Find a range of composite numbers of given length
Given an integer n, we need to find a range of positive integers such that all the number in that range are composite and length of that range is n. You may print anyone range in the case of more than one answer. A composite number is a positive integer that has at least one divisor other than 1 and itself (Source : wiki )
Input : 3 Output : [122, 124] Explanation 122, 123, 124 are all composite numbers
The solution is little tricky. Since there are many possible answers, we discuss a generalized solution here.
Let the length of range be n and range starts from a then, a, a+1, a+2, ...., a+n-1 all should be composite. So the problem boils down to finding such 'a'. If we closely observe p! (where p is a positive integers) then we will find that, p! has factors of 2, 3, 4, ..., p-1, Hence if we add i to p! such that 1 < i < p, then p! + i has a factor i, so p! + i must be composite. So we end up finding p! + 2, p! + 3, .... p! + p-1 are all composite and continuous integers forming a range [p! + 2, p! + p-1] The above range consists of p-2 elements. For a range of n elements we need to consider (n+2)! If we take a = (n+2)! + 2, Then, a + 1 = (n+2)! + 3 Then, a + 2 = (n+2)! + 4 ... Then, a + n-1 = (n+2)! + n+1 Hence, a = (n+2)! + 2 = 2*3*....*(n+2) + 2 a has 2 as its divisor because (n+2)! and 2 both divides 2 a + 1 = 2*3*....*(n+2) + 3 a + 1 has 3 as its divisor because (n+2)! and 3 both divides 3 ... a + n-1 = 2*3*....*(n+2) + n+1 a + n-1 has n+1 as its divisor because (n+2)! and n+1 both divides n+1 Therefore range will be [ (n+2)! + 2, ( (n+2)! + 2 ) + n-1]
Example for above algorithm
n = 3 Then a = (n+2)! + 2 a = 5! + 2 a + 1 = 5! + 3 a + 2 = 5! + 4 Here a is divisible by 2 Here a + 1 is divisible by 3 Here a + 2 is divisible by 4 Hence a, a+1, a+2 are all composites
Analysis of above algorithm
Time Complexity : O(n)
Auxiliary Space : O(1)
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