Consecutive Prime Sum in JavaThe sum of consecutive prime numbers refers to the total obtained by adding up a series of prime numbers that follow each other in sequence. Consecutive PrimeTo find consecutive prime numbers that sum up to a given value in Java, we can use a sliding window approach. Some prime numbers can be expressed as a sum of other consecutive prime numbers. For example
Input Format: First line contains a number N. Output: Print the total number of all such prime numbers which are less or equal to N. The time complexity of the code is O(n^{2} * b + k) The actual time complexity may vary depending on the values of n and b. ConsecutiveSum.java Output: Enter no 10 4 Let's see another approach. Time complexity: O(sqrt(N)) Space complexity: O(1) ConsecutivePrimeSum.java Output: 50 3 Let's see another approach for the same. Assume N = (x+1)+(x+2)+...+(x+k), where x >= 0, k >= 1. We have 2N = k(2x+k+1), which has two factors, k and 2x+k+1, one of which is odd number and another is even. So the question now is how many ways to factor 2N into one even and one odd number. Assume 2N = 2^t * M, where M is odd. If we factor M = a * b, then multiply 2^t to one of them will yield the even number. So, the question now becomes how many ways to factor the odd part of N. If N = 3*3*3*5*5, how many ways to factor N into two numbers? For one number, we can pick 0 to 3 3s, and 0 to 2 5s, so we have 4 * 3 = 12 options. Here you can see the answer is multiplication of 1 + {count of a factor}. But what if there is a factor greater than sqrt(N)? In this case it have to be a prime number and there will be only one of it. So, if eventually N > 1, we multiply 2 to answer. ConsecutivePrimeSum.java Output: 6 Time complexity: O(sqrt(N) Space complexity: O(1) |