Minimum size of Array possible with given sum and product values

• Last Updated : 08 Oct, 2021

Given two positive integers S and P, the task is to find the minimum possible size of the array such that the sum of elements is S and the product of the elements is P. If there doesn’t exist any such array, then print “-1”.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: S = 5, P = 6
Output: 2
Explanation: The valid array can be {2, 3}, which is of minimum size.

Input: S = 5, P = 100
Output: -1

Approach: The given problem can be solved based on the following observations:

• Using N numbers, an array can be formed of size N having sum S.
• Any product values can be achieved when the value of P is between [0, (S/N)N].

Follow the steps below to solve the given problem:

• Initially check if the value of S and P are the same, then return 1 as the S value itself is used to make a minimum size array.
• Iterate over the range [2, S] using the variable i, and if the value of (S/i) >= pow(P, 1/i) then print the value of i as the resultant minimum size of the array formed.
• After completing the above steps, if there doesn’t any possible value i satisfying the above criteria, then print “-1”.

Below is the implementation of the above approach:

C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum size``// of array  with sum S and product P``int` `minimumSizeArray(``int` `S, ``int` `P)``{``    ``// Base Case``    ``if` `(S == P) {` `        ``return` `1;``    ``}` `    ``// Iterate through all values of S``    ``// and check the mentioned condition``    ``for` `(``int` `i = 2; i <= S; i++) {` `        ``double` `d = i;``        ``if` `((S / d) >= ``pow``(P, 1.0 / d)) {``            ``return` `i;``        ``}``    ``}` `    ``// Otherwise, print "-1"``    ``return` `-1;``}` `// Driver Code``int` `main()``{``    ``int` `S = 5, P = 6;``    ``cout << minimumSizeArray(S, P);` `    ``return` `0;``}`

Java

 `// Java program for the above approach``class` `GFG{` `// Function to find the minimum size``// of array  with sum S and product P``static` `int` `minimumSizeArray(``int` `S, ``int` `P)``{``    ``// Base Case``    ``if` `(S == P) {` `        ``return` `1``;``    ``}` `    ``// Iterate through all values of S``    ``// and check the mentioned condition``    ``for` `(``int` `i = ``2``; i <= S; i++) {` `        ``double` `d = i;``        ``if` `((S / d) >= Math.pow(P, ``1.0` `/ d)) {``            ``return` `i;``        ``}``    ``}` `    ``// Otherwise, print "-1"``    ``return` `-``1``;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `S = ``5``, P = ``6``;``       ``System.out.println(minimumSizeArray(S, P));``}``}` `// This code is contributed by AnkThon`

Python3

 `# python program for the above approach` `# Function to find the minimum size``# of array with sum S and product P``def` `minimumSizeArray(S, P):` `    ``# Base Case``    ``if` `(S ``=``=` `P):``        ``return` `1` `    ``# Iterate through all values of S``    ``# and check the mentioned condition``    ``for` `i ``in` `range``(``2``, S``+``1``):` `        ``d ``=` `i``        ``if` `((S ``/` `d) >``=` `pow``(P, ``1.0` `/` `d)):``            ``return` `i` `    ``# Otherwise, print "-1"``    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``S ``=` `5``    ``P ``=` `6``    ``print``(minimumSizeArray(S, P))` `    ``# This code is contributed by rakeshsahni`

C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the minimum size``// of array  with sum S and product P``static` `int` `minimumSizeArray(``int` `S, ``int` `P)``{``    ``// Base Case``    ``if` `(S == P) {` `        ``return` `1;``    ``}` `    ``// Iterate through all values of S``    ``// and check the mentioned condition``    ``for` `(``int` `i = 2; i <= S; i++) {` `        ``double` `d = i;``        ``if` `((S / d) >= Math.Pow(P, 1.0 / d)) {``            ``return` `i;``        ``}``    ``}` `    ``// Otherwise, print "-1"``    ``return` `-1;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `S = 5, P = 6;``    ``Console.Write(minimumSizeArray(S, P));``}``}` `// This code is contributed by SURENDRA_GANGWAR.`

Javascript

 ``

Output:
`2`

Time Complexity: O(log P)
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up