# Number of array elements derivable from D after performing certain operations

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; ` `} ` |

*chevron_right*

*filter_none*

## 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.. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## PHP

**Output:**

4 3

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

## Recommended Posts:

- Maximum sum of all elements of array after performing given operations
- Minimum steps to reach end from start by performing multiplication and mod operations with array elements
- Maximum Possible Product in Array after performing given Operations
- Reduce a number to 1 by performing given operations
- Check if at least half array is reducible to zero by performing some operations
- Find the modified array after performing k operations of given type
- Maximum count of equal numbers in an array after performing given operations
- Minimum number of operations on an array to make all elements 0
- Minimum number of increment/decrement operations such that array contains all elements from 1 to N
- Minimum number of increment-other operations to make all array elements equal.
- Find the number of operations required to make all array elements Equal
- Minimum possible sum of array elements after performing the given operation
- Minimum gcd operations to make all array elements one
- Minimum no. of operations required to make all Array Elements Zero
- Minimum delete operations to make all elements of array same

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.