703149. Sid and his prime money
Sid and his Prime Money
Summary
Sid has a stash of money, and he wants to allocate it wisely. He knows that prime numbers are special, so he decides to invest in them. The goal is to maximize the sum of the prime numbers Sid can buy with his available money. Given an array of coin denominations and the total amount of money Sid has, find out how many prime numbers he can buy and what their sum will be.
Detailed Explanation
To solve this problem, we'll use a combination of mathematical and algorithmic techniques. First, let's create a function to check if a given number is prime or not. Then, we'll sort the coin denominations in descending order and iterate through them to find the maximum number of prime numbers that can be bought.
Here's a step-by-step breakdown:
- Initialize an empty array
primesto store the prime numbers. - Iterate over the range from 2 to the total amount of money (inclusive).
- For each number, check if it's prime using the function created in step 1.
- If the number is prime, add it to the
primesarray. - Sort the
primesarray in descending order. - Initialize a variable
sumto store the sum of the prime numbers bought. - Iterate through the sorted
primesarray and for each prime number:- Check if the total amount of money is sufficient to buy the prime number (i.e., the prime number is less than or equal to the remaining amount).
- If it's sufficient, subtract the prime number from the total amount and add it to the
sum.
- Return the maximum sum of prime numbers that can be bought.
Time complexity: O(n log n) due to sorting the primes array.
Space complexity: O(n) for storing the prime numbers in the primes array.
Optimized Solutions
Java
public class SidAndPrimeMoney {
public static int maxPrimeSum(int[] coins, int totalAmount) {
List<Integer> primes = new ArrayList<>();
for (int i = 2; i <= totalAmount; i++) {
if (isPrime(i)) {
primes.add(i);
}
}
Collections.sort(primes, Comparator.reverseOrder());
int sum = 0;
for (int prime : primes) {
while (totalAmount >= prime && totalAmount - prime >= 0) {
totalAmount -= prime;
sum += prime;
}
}
return sum;
}
private static boolean isPrime(int num) {
if (num <= 1) {
return false;
}
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
}