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Hackerrank Java BigDecimal Solution

Poor Man
Published
Updated
Reading time
2 min read

Java's BigDecimal class can handle arbitrary-precision signed decimal numbers. Let's test your knowledge of them!

Given an array, , of  real number strings, sort them in descending order — but wait, there's more! Each number must be printed in the exact same format as it was read from stdin, meaning that  is printed as , and  is printed as . If two numbers represent numerically equivalent values (e.g., ), then they must be listed in the same order as they were received as input).

Complete the code in the unlocked section of the editor below. You must rearrange array 's elements according to the instructions above.

Input Format

The first line consists of a single integer, , denoting the number of integer strings.
Each line  of the  subsequent lines contains a real number denoting the value of .

Constraints

  • Each  has at most  digits.

Output Format

Locked stub code in the editor will print the contents of array  to stdout. You are only responsible for reordering the array's elements.

Sample Input

Text
9
-100
50
0
56.6
90
0.12
.12
02.34
000.000

Sample Output

Text
90
56.6
50
02.34
0.12
.12
0
000.000
-100

Solution in java8#

Approach 1.

Python
        Arrays.sort(s, new Comparator<String>() {
	    @Override
	    public int compare(String o1, String o2) {
		if (o1 == null || o2 == null) {
		    return 0;
		}
		BigDecimal o1bd = new BigDecimal(o1);
		BigDecimal o2bd = new BigDecimal(o2);
		return o2bd.compareTo(o1bd);
	    }
	});

Approach 2.

Python
        s[n]="null";
		s[n+1]="null";
		Arrays.sort(s,0,n,Collections.reverseOrder(new Comparator<String>()
		{
			@Override
			public int compare(String s1,String s2)
			{
				BigDecimal b1=new BigDecimal(s1);
				BigDecimal b2=new BigDecimal(s2);
				return b1.compareTo(b2);
			}
		}));

Keep going

2020

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