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2979. Most Expensive Item That Can Not Be Bought

Description

You are given two distinct prime numbers primeOne and primeTwo.

Alice and Bob are visiting a market. The market has an infinite number of items, for any positive integer x there exists an item whose price is x. Alice wants to buy some items from the market to gift to Bob. She has an infinite number of coins in the denomination primeOne and primeTwo. She wants to know the most expensive item she can not buy to gift to Bob.

Return the price of the most expensive item which Alice can not gift to Bob.

 

Example 1:

Input: primeOne = 2, primeTwo = 5
Output: 3
Explanation: The prices of items which cannot be bought are [1,3]. It can be shown that all items with a price greater than 3 can be bought using a combination of coins of denominations 2 and 5.

Example 2:

Input: primeOne = 5, primeTwo = 7
Output: 23
Explanation: The prices of items which cannot be bought are [1,2,3,4,6,8,9,11,13,16,18,23]. It can be shown that all items with a price greater than 23 can be bought.

 

Constraints:

  • 1 < primeOne, primeTwo < 104
  • primeOne, primeTwo are prime numbers.
  • primeOne * primeTwo < 105

Solutions

Solution 1: Chicken McNugget Theorem

According to the Chicken McNugget Theorem, for two coprime positive integers $a$ and $b$, the largest number that cannot be expressed as a combination of $a$ and $b$ is $ab - a - b$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • class Solution {
    public int mostExpensiveItem(int primeOne, int primeTwo) {
        return primeOne * primeTwo - primeOne - primeTwo;
    }
}

  • class Solution {
public:
    int mostExpensiveItem(int primeOne, int primeTwo) {
        return primeOne * primeTwo - primeOne - primeTwo;
    }
};

  • class Solution:
    def mostExpensiveItem(self, primeOne: int, primeTwo: int) -> int:
        return primeOne * primeTwo - primeOne - primeTwo


  • func mostExpensiveItem(primeOne int, primeTwo int) int {
	return primeOne*primeTwo - primeOne - primeTwo
}

  • function mostExpensiveItem(primeOne: number, primeTwo: number): number {
    return primeOne * primeTwo - primeOne - primeTwo;
}


  • impl Solution {
    pub fn most_expensive_item(prime_one: i32, prime_two: i32) -> i32 {
        prime_one * prime_two - prime_one - prime_two
    }
}

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