Given n, find x, y, z such that x, y, z satisfy the equation “2/n = 1/x + 1/y + 1/z”
There are multiple x, y and z that satisfy the equation print anyone of them, if not possible then print -1.
Input : 3 Output : 3 4 12 Explanation: here 3 4 and 12 satisfy the given equation Input : 7 Output : 7 8 56
Note that for n = 1 there is no solution, and for n > 1 there is solution x = n, y = n+1, z = n·(n+1).
To come to this solution, represent 2/n as 1/n+1/n and reduce the problem to represent 1/n as a sum of two fractions. Let’s find the difference between 1/n and 1/(n+1) and get a fraction 1/(n*(n+1)), so the solution is
2/n = 1/n + 1/(n+1) + 1/(n*(n+1))
x is 7 y is 8 z is 56
Time complexity: O(1)
We can write 2/n = 1/n + 1/n. And further as 2/n = 1/n + 1/2n + 1/2n.
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