[BUGFIX] [Hacker Rank]: Project Euler #1: Multiples of 3 and 5. Types changed to reach problem limits. Solved ✓

Author: Gonzalo Diaz

src/lib/exercises/include/exercises/hackerrank/projecteuler/euler001.hpp

@@ -2,6 +2,6 @@
 
 namespace hackerrank::projecteuler {
 
-int euler001(int a, int b, int n);
+unsigned long euler001(int a, int b, int n);
 
 }

src/lib/exercises/src/hackerrank/projecteuler/euler001.cpp

@@ -8,10 +8,18 @@
 
 namespace hackerrank::projecteuler {
 
+unsigned long min(unsigned long a, unsigned long b) {
+  if (a > b) {
+    return b;
+  }
+
+  return a;
+}
+
 // Function to return gcd of a and b
-int gcd(int a, int b) {
+unsigned long gcd(unsigned long a, unsigned long b) {
   // Find Minimum of a and b
-  int result = std::min(a, b);
+  unsigned long result = min(a, b);
   while (result > 0) {
     if (a % result == 0 && b % result == 0) {
       break;
@@ -24,18 +32,18 @@ int gcd(int a, int b) {
 }
 
 // Function to find sum of Arithmetic Progression series
-int sum_of_arithmetic_progression(int n, int d) {
+unsigned long sum_of_arithmetic_progression(unsigned long n, unsigned long d) {
   // Number of terms
   n = n / d;
   return n * (1 + n) * d / 2;
 }
 
 // Function to find the sum of all multiples of a and b below n
-int euler001(int a, int b, int n) {
+unsigned long euler001(int a, int b, int n) {
   // Since, we need the sum of multiples less than N
   n = n - 1;
 
-  int lcm = (a * b) / gcd(a, b);
+  unsigned long lcm = (a * b) / gcd(a, b);
 
   return sum_of_arithmetic_progression(n, a) +
          sum_of_arithmetic_progression(n, b) -