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:
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++ program for the above approach #include <bits/stdc++.h> 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 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 |
# 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# 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. |
<script> // JavaScript Program to implement
// the above approach
// Function to find the minimum size
// of array with sum S and product P
function minimumSizeArray(S, P)
{
// Base Case
if (S == P) {
return 1;
}
// Iterate through all values of S
// and check the mentioned condition
for (let i = 2; i <= S; i++) {
let d = i;
if ((S / d) >= Math.pow(P, 1.0 / d)) {
return i;
}
}
// Otherwise, print "-1"
return -1;
}
// Driver Code
let S = 5, P = 6;
document.write(minimumSizeArray(S, P));
// This code is contributed by Potta Lokesh </script>
|
2
Time Complexity: O(log P)
Auxiliary Space: O(1)