Given a positive integer N, the task is to find the sum of the product of all the integers in the range [1, N] with their count of divisors.
Input: N = 3
Number of positive divisors of 1 is 1( i.e. 1 itself). Therefore, f(1) = 1.
Number of positive divisors of 2 is 2( i.e. 1, 2). Therefore, f(2) = 2.
Number of positive divisors of 3 is 2( i.e. 1, 3). Therefore, f(3) = 2.
So the answer is 1*f(1) + 2*f(2) + 3*f(3) = 1*1 + 2*2 + 3*2 = 11.
Input: N = 4
Here f(1) = 1, f(2) = 2, f(3) = 2 and f(4) = 3. So, the answer is 1*1 + 2*2 + 3*2 + 4*3 = 23.
Naive Approach: The naive approach is to traverse from 1 to N and find the sum of all the numbers with their count of divisors.
Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to consider the total contribution each number makes to the answer. Below are the steps:
- For any number X in the range [1, N], X contributes K to the sum for each K from 1 to N such that K is a multiple of X.
- Observe that the list of these K is of form i, 2i, 3i, …, Fi where F is the number of multiples of i between 1 and N.
- Hence, to find the sum of these numbers generated above is given by
i*(1+2+3 + … F) = i*(F*(F+1))/2.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(1)
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