Minimize the cost of selecting two numbers whose product is X

• Difficulty Level : Medium
• Last Updated : 24 Nov, 2021

Prerequisite: To Find Maximum and Minimum Prime Number
Given four integers A, B, C and X, the task is to minimize the cost of selecting two numbers N and M such that the product of N and M is equal to X, i.e. N * M = X. The cost to select the numbers N and M is decided as follows:

1. For the first number N:
• The cost is A if N is a prime number.
• The cost is B if N is a composite number.
• The cost is C if N is 1.
2. For the second number M (!= 1), the cost is M.

Examples:

Input: A = 7, B = 11, C = 2, X = 20
Output: 11
Explanation:
The following are the possible values and the cost for each pair:
Let N = 1 and M = 20, cost used in selecting N as 1 is C = 2, Total cost = 2 + 20 = 22.
Let N = 2 and M = 10, cost used in selecting N as prime number is A = 7, Total cost = 7 + 10 = 17
Let N = 4 and M = 5, cost used in selecting N as composite number is B = 11, Total cost = 11 + 5 = 15
Let N = 5 and M = 4, cost used in selecting N as prime number is A = 7, Total cost = 7 + 4 = 11
Let N = 10 and M = 2, cost used in selecting N as composite number is B = 11, Total cost = 11 + 2 = 13
Minimum among all the above is 11.
Input: A = 1, B = 1, C = 1, X = 40
Output:
Explanation:
The minimum cost is when N = 20 and M = 2, Total cost = 1 + 2 = 3.

Naive Approach: Find all the factors of the number and check if the factor is prime or not, accordingly find the cost and select minimum from them.
Efficient Approach: There can be three possible cases for N as follows:

1. N is Prime: Then the first number cost is fixed. In order to minimize the cost, the highest prime number possible is chosen such that N ≠ X.
2. N is Composite: Similar to the above case, the maximum composite number that divides the number, but not the number itself has to be found. In order to do this, the minimum prime number that divides X is found and this is considered as M and cost is computed.
3. N is 1: Any number can be formed(except when X = 1) and M = X for this case.

Therefore, the idea is to compute the cost for all the three cases and find the minimum among all. In order to do this, the list of all the minimum prime factors and maximum prime factors is precomputed using a slight variation of Sieve of Eratosthenes and stored in an array. The cost for all the three cases can be easily computed from these arrays.
Below is the implementation of the above approach:

C++

 // C++ implementation of// the above approach #include using namespace std; const int MAX = 100000; // max_prime[i] represents maximum prime// number that divides the number iint max_prime[MAX]; // min_prime[i] represents minimum prime// number that divides the number iint min_prime[MAX]; // Function to store the minimum// prime factor and the maximum// prime factor in two arraysvoid sieve(int n){    for (int i = 2; i <= n; ++i) {         // Check for prime number        // if min_prime[i] > 0,        // then it is not a prime number        if (min_prime[i] > 0) {            continue;        }         // If i is a prime number,        // then both minimum and maximum        // prime numbers that divide        // the number is the number itself        min_prime[i] = i;        max_prime[i] = i;         int j = i + i;         while (j <= n) {            if (min_prime[j] == 0) {                 // If this number is being visited                // for first time then this divisor                // must be the smallest prime number                // that divides this number                min_prime[j] = i;            }             // Update prime number till the last            // prime number that divides this number             // The last prime number that            // divides this number will be maximum.            max_prime[j] = i;            j += i;        }    }} // Function to minimize the cost of finding// two numbers for every number such that// the product of those two is equal to Xint findCost(int A, int B, int C, int X){    // Pre-calculation    sieve(MAX);     int N, M;     // If X == 1, then there is no way to    // find N and M. Print -1    if (X == 1) {        return -1;    }     // Case 3 is always valid and cost for that    // is C + X C for choosing 1 and M = X/1    int min_cost = C + X;     // Case 1    // N is prime, first number cost is fixed    // N is max_prime number divides this number    int cost_for_prime = A;    N = max_prime[X];     // If X is prime then the maximum prime number    // is the number itself. For this case,    // M becomes 1 and this shouldn't be considered.    if (N != X) {         // Find M for this case        M = X / N;         // Add cost for the second number also        cost_for_prime += M;         // Update min_cost, if the        // cost for prime is minimum        min_cost = min(min_cost, cost_for_prime);    }     // Case 2    // If N is composite    // For this find the minimum prime number    // that divides A[i] and consider this as M    M = min_prime[X];     // Find N for that number    N = X / M;     // Check if this number is composite or not    // if N is prime then there is no way    // to find any composite number that divides X    // If N = min_prime[N] then N is prime    if (N != min_prime[N]) {        int cost_for_comp = B + M;         // Update min_cost, if the        // cost for the composite is minimum        min_cost = min(min_cost, cost_for_comp);    }     return min_cost;} // Driver codeint main(){    int A = 7, B = 11, C = 2, X = 20;     cout << findCost(A, B, C, X) << " ";    return 0;}

Java

 // Java implementation of the above approachclass GFG {         static final int MAX = 1000;         // max_prime[i] represents maximum prime    // number that divides the number i    static int max_prime[] = new int[MAX];         // min_prime[i] represents minimum prime    // number that divides the number i    static int min_prime[] = new int[MAX];         // Function to store the minimum    // prime factor and the maximum    // prime factor in two arrays    static void sieve(int n)    {        for (int i = 2; i < n; ++i) {                 // Check for prime number            // if min_prime[i] > 0,            // then it is not a prime number            if (min_prime[i] > 0) {                continue;            }                 // If i is a prime number,            // then both minimum and maximum            // prime numbers that divide            // the number is the number itself            min_prime[i] = i;            max_prime[i] = i;                 int j = i + i;                 while (j < n) {                if (min_prime[j] == 0) {                         // If this number is being visited                    // for first time then this divisor                    // must be the smallest prime number                    // that divides this number                    min_prime[j] = i;                }                     // Update prime number till the last                // prime number that divides this number                     // The last prime number that                // divides this number will be maximum.                max_prime[j] = i;                j += i;            }        }    }         // Function to minimize the cost of finding    // two numbers for every number such that    // the product of those two is equal to X    static int findCost(int A, int B, int C, int X)    {        // Pre-calculation        sieve(MAX);             int N, M;             // If X == 1, then there is no way to        // find N and M. Print -1        if (X == 1) {            return -1;        }             // Case 3 is always valid and cost for that        // is C + X C for choosing 1 and M = X/1        int min_cost = C + X;             // Case 1        // N is prime, first number cost is fixed        // N is max_prime number divides this number        int cost_for_prime = A;        N = max_prime[X];             // If X is prime then the maximum prime number        // is the number itself. For this case,        // M becomes 1 and this shouldn't be considered.        if (N != X) {                 // Find M for this case            M = X / N;                 // Add cost for the second number also            cost_for_prime += M;                 // Update min_cost, if the            // cost for prime is minimum            min_cost = Math.min(min_cost, cost_for_prime);        }             // Case 2        // If N is composite        // For this find the minimum prime number        // that divides A[i] and consider this as M        M = min_prime[X];             // Find N for that number        N = X / M;             // Check if this number is composite or not        // if N is prime then there is no way        // to find any composite number that divides X        // If N = min_prime[N] then N is prime        if (N != min_prime[N]) {            int cost_for_comp = B + M;                 // Update min_cost, if the            // cost for the composite is minimum            min_cost = Math.min(min_cost, cost_for_comp);        }             return min_cost;    }         // Driver code    public static void main (String[] args)    {        int A = 7, B = 11, C = 2, X = 20;             System.out.println(findCost(A, B, C, X));    } } // This code is contributed by AnkitRai01

Python3

 # Python3 implementation of# the above approach MAX = 10000; # max_prime[i] represents maximum prime# number that divides the number imax_prime = *MAX; # min_prime[i] represents minimum prime# number that divides the number imin_prime = *MAX; # Function to store the minimum# prime factor and the maximum# prime factor in two arraysdef sieve(n) :     for i in range(2, n) :         # Check for prime number        # if min_prime[i] > 0,        # then it is not a prime number        if (min_prime[i] > 0) :            continue;         # If i is a prime number,        # then both minimum and maximum        # prime numbers that divide        # the number is the number itself        min_prime[i] = i;        max_prime[i] = i;         j = i + i;         while (j < n) :            if (min_prime[j] == 0) :                 # If this number is being visited                # for first time then this divisor                # must be the smallest prime number                # that divides this number                min_prime[j] = i;             # Update prime number till the last            # prime number that divides this number             # The last prime number that            # divides this number will be maximum.            max_prime[j] = i;            j += i; # Function to minimize the cost of finding# two numbers for every number such that# the product of those two is equal to Xdef findCost(A, B, C, X) :     # Pre-calculation    sieve(MAX);     # If X == 1, then there is no way to    # find N and M. Print -1    if (X == 1) :        return -1;     # Case 3 is always valid and cost for that    # is C + X C for choosing 1 and M = X/1    min_cost = C + X;     # Case 1    # N is prime, first number cost is fixed    # N is max_prime number divides this number    cost_for_prime = A;    N = max_prime[X];     # If X is prime then the maximum prime number    # is the number itself. For this case,    # M becomes 1 and this shouldn't be considered.    if (N != X) :         # Find M for this case        M = X // N;         # Add cost for the second number also        cost_for_prime += M;         # Update min_cost, if the        # cost for prime is minimum        min_cost = min(min_cost, cost_for_prime);     # Case 2    # If N is composite    # For this find the minimum prime number    # that divides A[i] and consider this as M    M = min_prime[X];     # Find N for that number    N = X // M;     # Check if this number is composite or not    # if N is prime then there is no way    # to find any composite number that divides X    # If N = min_prime[N] then N is prime    if (N != min_prime[N]) :        cost_for_comp = B + M;         # Update min_cost, if the        # cost for the composite is minimum        min_cost = min(min_cost, cost_for_comp);         return min_cost; # Driver codeif __name__ == "__main__" :     A = 7; B = 11; C = 2; X = 20;     print(findCost(A, B, C, X)) ;     # This code is contributed by AnkitRai01

C#

 // C# implementation of the above approachusing System; class GFG {         static int MAX = 1000;         // max_prime[i] represents maximum prime    // number that divides the number i    static int []max_prime = new int[MAX];         // min_prime[i] represents minimum prime    // number that divides the number i    static int []min_prime = new int[MAX];         // Function to store the minimum    // prime factor and the maximum    // prime factor in two arrays    static void sieve(int n)    {        for (int i = 2; i < n; ++i) {                 // Check for prime number            // if min_prime[i] > 0,            // then it is not a prime number            if (min_prime[i] > 0) {                continue;            }                 // If i is a prime number,            // then both minimum and maximum            // prime numbers that divide            // the number is the number itself            min_prime[i] = i;            max_prime[i] = i;                 int j = i + i;                 while (j < n) {                if (min_prime[j] == 0) {                         // If this number is being visited                    // for first time then this divisor                    // must be the smallest prime number                    // that divides this number                    min_prime[j] = i;                }                     // Update prime number till the last                // prime number that divides this number                     // The last prime number that                // divides this number will be maximum.                max_prime[j] = i;                j += i;            }        }    }         // Function to minimize the cost of finding    // two numbers for every number such that    // the product of those two is equal to X    static int findCost(int A, int B, int C, int X)    {        // Pre-calculation        sieve(MAX);             int N, M;             // If X == 1, then there is no way to        // find N and M. Print -1        if (X == 1) {            return -1;        }             // Case 3 is always valid and cost for that        // is C + X C for choosing 1 and M = X/1        int min_cost = C + X;             // Case 1        // N is prime, first number cost is fixed        // N is max_prime number divides this number        int cost_for_prime = A;        N = max_prime[X];             // If X is prime then the maximum prime number        // is the number itself. For this case,        // M becomes 1 and this shouldn't be considered.        if (N != X) {                 // Find M for this case            M = X / N;                 // Add cost for the second number also            cost_for_prime += M;                 // Update min_cost, if the            // cost for prime is minimum            min_cost = Math.Min(min_cost, cost_for_prime);        }             // Case 2        // If N is composite        // For this find the minimum prime number        // that divides A[i] and consider this as M        M = min_prime[X];             // Find N for that number        N = X / M;             // Check if this number is composite or not        // if N is prime then there is no way        // to find any composite number that divides X        // If N = min_prime[N] then N is prime        if (N != min_prime[N]) {            int cost_for_comp = B + M;                 // Update min_cost, if the            // cost for the composite is minimum            min_cost = Math.Min(min_cost, cost_for_comp);        }             return min_cost;    }         // Driver code    public static void Main (string[] args)    {        int A = 7, B = 11, C = 2, X = 20;             Console.WriteLine(findCost(A, B, C, X));    }} // This code is contributed by AnkitRai01

Javascript


Output:
11

Time Complexity: O(MAX)2

Auxiliary Space: O(MAX)

My Personal Notes arrow_drop_up