Given an integer N. The task is to count the number of ordered pairs (a, b) such that .
Input: N = 5 Output: 8 Ordered Pairs are = (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1), (4, 1). Input: N = 1000 Output: 7053
Naive Approach: Run two for loops upto N – 1 and count ordered pairs whose product are less than N.
Efficient Approach: Let’s considered an ordered pair (a, b). Then, if the product of two numbers is less than n i:e a * b < n, then at least one of them must be less then square root of n. We can proof by contradiction that if both numbers are greater then square root of n the their product is not less than n.
So, you can take all integers a up to sqrt(n – 1) instead of all integers up to n. For each a, count the number of b >= a such that a * b < n. Then multiply the result by two to count the pair (b, a) for each pair (a, b) you saw. After that, subtract the integer part of sqrt(n – 1) to ensure the pairs (a, a) were counted exactly once.
Below is the implementation of the above approach:
Time Complexity: O(N*sqrt(N))
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