# Minimum possible sum of array B such that AiBi = AjBj for all 1 ≤ i < j ≤ N

Given an array a[] of size N. The task is to find the minimum possible sum of the elements of array b[] such that a[i] * b[i] = a[j] * b[j] for all 1 ≤ i < j ≤ N. The answer could be large. So, print the answer modoulo 109 + 7.

Examples:

Input: a[] = {2, 3, 4}
Output: 13
b[] = {6, 4, 3}

Input: a[] = {5, 5, 5}
Output: 3
b = {1, 1, 1}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Assume that Bi satisfying the given conditions are determined. Then let K = A1*B1, then the constraints K = A1*B1 = Aj*Bj hold for all j > 1. Therefore K is a common multiple of A1, …, AN.
Conversely, let lcm be the least common multiple of A1, …, AN, and let Bi = lcm / Ai then such B satisfies the conditions.
Therefore, the desired answer is ∑lcm/Ai. However, lcm can be a very big number, so it can’t be calculated directly. Now, let’s consider calculating, holding lcm in a factorized form. Let pi be primes, and assume that factorizations are given by X = ∏pigi, Y = ∏pifi (either of gi, fi may be 0). Then the least common multiple of X and Y is given by ∏pimax(gi, fi). By using this, the least common multiple of A1, …, AN can be obtained in a factorized form. Therefore, this problem can be solved in a total of O(N*sqrt(A)) time, where A = max(Ai). Also, by speeding up the prime factorization with proper precalculations, the answer can also be obtained in a total of O(A + N*logA) time.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define mod (int)(1e9 + 7) ` `#define N 1000005 ` ` `  `// To store least prime factors ` `// of all the numbers ` `int` `lpf[N]; ` ` `  `// Function to find the least prime ` `// factor of all the numbers ` `void` `least_prime_factor() ` `{ ` `    ``for` `(``int` `i = 1; i < N; i++) ` `        ``lpf[i] = i; ` ` `  `    ``for` `(``int` `i = 2; i < N; i++) ` `        ``if` `(lpf[i] == i) ` `            ``for` `(``int` `j = i * 2; j < N; j += i) ` `                ``if` `(lpf[j] == j) ` `                    ``lpf[j] = i; ` `} ` ` `  `// Function to return the ((a^m1) % mod) ` `int` `power(``int` `a, ``int` `m1) ` `{ ` `    ``if` `(m1 == 0) ` `        ``return` `1; ` `    ``else` `if` `(m1 == 1) ` `        ``return` `a; ` `    ``else` `if` `(m1 == 2) ` `        ``return` `(1LL * a * a) % mod; ` `    ``else` `if` `(m1 & 1) ` `        ``return` `(1LL * a * power(power(a, m1 / 2), 2)) % mod; ` `    ``else` `        ``return` `power(power(a, m1 / 2), 2) % mod; ` `} ` ` `  `// Function to return the sum of ` `// elements of array B ` `long` `long` `sum_of_elements(``int` `a[], ``int` `n) ` `{ ` `    ``// Find the prime factors of ` `    ``// all the numbers ` `    ``least_prime_factor(); ` ` `  `    ``// To store each prime count in lcm ` `    ``map<``int``, ``int``> prime_factor; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Current number ` `        ``int` `temp = a[i]; ` ` `  `        ``// Map to store the prime count ` `        ``// of a single number ` `        ``map<``int``, ``int``> single_number; ` ` `  `        ``// Basic way to calculate all prime factors ` `        ``while` `(temp > 1) { ` `            ``int` `x = lpf[temp]; ` `            ``single_number[x]++; ` `            ``temp /= x; ` `        ``} ` ` `  `        ``// If it is the first number in the array ` `        ``if` `(i == 0) ` `            ``prime_factor = single_number; ` ` `  `        ``// Take the maximum count of  ` `        ``// prime in a number ` `        ``else` `{ ` `            ``for` `(``auto` `x : single_number) ` `                ``prime_factor[x.first] = max(x.second,  ` `                                ``prime_factor[x.first]); ` `        ``} ` `    ``} ` ` `  `    ``long` `long` `ans = 0, lcm = 1; ` ` `  `    ``// Calculate lcm of given array ` `    ``for` `(``auto` `x : prime_factor) ` `        ``lcm = (lcm * power(x.first, x.second)) % mod; ` ` `  `    ``// Calculate sum of elements of array B ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans = (ans + (lcm * power(a[i],  ` `                      ``mod - 2)) % mod) % mod; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 2, 3, 4 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(``int``); ` ` `  `    ``cout << sum_of_elements(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `mod = ``1000000007``; ` `static` `int` `N = ``1000005``; ` ` `  `// To store least prime factors ` `// of all the numbers ` `static` `int` `lpf[] = ``new` `int``[N]; ` ` `  `// Function to find the least prime ` `// factor of all the numbers ` `static` `void` `least_prime_factor() ` `{ ` `    ``for` `(``int` `i = ``1``; i < N; i++) ` `        ``lpf[i] = i; ` ` `  `    ``for` `(``int` `i = ``2``; i < N; i++) ` `        ``if` `(lpf[i] == i) ` `            ``for` `(``int` `j = i * ``2``; j < N; j += i) ` `                ``if` `(lpf[j] == j) ` `                    ``lpf[j] = i; ` `} ` ` `  `// Function to return the ((a^m1) % mod) ` `static` `long` `power(``long` `a, ``long` `m1) ` `{ ` `    ``if` `(m1 == ``0``) ` `        ``return` `1``; ` `    ``else` `if` `(m1 == ``1``) ` `        ``return` `a; ` `    ``else` `if` `(m1 == ``2``) ` `        ``return` `(1l * a * a) % mod; ` `    ``else` `if` `((m1 & ``1``) != ``0``) ` `        ``return` `(1l * a * power(power(a, m1 / ``2``), ``2``)) % mod; ` `    ``else` `        ``return` `power(power(a, m1 / ``2``), ``2``) % mod; ` `} ` ` `  `// Function to return the sum of ` `// elements of array B ` `static` `long` `sum_of_elements(``long` `a[], ``int` `n) ` `{ ` `    ``// Find the prime factors of ` `    ``// all the numbers ` `    ``least_prime_factor(); ` ` `  `    ``// To store each prime count in lcm ` `    ``HashMap prime_factor  ` `            ``= ``new` `HashMap<>(); ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// Current number ` `        ``long` `temp = a[i]; ` ` `  `        ``// Map to store the prime count ` `        ``// of a single number ` `        ``HashMap single_number ` `            ``= ``new` `HashMap<>(); ` ` `  `        ``// Basic way to calculate all prime factors ` `        ``while` `(temp > ``1``)  ` `        ``{ ` `            ``long` `x = lpf[(``int``)temp]; ` `            ``single_number.put(x,(single_number.get(x) ==  ` `                        ``null` `? ``1``:single_number.get(x) + ``1``)); ` `            ``temp /= x; ` `        ``} ` ` `  `        ``// If it is the first number in the array ` `        ``if` `(i == ``0``) ` `            ``prime_factor = single_number; ` ` `  `        ``// Take the maximum count of  ` `        ``// prime in a number ` `        ``else` `{ ` `            ``for` `(Map.Entry x : single_number.entrySet() ) ` `                ``prime_factor.put(x.getKey(), Math.max(x.getValue(),  ` `                                ``(prime_factor.get(x.getKey()) ==  ` `                                ``null` `? ``0``:prime_factor.get(x.getKey())))); ` `        ``} ` `    ``} ` ` `  `    ``long` `ans = ``0``, lcm = ``1``; ` ` `  `    ``// Calculate lcm of given array ` `    ``for` `(Map.Entry x : prime_factor.entrySet()) ` `        ``lcm = (lcm * power(x.getKey(), x.getValue())) % mod; ` ` `  `    ``// Calculate sum of elements of array B ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``ans = (ans + (lcm * power(a[i],  ` `                    ``mod - ``2``)) % mod) % mod; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``long` `a[] = { ``2``, ``3``, ``4` `}; ` `    ``int` `n = a.length; ` ` `  `    ``System.out.println(sum_of_elements(a, n)); ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of the approach ` `mod ``=` `10` `*``*` `9` `+` `7` `N ``=` `1000005` ` `  `# To store least prime factors ` `# of all the numbers ` `lpf ``=` `[``0` `for` `i ``in` `range``(N)] ` ` `  `# Function to find the least prime ` `# factor of all the numbers ` `def` `least_prime_factor(): ` `    ``for` `i ``in` `range``(``1``, N): ` `        ``lpf[i] ``=` `i ` ` `  `    ``for` `i ``in` `range``(``2``,N): ` `        ``if` `(lpf[i] ``=``=` `i): ` `            ``for` `j ``in` `range``(i ``*` `2``, N, i): ` `                ``if` `(lpf[j] ``=``=` `j): ` `                    ``lpf[j] ``=` `i ` ` `  `# Function to return the sum of ` `# elements of array B ` `def` `sum_of_elements(a, n): ` `     `  `    ``# Find the prime factors of ` `    ``# all the numbers ` `    ``least_prime_factor() ` ` `  `    ``# To store each prime count in lcm ` `    ``prime_factor``=``dict``() ` ` `  `    ``for` `i ``in` `range``(n): ` ` `  `        ``# Current number ` `        ``temp ``=` `a[i] ` ` `  `        ``# Map to store the prime count ` `        ``# of a single number ` `        ``single_number ``=` `dict``() ` ` `  `        ``# Basic way to calculate all prime factors ` `        ``while` `(temp > ``1``): ` `            ``x ``=` `lpf[temp] ` `            ``single_number[x] ``=` `single_number.get(x, ``0``) ``+` `1` `            ``temp ``/``/``=` `x ` ` `  ` `  `        ``# If it is the first number in the array ` `        ``if` `(i ``=``=` `0``): ` `            ``prime_factor ``=` `single_number ` ` `  `        ``# Take the maximum count of ` `        ``# prime in a number ` `        ``else``: ` `            ``for` `x ``in` `single_number: ` `                ``if` `x ``in` `prime_factor: ` `                    ``prime_factor[x] ``=` `max``(prime_factor[x],  ` `                                           ``single_number[x]) ` `                ``else``: ` `                    ``prime_factor[x] ``=` `single_number[x] ` ` `  `    ``ans, lcm ``=` `0``, ``1` ` `  `    ``# Calculate lcm of given array ` `    ``for` `x ``in` `prime_factor: ` `        ``lcm ``=` `(lcm ``*` `pow``(x, prime_factor[x],mod)) ``%` `mod ` ` `  `    ``# Calculate sum of elements of array B ` `    ``for` `i ``in` `range``(n): ` `        ``ans ``=` `(ans ``+` `(lcm ``*` `pow``(a[i],  ` `                ``mod ``-` `2``,mod)) ``%` `mod) ``%` `mod ` ` `  `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``a ``=` `[``2``, ``3``, ``4``] ` `    ``n ``=` `len``(a) ` `    ``print``(sum_of_elements(a, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```13
```

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