pystardust
commented
3 years ago python3
#!/bin/python3
coeff = [ 2, 2, 3 ]
rhs = 4
# the values can't go negative, so we can try to iterate over each term and try
# figuring out the linear equation in the remaining terms (1 less term)
# if there is only one term, check divisibility and return 1 and 0 respectively
def get_sol_num(coeff: list[int], rhs: int) -> int:
term_coeff = coeff[0]
# when only a single term is coeffs, then check divisibility
if len(coeff) == 1:
if rhs % coeff[0] == 0:
return 1
else:
return 0
term = 0
count = 0
# when more then one term then, recursively call and add the counts for
# each value this term can take without going negative
while (rhs - term_coeff * term) >= 0:
count+=get_sol_num(coeff[1:], rhs - term_coeff * term)
term+=1
return count
if __name__ == '__main__':
print(get_sol_num(coeff,rhs))
Coding Problems | Linear equation solution
Given a linear equation of n variables...
https://codingproblems.netlify.app/post/linear_solution/