Given a perimeter P, the task is to find the number of right triangles possible with perimeter equal to p.
Examples:
Input: P = 12 Output: number of right triangles = 1 The only right angle possible is with sides hypotenuse = 5, perpendicular = 4 and base = 3. Input: p = 840 Output: number of right triangles = 8
So the aim is to find the number of solutions which satisfy equations a + b + c = p and a^{2} + b^{2} = c^{2}.
A naive approach is to run two loops for a(1 to p/2) and b(a+1 to p/3) then make c=pab and increase count by one if
An efficient approach can be found by little algebraic manipulation :
Since a + c > b or, p – b > b or, b < p/2. Thus iterating b from 1 to p/2, calculating a and storing only the whole number a would give all solutions for a given p. There are no right triangles are possible for odd p as right angle triangles follow the Pythagoras theorem. Use a list of pairs to store the values of a and band return the count at the end.
Below is the implementation of the above approach.
# python program to find the number of # right triangles with given perimeter # Function to return the count def countTriangles(p):
# making a list to store (a, b) pairs
store = []
# no triangle if p is odd
if p % 2 ! = 0 : return 0
else :
count = 0
for b in range ( 1 , p / / 2 ):
a = p / 2 * ((p  2 * b) / (p  b))
inta = int (a)
if (a = = inta ):
# make (a, b) pair in sorted order
ab = tuple ( sorted ((inta, b)))
# check to avoid duplicates
if ab not in store :
count + = 1
# store the new pair
store.append(ab)
return count
# Driver Code p = 840
print ( "number of right triangles = " + str (countTriangles(p)))

number of right triangles = 8
Time complexity: O(P)
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Improved By : siddhanthapliyal