Question

This looks very trivial and is not a home work question.

public void sum(int[] arr){
  for(int i=0;i<arr.length;i++)
  {
    for(int j=0;j<arr.length;j++)
      System.out.println(arr[i]+"+"+arr[j]+"+"+"="+(arr[i]+arr[j]));
  }   
}//end of sum function

This prints all the sum of each elements. This is O(n^2).

I want to know if this could be solved more efficiently.

Answer 1

Since A + B is equal to B + A, you could just check the elements after the initial element in index i:

public void sum(int[] arr){
  for(int i=0;i<arr.length;i++)
  {
    for(int j=i;j<arr.length;j++) //Note: j = i, not j = 0
      System.out.println(arr[i]+"+"+arr[j]+"+"+"="+(arr[i]+arr[j]));
  }   
}//end of sum function

It’s still O(n^2)/2, so the complexity is still basically quadratic.