556. Next Greater Element III

Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1.
Example 1:
Input: 12
Output: 21
 
Example 2:
Input: 21
Output: -1


// not my code, kinda know how to do it,
// need to do this later, get the logic right 
// and then implement 

class Solution {
    public int nextGreaterElement(int n) {
        char[] number = (n + "").toCharArray();
        
        int i, j;
        // I) Start from the right most digit and 
        // find the first digit that is
        // smaller than the digit next to it.
        for (i = number.length-1; i > 0; i--)
            if (number[i-1] < number[i])
               break;

        // If no such digit is found, its the edge case 1.
        if (i == 0)
            return -1;
            
         // II) Find the smallest digit on right side of (i-1)'th 
         // digit that is greater than number[i-1]
        int x = number[i-1], smallest = i;
        for (j = i+1; j < number.length; j++)
            if (number[j] > x && number[j] <= number[smallest])
                smallest = j;
        
        // III) Swap the above found smallest digit with 
        // number[i-1]
        char temp = number[i-1];
        number[i-1] = number[smallest];
        number[smallest] = temp;
        
        // IV) Sort the digits after (i-1) in ascending order
        Arrays.sort(number, i, number.length);
        
        long val = Long.parseLong(new String(number));
        return (val <= Integer.MAX_VALUE) ? (int) val : -1;
    }
}