how to write it in efficient way

Issue

This Content is from Stack Overflow. Question asked by Akash Sharma

Accept four integers as input and write a program to print these integers in non-decreasing order.

The input will be four integers in four lines. The output should be a single line with all the integers separated by a single space in non-decreasing order.

Note: There is no space after the fourth integer.

a = int(input())
b = int(input())
c = int(input())
d = int(input())

if (a<=b and a<=c and a<=d):
    if (b<=c and b<=d):
        if(c<=d):
            print(a,b,c,d)
        else:
            print(a,b,d,c)
    elif(c<=b and c<=d):
        if(b<=d):
            print(a,c,b,d)
        else:
            print(a,c,d,b)
    else:
        if(b<=c):
            print(a,d,b,c)
        else:
            print(a,d,c,b)
elif(b<=a and b<=c and b<=d):
    if(a<=c and a<=d):
        if(c<=d):
            print(b,a,c,d)
        else:
            print(b,a,d,c)
    elif(c<=a and c<=d):
        if(a<=d):
            print(b,c,a,d)
        else:
            print(b,c,d,a)
    else:
        if(a<=c):
            print(b,d,a,c)
        else:
            print(b,d,c,a)
elif(c<=a and c<=b and c<=d):
    if(a<=b and a<=d):
        if(a<=d):
            print(c,a,b,d)
        else:
            print(c,a,d,b)
    elif(b<=a and b<=d):
        if(a<=d):
            print(c,b,a,d)
        else:
            print(c,b,d,a)
    else:
        if(a<=b):
            print(c,d,a,b)
        else:
            print(c,d,b,a)
else:
    if(a<=b and a<=c):
        if(b<=c):
            print(d,a.b,c)
        else:
            print(d,a,c,b)
    elif(b<=a and b<=c):
        if(a<=c):
            print(d,b,a,c)
        else:
            print(d,b,c,a)
    else:
        if(a<=b):
            print(d,c,a,b)
        else:
            print(d,c,b,a)



Solution

This question is not yet answered, be the first one who answer using the comment. Later the confirmed answer will be published as the solution.

This Question and Answer are collected from stackoverflow and tested by JTuto community, is licensed under the terms of CC BY-SA 2.5. - CC BY-SA 3.0. - CC BY-SA 4.0.

people found this article helpful. What about you?