Find the next greater number

Given a number find the smallest number that is greater than n, and has same number of digits as n.

 
Input:  n = "218765"
Output: "251678"

Input:  n = "1234"
Output: "1243"

Input: n = "4321"
Output: "Not Possible"

Input: n = "534976"
Output: "536479"

 Steps:

 Consider  3,2,8,4,1

 1. Start from right and find the digit that is smaller than its next digit.
    3,2 < 8 >4 >1

2. Swap that digit with its next higher number in the right.

     3, 2, 8, 4 , 1    (next higher digit of 2 is 4)

     3,4,8,2,1

 3. Reverse the digits to the right of that digit (4).

    3,4,1,2,8
Code:

 import java.util.*;

 public class Solution {
    
    static void findNextGreater(int[] a){
        int n = a.length;
        int position = -1;
        
        for(int i=n-1; i>0; i--){
            if(a[i]>a[i-1]){
                position = i-1;
                break;
            }
        }
        
        int smallest = position+1;
        
        //here we have to find the next higher number. The number has to be higher than a[position]
        //and smaller than other digits, we initialize smallest to a[poisition+1] as it is
        // higher than a[poistion]
        for(int i=position+2; i<n; i++){
            if(a[i]>a[position] && a[i] < a[smallest])
                smallest = i;
        }
        
        swap(a, position,smallest);
        reverse(a, position+1);
        
    }
    
    static void swap(int[] a, int position, int nexthigh){
        int temp = a[position];
        a[position] = a[nexthigh];
        a[nexthigh] = temp;
    }
    
    static void reverse(int[] a, int start){
        int n = a.length;
        for(int i=start, j=n-1; i<j; i++,j--){
            int temp = a[i];
            a[i] = a[j];
            a[j] = temp;
        }
    }
    
    public static void main(String[] args) {
        int[] a = {3,2,8,4,1} ;
        findNextGreater(a);
        for(int i=0; i<a.length; i++){
            System.out.print(a[i]);
        }
     }
}
Output:

 
34128



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