Finding minimum possible degree of a graph(n,e)

The following code snippet is to find the minimum possible degree of a graph whose number of vertices(n) and edges(e) are given.

The algorithm is such that a vertex can have maximum degree equal to (n-1) but if it is having degree 1 it means there is one more vertex whose degree is one and if its degree is 2 then there are 2 vertices whose degree is 1.

The here is the code for finding minimum degree of a graph

import java.io.*;
import java.util.Arrays;
public class MinDegree {
        //function to find sum of an array
 static int sum(int[] arr){
  int sum = 0;
  for(int a: arr){
   sum += a;
  }
  return sum;
 }
        //function to find minimum of an array
 static int minimum(int[] arr){
  int min = Integer.MAX_VALUE;
  for(int i=0; i<arr.length; i++){
    if(arr[i] < min)
       min = arr[i];
   }
  }
  return min;
 }
 public static void main(String[] ar) throws Exception{
  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                //number of edges should not exceed n(n-1)/2
                System.out.println("Enter no of vertices and edges: n e");
  String[] in = br.readLine().split(" ");//input should be like 5 7
  int vertices = Integer.parseInt(in[0]);//parsing string into integers
  int edges = Integer.parseInt(in[1]);
  int[] degree = new int[vertices];//buffer to store degrees 
  int counter = 0;
  int round = 0;
  while(sum(degree) < (2*edges)){//sum of degrees must be less than 2*edges
   if(degree[counter] < (vertices-1)){
    degree[counter] += 1;
    degree[round+counter+1] += 1;//it will increase neighbours degree
    round++;
   }
   else{
    counter++;
    round = 0;
   }
  }
  System.out.println(minimum(degree));
 }
}

Happy Coding :)