Find the maximum integer, that can be obtained by permutation of numbers of an arbitrary three-digit positive integer n (100<=n<=999)

Question

Find the maximum integer, that can be obtained by permutation of numbers of an arbitrary three-digit positive integer n (100<=n<=999).

public static int maxNum(int n) {
int j , k ,l, result;
String  q, z, y ,w;

    do {
        j = n % 10;
        k = n % 100/ 10;
        l = n % 1000 /100;

        q =Integer.toString(k);
        z =Integer.toString(j);
        y =Integer.toString(l);
        w= (q+z+y);
        result =Integer.parseInt(w);

        break;
    } while (n >=100 && n<=999);

    return result;
}


``` input 

165

output 

651    

input 123 

output 
321

Answer 1

You need to order the digits in descending order and then build up the 3 digit value accordingly.

- 152 - 1, 5, 2
- in ascending order is 5, 2, 1
- ( 10 * 5) + 2) * 10) + 1 = 521

Here is one way to do it with streams.

Random r = new Random();

• generate a value between 100 and 999 inclusive
• convert to a String and split the digit characters
• sort in descending order
• convert each integer to a numeric digit
• and use reduce to get the largest number

for (int i = 0; i < 10; i++) {
     int value = r.nextInt(100, 1000);
     int max = Arrays.stream(Integer.toString(value).split(""))
             .sorted(Comparator.reverseOrder())
             .mapToInt(Integer::parseInt)
             .reduce(0, (a, b) -> a * 10 + b);
     System.out.println(value + " -> " + max);
}

prints something like the following:

105 -> 510
694 -> 964
983 -> 983
432 -> 432
792 -> 972
868 -> 886
979 -> 997
839 -> 983
803 -> 830
911 -> 911

You can also do it without sorting and using just a single while loop.

• the while loop simply replaces the sorting by ordering the 3 digits in decreasing order.
• then max1, max2, and max3 are combined as previously explained.

for (int i = 0; i < 10; i++) {
     int value = r.nextInt(100,1000);
     int max1 = 0,max2 = 0,max3 = 0;
     int temp = value; // save for later printing.
     while (value > 0) {
         int d = value % 10; // get last digit of value
      
         if (max1 < d) {
             max3 = max2;
             max2 = max1;
             max1 = d;
         }
         else if (max2 < d) {
             max3 = max2;
             max2 = d;
         }
         else if (max3 < d) {
             max3 = d;
         }
         value /= 10;  // expose next digit
     }
     
     int max = (((max1*10)+max2)*10)+max3;
     System.out.println(temp + " -> " + max);     
 }