Max / Min calculation for a list (Python)-CodePudding

I guess I didn’t search for the right keyword so ended not founding an existing answer…. I’m trying to get the best way of getting the max or min of a 2 lists with same length, taking the max or min of each position.

I got to a point as below

Input A = 725324
Input B = 9341

Expected Max calculation of A & B would be 729344 (taking the max number on that position)

I got to the point only aligning 2 numbers to have same digits, and zipped the 2 strings, then got stuck with below list… the way I’m going towards will be Max(L[1]), Max(L[2])…all the way to Max(L[digits - 1]),but not sure what’s the best way of expressing it in Python and if that’s the right approach.

List = [(7, 0), (2, 0), (5, 9), (3, 3), (2, 4), (4, 1)]

CodePudding user response：

You can use string padding str.rjust() to properly pad the "shorter" number:

a = 725324
b = 9341

a, b = map(str, (a, b))
w = max(map(len, (a, b)))

result = ""
for t in zip(a.rjust(w, "0"), b.rjust(w, "0")):
    result  = max(t)

result = int(result)

Output:

CodePudding user response：

One approach we can use for this problem is to use math (division and remainders):

a = 725324
b = 9341

digits = 0
num = 0
while a != 0 or b != 0:
  if a != 0 and b != 0:
    num  = 10 ** digits * max(a % 10, b % 10)
    a //= 10
    b //= 10
  elif a != 0:
    num  = 10 ** digits * (a % 10)
    a //= 10
  else:
    num  = 10 ** digits * (b % 10)
    b //= 10
  digits  = 1
print(num)

With this code, we will enter a while loop while either a or b is nonzero.

From there, if both a and b are nonzero, then we find which number has the greatest last digit, which we calculate by doing % 10. We then scale it up so that we can add it to a total variable, num by storing the number of digits we have added, digits, and raising 10 to the power of digits and multiplying that value by the remainder.

We then do this same process for if only a or if only b are nonzero, but without taking the other variable into account.

I hope this helped! Please let me know if you need any further clarification or details :)

CodePudding user response：

Here is another pure numerical solution using divmod and powers of 10:

def max_digit(A,B):
    out = 0
    exp = 0
    if A < B: # ensure looping on the largest number
        B,A = A,B
    while A>0:
        A,a = divmod(A,10) # get unit and shift
        B,b = divmod(B,10)
        out  = (10**exp)*max(a,b)
        exp  = 1
    return out

A = 725324
B =   9341

max_digit(A,B)
# 729344

CodePudding user response：

If I understand your problem well so here is my solution

def maximum_digits():
    a = input("nummber a :")
    b = input("nummber b :")

    # finding the difference lenght between two number
    c = len(a) - len(b)

    # making a list of each digits in a number so 725324 >>> [7,2,5,3,2,4]
    list_a = []
    a=int(a)
    while a > 0:
        list_a.append(a % 10)
        a //= 10

    list_b = []
    b=int(b)
    while b > 0:
        list_b.append(b % 10)
        b //= 10

    # making the two number have same lenght by adding zero
    if a >= b:
        for k in range(c):
            list_b.append(0)

    else:
        for k in range(c):
            list_a.append(0)

    final_list = []
    for i, j in zip(list_a, list_b):
        if i > j:
            final_list.append(i)
        else:
            final_list.append(j)

    number = ""
    for i in final_list[::-1]:
        number  = str(i)

    return number

print(maximum_digits())

CodePudding user response：

The solution by @ddejohn written even shorter:

a = 725324
b = 9341

a, b = map(str, (a, b))
w = max(map(len, (a, b)))

result = int("".join(max(t) for t in zip(a.rjust(w, "0"), b.rjust(w, "0"))))