Given an array of N integers and 3 integers D, A and B. The task is to find the number of array elements that we can convert D into by performing the following operations on D:

- Add A (+A)
- Subtract A (-A)
- Add B (+B)
- Subtract B (-B)

**Note**: It is allowed to perform any number of operations of any type.

**Examples:**

Input :arr = {1, 2, 3}, D = 6, A = 3, B = 2Output :3Explanation:We can derive 1 from D by performing (6 - 3(A) - 2(B)) We can derive 2 from D by performing (6 - 2(A) - 2(A)) We can derive 3 from D by performing (6 - 3(A)) Thus,Allarray elements can be derived from D.Input :arr = {1, 2, 3}, D = 7, A = 4, B = 2Output :2Explanation:We can derive 1 from D by performing (7 - 4(A) - 2(B)) We can derive 3 from D by performing (7 - 4(A)) Thus, we can derive{1, 3}

Lets say the we want to check if the element **a _{i}** can be derived from D:

Suppose we perform:

- The operation of type1(i.e Add A) P times.
- The operation of type 2(i.e Subtract A) Q times.
- The operation of type 3(i.e Add B) R times.
- The operation of type 4(i.e Subtract B) S times.

Let the value we get after performing these operations be X, then,

-> X = P*A – Q*A + R*B – S*B

-> X = (P – Q) * A + (R – S) * BSuppose we successfully derive A

_{i}from D, i.e X = |A_{i}– D|,-> |A

_{i}– D| = (P – Q) * A + (R – S) * BLet (P – Q) = some constant say, U

and similarly let (R – S) be a constant, V-> |A

_{i}– D| = U * A + V * BThis is in the form of the Linear Diophantine Equation and the solution exists only when |A

_{i}– D| is divisible by gcd(A, B).

Thus now we can simply iterate over the array and count all such A_{i} for which |A_{i} – D| is divisible by gcd(a, b).

Below is the implementation of the above approach:

## C++

`// CPP program to find the number of array elements ` `// which can be derived by perming (+A, -A, +B, -B) ` `// operations on D ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `// Function to return ` `// gcd of a and b ` `int` `gcd(` `int` `a, ` `int` `b) ` `{ ` ` ` `if` `(a == 0) ` ` ` `return` `b; ` ` ` `return` `gcd(b % a, a); ` `} ` ` ` `/* Function to Return the number of elements ` ` ` `of arr[] which can be derived from D by ` ` ` `performing (+A, -A, +B, -B) */` `int` `findPossibleDerivables(` `int` `arr[], ` `int` `n, ` `int` `D, ` ` ` `int` `A, ` `int` `B) ` `{ ` ` ` `// find the gcd of A and B ` ` ` `int` `gcdAB = gcd(A, B); ` ` ` ` ` `// counter stores the number of ` ` ` `// array elements which ` ` ` `// can be derived from D ` ` ` `int` `counter = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `// arr[i] can be derived from D only if ` ` ` `// |arr[i] - D| is divisible by gcd of A and B ` ` ` `if` `((` `abs` `(arr[i] - D) % gcdAB) == 0) { ` ` ` `counter++; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `counter; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 2, 3, 4, 7, 13 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `int` `D = 5, A = 4, B = 2; ` ` ` `cout << findPossibleDerivables(arr, n, D, A, B) <<` `"\n"` `; ` ` ` ` ` `int` `a[] = { 1, 2, 3 }; ` ` ` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `D = 6, A = 3, B = 2; ` ` ` `cout << findPossibleDerivables(a, n, D, A, B) <<` `"\n"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find the number of array elements ` `// which can be derived by perming (+A, -A, +B, -B) ` `// operations on D ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` ` ` `// Function to return ` `// gcd of a and b ` ` ` `static` `int` `gcd(` `int` `a, ` `int` `b) ` `{ ` ` ` `if` `(a == ` `0` `) ` ` ` `return` `b; ` ` ` `return` `gcd(b % a, a); ` `} ` ` ` `/* Function to Return the number of elements ` `of arr[] which can be derived from D by ` `performing (+A, -A, +B, -B) */` `static` `int` `findPossibleDerivables(` `int` `arr[], ` `int` `n, ` `int` `D, ` ` ` `int` `A, ` `int` `B) ` `{ ` ` ` `// find the gcd of A and B ` ` ` `int` `gcdAB = gcd(A, B); ` ` ` ` ` `// counter stores the number of ` ` ` `// array elements which ` ` ` `// can be derived from D ` ` ` `int` `counter = ` `0` `; ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` `// arr[i] can be derived from D only if ` ` ` `// |arr[i] - D| is divisible by gcd of A and B ` ` ` `if` `((Math.abs(arr[i] - D) % gcdAB) == ` `0` `) { ` ` ` `counter++; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `counter; ` `} ` ` ` `// Driver Code ` ` ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` `int` `arr[] = { ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `7` `, ` `13` `}; ` ` ` `int` `n = arr.length; ` ` ` `int` `D = ` `5` `, A = ` `4` `, B = ` `2` `; ` ` ` `System.out.println( findPossibleDerivables(arr, n, D, A, B)); ` ` ` ` ` `int` `a[] = { ` `1` `, ` `2` `, ` `3` `}; ` ` ` `n = a.length; ` ` ` `D = ` `6` `; ` ` ` `A = ` `3` `; ` ` ` `B = ` `2` `; ` ` ` `System.out.println( findPossibleDerivables(a, n, D, A, B)); ` ` ` `} ` `} ` `// This code is contributed by anuj_67.. ` |

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## Python3

`# Python3 program to find the number of array ` `# elements which can be derived by perming ` `# (+A, -A, +B, -B) operations on D ` ` ` `# Function to return gcd of a and b ` `def` `gcd(a, b) : ` ` ` ` ` `if` `(a ` `=` `=` `0` `) : ` ` ` `return` `b ` ` ` ` ` `return` `gcd(b ` `%` `a, a); ` ` ` `""" Function to Return the number of elements ` `of arr[] which can be derived from D by ` `performing (+A, -A, +B, -B) """` `def` `findPossibleDerivables(arr, n, D, A, B) : ` ` ` ` ` `# find the gcd of A and B ` ` ` `gcdAB ` `=` `gcd(A, B) ` ` ` ` ` `# counter stores the number of ` ` ` `# array elements which ` ` ` `# can be derived from D ` ` ` `counter ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(n) : ` ` ` ` ` `# arr[i] can be derived from D only ` ` ` `# if |arr[i] - D| is divisible by ` ` ` `# gcd of A and B ` ` ` `if` `((` `abs` `(arr[i] ` `-` `D) ` `%` `gcdAB) ` `=` `=` `0` `) : ` ` ` `counter ` `+` `=` `1` ` ` ` ` `return` `counter ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `arr ` `=` `[ ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `7` `, ` `13` `] ` ` ` `n ` `=` `len` `(arr) ` ` ` `D, A, B ` `=` `5` `, ` `4` `, ` `2` ` ` ` ` `print` `(findPossibleDerivables(arr, n, D, A, B)) ` ` ` ` ` `a ` `=` `[ ` `1` `, ` `2` `, ` `3` `] ` ` ` `n ` `=` `len` `(a) ` ` ` `D, A, B ` `=` `6` `, ` `3` `, ` `2` ` ` ` ` `print` `(findPossibleDerivables(a, n, D, A, B)) ` ` ` `# This code is contributed by Ryuga ` |

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## C#

`// C# program to find the number of array elements ` `// which can be derived by perming (+A, -A, +B, -B) ` `// operations on D ` `using` `System; ` `public` `class` `GFG { ` ` ` ` ` `// Function to return ` ` ` `// gcd of a and b ` ` ` `static` `int` `gcd(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `if` `(a == 0) ` ` ` `return` `b; ` ` ` `return` `gcd(b % a, a); ` ` ` `} ` ` ` ` ` `/* Function to Return the number of elements ` ` ` `of arr[] which can be derived from D by ` ` ` `performing (+A, -A, +B, -B) */` ` ` `static` `int` `findPossibleDerivables(` `int` `[]arr, ` `int` `n, ` `int` `D, ` ` ` `int` `A, ` `int` `B) ` ` ` `{ ` ` ` `// find the gcd of A and B ` ` ` `int` `gcdAB = gcd(A, B); ` ` ` ` ` `// counter stores the number of ` ` ` `// array elements which ` ` ` `// can be derived from D ` ` ` `int` `counter = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `// arr[i] can be derived from D only if ` ` ` `// |arr[i] - D| is divisible by gcd of A and B ` ` ` `if` `((Math.Abs(arr[i] - D) % gcdAB) == 0) { ` ` ` `counter++; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `counter; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` ` ` `public` `static` `void` `Main () { ` ` ` `int` `[]arr = { 1, 2, 3, 4, 7, 13 }; ` ` ` `int` `n = arr.Length; ` ` ` `int` `D = 5, A = 4, B = 2; ` ` ` `Console.WriteLine( findPossibleDerivables(arr, n, D, A, B)); ` ` ` ` ` `int` `[]a = { 1, 2, 3 }; ` ` ` `n = a.Length; ` ` ` `D = 6; ` ` ` `A = 3; ` ` ` `B = 2; ` ` ` `Console.WriteLine( findPossibleDerivables(a, n, D, A, B)); ` ` ` `} ` `} ` `// This code is contributed by 29AjayKumar ` |

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## PHP

**Output:**

4 3

**Time Complexity: ** O(N), where N is the number of array elements.

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