
The compareSub API uses inclusive indices. A common mistake is miscalculating the end indices of the subarrays. For two adjacent subarrays of length len starting at left, the ranges are [left, left + len - 1] and [left + len, left + 2*len - 1]. Using left + len as the end of the first subarray causes overlapping or gaps.
After comparing subarrays, the search should continue in the half with the larger sum. A subtle bug is updating left incorrectly when the right subarray is larger. The new left should be left + length, not left + length + 1, since we are narrowing to exactly the right half.