# Program to count number of fraction pairs whose sum is 1 in python

Suppose we have a list of fractions where each fraction is individual lists [numerator, denominator] which represents the number (numerator / denominator). We have to find the number of pairs of fractions whose sum is 1.

So, if the input is like fractions = [[2, 7],[3, 12],[4, 14],[5, 7],[3, 4],[1, 4]], then the output will be 4, as (2/7 + 5/7), (3/12 + 3/4), (3/4 + 1/4), (4/14 + 5/7) are the four pairs which sum to 1.

To solve this, we will follow these steps:

• d := a new map
• ans := 0
• for each fraction i in fractions, do
• x := i[numerator]
• y := i[denominator]
• g := gcd of (x, y)
• x := x / g
• y := y / g
• temp_x := y - x
• temp_y := y
• if (temp_x, temp_y) is in d, then
• ans := ans + d[temp_x, temp_y]
• d[x, y] := 1 + (d[(x, y)] when it is available, otherwise 0)
• return ans

Let us see the following implementation to get better understanding:

## Example Code

Live Demo

class Solution:
def solve(self, fractions):
import math

d = {}
ans = 0
for i in fractions:
x = i[0]
y = i[1]
g = math.gcd(x, y)
x /= g
y /= g
temp_x = y - x
temp_y = y
if (temp_x, temp_y) in d:
ans += d[(temp_x, temp_y)]
d[(x, y)] = d.get((x, y), 0) + 1
return ans

ob = Solution()
fractions = [[2, 7],[3, 12],[4, 14],[5, 7],[3, 4],[1, 4]]
print(ob.solve(fractions))

## Input

[[2, 7],[3, 12],[4, 14],[5, 7],[3, 4],[1, 4]]

## Output

4