given a number p , find two elements in array whose product = P

Make a pass through the array, and add the elements to a Hashtable. For each element x added, check whether P/x already exists in the Hashtable - if it does then x and P/x is one of your solutions. This'd be about as optimal as you'll get.

You can try a sliding window approach. First sort all the numbers increasingly, and then use two integers begin and end to index the current pair of numbers. Initialize begin to 0 and end to the last position. Then compare the product of v[begin] and v[end] with P:

• If it is equal, you found the answer.
• If it is lower, you must find a bigger product, move begin forward.
• If it is higher, you must find a smaller product, move end backward.

Here is a C++ code with this idea implemented. This solution is O(n*log(n)) because of the sorting, if you can assume the data is sorted then you can skip the sorting for an O(n) solution.

pair<int, int> GetProductPair(vector<int>& v, int P) {
  sort(v.begin(), v.end());
  int begin = 0, end = static_cast<int>(v.size()) - 1;
  while (begin < end) {
    const int prod = v[begin] * v[end];
    if (prod == P) return make_pair(begin, end);
    if (prod < P) ++begin;
    else --end;
  }
  return make_pair(-1, -1);
}

This one would work only for integers: Decompose P as product of prime numbers. By dividing these in two groups you can obtain the pairs that gives P as product. Now you just have to check both of them are present in the array, this is where a hash table would be very useful. Also, while creating the hash table, you could also filter the array of repeating values, values that are greater than P, or even values that have prime factors not contained in P.