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LeetCode 303. Range Sum Query - Immutable

Assuming use of Java, use the Arrays.split(bound) function (O(1)) [twice but still O(1)]
Then call arrays.length() (O(1)) for the # of elements between.

Not sure if this is what you're looking for, but this is what I came up with.

Could you please elaborate the question a little
Thanks

if max maximum element is not too large(<=10^6). we can make dp array where
dp[i]=number of elements less than or equal to i
then answer for query [l,r] is dp[r]-dp[l-1]

here o(1) means space complexity or time complexity?

here o(1) in what space or time?