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 = 2
Output : 3
Explanation:
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, All array elements can be derived from D.
Input : arr = {1, 2, 3}, D = 7, A = 4, B = 2
Output : 2
Explanation:
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 ai 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) * B
Suppose we successfully derive Ai from D, i.e X = |Ai – D|,
-> |Ai – D| = (P – Q) * A + (R – S) * B
Let (P – Q) = some constant say, U
and similarly let (R – S) be a constant, V
-> |Ai – D| = U * A + V * B
This is in the form of the Linear Diophantine Equation and the solution exists only when |Ai – D| is divisible by gcd(A, B).
Thus now we can simply iterate over the array and count all such Ai for which |Ai – D| is divisible by gcd(a, b).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
int findPossibleDerivables(int arr[], int n, int D,
int A, int B)
{
int gcdAB = gcd(A, B);
int counter = 0;
for (int i = 0; i < n; i++) {
if ((abs(arr[i] - D) % gcdAB) == 0) {
counter++;
}
}
return counter;
}
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;
}
|
Java
import java.io.*;
class GFG {
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int findPossibleDerivables(int arr[], int n, int D,
int A, int B)
{
int gcdAB = gcd(A, B);
int counter = 0;
for (int i = 0; i < n; i++) {
if ((Math.abs(arr[i] - D) % gcdAB) == 0) {
counter++;
}
}
return counter;
}
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));
}
}
|
Python3
def gcd(a, b) :
if (a == 0) :
return b
return gcd(b % a, a);
def findPossibleDerivables(arr, n, D, A, B) :
gcdAB = gcd(A, B)
counter = 0
for i in range(n) :
if ((abs(arr[i] - D) % gcdAB) == 0) :
counter += 1
return counter
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))
|
C#
using System;
public class GFG {
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int findPossibleDerivables(int []arr, int n, int D,
int A, int B)
{
int gcdAB = gcd(A, B);
int counter = 0;
for (int i = 0; i < n; i++) {
if ((Math.Abs(arr[i] - D) % gcdAB) == 0) {
counter++;
}
}
return counter;
}
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));
}
}
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PHP
<?php
function gcd($a, $b)
{
if ($a == 0)
return $b;
return gcd($b % $a, $a);
}
function findPossibleDerivables($arr, $n,
$D, $A, $B)
{
$gcdAB = gcd($A, $B);
$counter = 0;
for ($i = 0; $i < $n; $i++)
{
if ((abs($arr[$i] - $D) % $gcdAB) == 0)
{
$counter++;
}
}
return $counter;
}
$arr = array( 1, 2, 3, 4, 7, 13 );
$n = sizeof($arr);
$D = 5;
$A = 4;
$B = 2;
echo findPossibleDerivables($arr, $n,
$D, $A, $B), "\n";
$a = array( 1, 2, 3 );
$n = sizeof($a);
$D = 6;
$A = 3;
$B = 2;
echo findPossibleDerivables($arr, $n,
$D, $A, $B), "\n";
?>
|
Javascript
<script>
function gcd(a , b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
function findPossibleDerivables(arr , n , D , A , B) {
var gcdAB = gcd(A, B);
var counter = 0;
for (i = 0; i < n; i++)
{
if ((Math.abs(arr[i] - D) % gcdAB) == 0) {
counter++;
}
}
return counter;
}
var arr = [ 1, 2, 3, 4, 7, 13 ];
var n = arr.length;
var D = 5, A = 4, B = 2;
document.write(findPossibleDerivables(arr, n, D, A, B)+"<br/>");
var a = [ 1, 2, 3 ];
n = a.length;
D = 6;
A = 3;
B = 2;
document.write(findPossibleDerivables(a, n, D, A, B));
</script>
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Complexity Analysis:
- Time Complexity: O(log(max(A, B) + N), where N is the number of array elements and A & B represents the value of given integers.
- Auxiliary Space: O(log(max(A, B)) due to recursive stack space .