Given an interval of integers [A, B]. For each number in this interval compute its greatest odd divisor. Output the sum of these divisors.
Input : A = 1, B = 3 Output : 5 1 + 1 + 3 = 5 Input : A = 3, B = 9 Output : 29 3 + 1 + 5 + 3 + 7 + 1 + 9 = 29
Naive Approach :
A simple approach is to iterate through all numbers in the range find their greatest odd divisor but this algorithm has time complexity of O(n).
Efficient Approach :
We need to find the answer in range [ A, B ] if we can find answer in range [ 1, B ] and subtract it from [ 1, A -1 ] then we will get our required answer.
Here we can see that –
- The answer for an odd number X is X itself.
- The answer for an even number X is equal to the answer for X/2. This is true because X and X/2 have the same odd divisors.( if X = 4 then 4 and 2 both have 1 as greatest odd divisor).
If we want to find answer in range [1, N], then first we need to determine the sum of all the odd numbers
( 1, 3, 5, …) using a simple formula: the sum of the first K odd numbers is equal to K2. Then we need to add the answers for the even numbers (2, 4, 6, …). But these are actually equal to the answers for 1, 2, 3, …, floor(N/2), so we can call our function recursively for floor(N/2).
The complexity of this algorithm is O(log( N)).
Below is the implementation of the above idea :
# Python3 program to find sum of greatest
# odd divisor of numbers in given range
# Function to return sum of first
# n odd numbers
return n * n;
# Recursive function to return sum
# of greatest odd divisor of numbers
# in range [1, n]
if (n == 0):
if (n % 2 == 1):
# Odd n
return (square(int((n + 1) / 2)) +
sum(int(n / 2)));
# Even n
return (square(int(n / 2)) +
sum(int(n / 2)));
# Function to return sum of greatest
# odd divisor of numbers in range [a, b]
def oddDivSum(a, b):
return sum(b) – sum(a – 1);
# Driver code
a, b = 3, 9;
# This code is contributed by mits
Time Complexity : O(log(N))
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