Hackerrank Classes: Dealing with Complex Numbers Solution

Hackerrank Classes: Dealing with Complex Numbers Solution

For this challenge, you are given two complex numbers, and you have to print the result of their addition, subtraction, multiplication, division and modulus operations.

The real and imaginary precision part should be correct up to two decimal places.

Input Format

One line of input: The real and imaginary part of a number separated by a space.

Output Format

For two complex numbers  and , the output should be in the following sequence on separate lines:


For complex numbers with non-zero real and complex part, the output should be in the following format:

Replace the plus symbol  with a minus symbol  when .

For complex numbers with a zero complex part i.e. real numbers, the output should be:

For complex numbers where the real part is zero and the complex part is non-zero, the output should be:

Sample Input

2 1
5 6

Sample Output

7.00+7.00i
-3.00-5.00i
4.00+17.00i
0.26-0.11i
2.24+0.00i
7.81+0.00i

Concept

Python is a fully object-oriented language like C++, Java, etc. For reading about classes, refer here.

Methods with a double underscore before and after their name are considered as built-in methods. They are used by interpreters and are generally used in the implementation of overloaded operators or other built-in functionality.

__add__-> Can be overloaded for + operation
__sub__ -> Can be overloaded for - operation



For more information on operator overloading in Python, refer here

Solution in python3

Approach 1.

python
parts = list(map(float, input().split())) a = complex(parts[0], parts[1]) parts = list(map(float, input().split())) b = complex(parts[0], parts[1]) def p(comp): if comp.real != 0 and comp.imag != 0: print('%.2f %s %.2fi' % (comp.real, '+' if comp.imag >= 0 else '-', abs(comp.imag))) elif comp.imag == 0: print("%.2f" % comp.real) elif comp.real == 0: print("%.2fi" % comp.imag) p(a+b) p(a-b) p(a*b) p(a/b) print("%.2f" % abs(a)) print("%.2f" % abs(b))

Approach 2.

python
p = list(map(float, input().split())) A = p[0] + p[1] * 1j p = list(map(float, input().split())) B = p[0] + p[1] * 1j def printc(A): sign = "" if A.imag >= 0: sign = "+" else: sign = "-" if A.imag == 0: print("%.2f" % A.real) elif A.real != 0: print("%.2f" % A.real, sign, "%.2fi" % abs(A.imag)) else: print("%.2fi" % A.imag) printc(A + B) printc(A - B) printc(A * B) printc(A / B) print("%.2f" % abs(A)) print("%.2f" % abs(B))

Approach 3.

python
import sys def output(num): real = num.real imag = num.imag if real == 0 and imag == 0: print('0.00') elif real == 0: print('{:.2f}i'.format(imag)) elif imag == 0: print('{:.2f}'.format(real)) else: print('{:.2f} {} {:.2f}i'.format(real, '-' if imag < 0 else '+', abs(imag))) real1, imag1 = [float(x) for x in input().split()] real2, imag2 = [float(x) for x in input().split()] C1 = complex(real1, imag1) C2 = complex(real2, imag2) output(C1 + C2) output(C1 - C2) output(C1 * C2) output(C1 / C2) print('{:.2f}'.format(abs(C1))) print('{:.2f}'.format(abs(C2)))

Solution in python

Approach 1.

python
import math def get_num(): return map(float, raw_input().strip().split()) c = get_num() d = get_num() def output(a, b): sa = "%.2f" % (a,) sb = "%.2f" % (abs(b),) if 0 == b: print sa elif 0 == a: print "%.2fi" % (b,) else: c = "+" if b < 0: c = "-" print "%s %s %si" % (sa, c, sb) def mod(lst): return math.sqrt(lst[0]*lst[0] + lst[1]*lst[1]) output(c[0]+d[0], c[1]+d[1]) output(c[0]-d[0], c[1]-d[1]) output(c[0]*d[0] - c[1]*d[1], c[0]*d[1] + c[1]*d[0]) output((c[0]*d[0] + c[1]*d[1])/(mod(d) * mod(d)), (- c[0]*d[1] + c[1]*d[0])/(mod(d) * mod(d))) output(mod(c), 0) output(mod(d), 0)

Approach 2.

python
import math class Complex: def __init__(self, re, im): self.re = re self.im = im def mod(self): return math.sqrt(self.re*self.re + self.im*self.im) def __add__(self, c2): return Complex(self.re + c2.re, self.im + c2.im) def __sub__(self, c2): return Complex(self.re - c2.re, self.im - c2.im) def __mul__(self, c2): return Complex(self.re*c2.re - self.im*c2.im, self.im*c2.re + self.re*c2.im) def __div__(self, c2): c2_mod2 = float(c2.re*c2.re + c2.im*c2.im) return Complex((self.re*c2.re + self.im*c2.im)/c2_mod2, (self.im*c2.re - self.re*c2.im)/c2_mod2); def conjugate(self): return Complex(self.re, -self.im) def __str__(self): if self.im == 0: return '%.2f' % self.re if self.re == 0: return '%.2fi' % self.im if self.im > 0: return '%.2f + %.2fi' % (self.re, self.im) else: return '%.2f - %.2fi' % (self.re, -self.im) line = raw_input().split(' ') c1 = Complex(float(line[0]), float(line[1])) line = raw_input().split(' ') c2 = Complex(float(line[0]), float(line[1])) print c1 + c2 print c1 - c2 print c1 * c2 print c1 / c2 print '%.2f' % c1.mod() print '%.2f' % c2.mod()

Approach 3.

python
# Enter your code here. Read input from STDIN. Print output to STDOUT import math class Calc(object): def __init__(self,r,i): self.r = r self.i = i [a,b] = [float(j) for j in raw_input().split()] A = Calc(a,b) [a,b] = [float(j) for j in raw_input().split()] B = Calc(a,b) #Addition addr = (A.r+B.r) addi = (A.i+B.i) if addi == 0: print "%.2f"%addr elif addr == 0: print "%.2fi"%addi else: if addi > 0: print "%.2f"%addr+" + "+"%.2fi"%addi else: print "%.2f"%addr+" - "+"%.2fi"%(addi*-1) #Subtraction addr = (A.r-B.r) addi = (A.i-B.i) if addi == 0: print "%.2f"%addr elif addr == 0: print "%.2fi"%addi else: if addi > 0: print "%.2f"%addr+" + "+"%.2fi"%addi else: print "%.2f"%addr+" - "+"%.2fi"%(addi*-1) #Multiplication addr = (A.r*B.r - A.i*B.i) addi = (A.r*B.i+B.r*A.i) if addi == 0: print "%.2f"%addr elif addr == 0: print "%.2fi"%addi else: if addi > 0: print "%.2f"%addr+" + "+"%.2fi"%addi else: print "%.2f"%addr+" - "+"%.2fi"%(addi*-1) #Division addr = (A.r*B.r + A.i*B.i)/(B.r**2 + B.i**2) addi = (B.r*A.i - A.r*B.i)/(B.r**2 + B.i**2) if addi == 0: print "%.2f"%addr elif addr == 0: print "%.2fi"%addi else: if addi > 0: print "%.2f"%addr+" + "+"%.2fi"%addi else: print "%.2f"%addr+" - "+"%.2fi"%(addi*-1) #Mod print "%.2f"%math.sqrt(A.r**2+A.i**2) print "%.2f"%math.sqrt(B.r**2+B.i**2)

Subscribe to The Poor Coder | Algorithm Solutions

Don’t miss out on the latest issues. Sign up now to get access to the library of members-only issues.
jamie@example.com
Subscribe