Question

You are given a set of n numbers. Give an O(n^2) algorithm (NOT O(n^3), O(n^2)) to decide if there exist three numbers a, b and c are in the set such that a + b = c

(Hint: sort the numbers first).

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

Efficient approach : The idea is similar to Find a triplet that sum to a given value.

• Sort the given array first.
• Start fixing the greatest element of three from back and traverse the array to find other two numbers which sum upto the third element.
• Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
• If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increase the j pointer, so as to increase the value of A[j] + A[k].
• If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k].

Below is the algorithm in coding style

sort(a); // non-descending
for (i = 0; i < n; i++) {
  j = i; k = i + 1;
  while (j < n && k < n) {
    if (a[i] + a[j] == a[k])
      return true;
    else if (a[i] + a[k] < a[j])
      k++;
    else
      j++;
  }
}
return false;

Since there are 2 nested loops involved. So, it is O(n^2)

Time complexity: O(N^2)

Kindly revert for any queries

Thanks.