Number Of Ways To Buy Pens And Pencils - Leetcode Solution

Problem Statement: You are given an array of prices for pens and pencils. You have been given some money to buy pens and pencils with the condition that you have to buy at least one pen and one pencil. You have to find all the possible ways you can spend the money to buy the pens and pencils.

Example: Input: prices = [2,3] money = 5

Output: 3

Explanation: There are three ways to spend the money to buy pens and pencils:

1. Buy 1 pen and 2 pencils for a total cost of (21 + 32) = 8, which is greater than 5.
2. Buy 2 pens and 1 pencil for a total cost of (22 + 31) = 7, which is less than 5.
3. Buy 1 pen and 1 pencil for a total cost of (21 + 31) = 5, which is equal to 5.

Solution: This problem can be solved using the brute-force approach by generating all the possible combinations of pens and pencils and checking their total cost. But this approach is not efficient as it will take a lot of time for the larger inputs.

A better approach is to use recursion with memoization to solve this problem. We can start with the first pen and try to buy as many pencils as possible with the remaining money. Then, we can move to the second pen and repeat this process until we have reached the end of the array or we have spent all the money.

We can keep track of the remaining money and the current index of the array in the recursive function. We can also use memoization to avoid recalculating the same case multiple times.

Algorithm:

1. Declare the function 'dfs' to count the number of possible ways to buy pens and pencils.
2. Initialize the memoization table with -1.
3. If the remaining money is less than zero, return 0 as no valid combination can be found.
4. If the remaining money is zero, check the memoization table. If the result is already calculated, return the memoized result.
5. Loop through the prices array and count the number of ways to buy pens and pencils with the remaining money after buying a pen and some pencils using the recursive approach.
6. Memoize the result and return the total count of ways to buy pens and pencils.

Python Code:

def dfs(prices, i, money, memo):
    if money < 0:
        return 0
    if money == 0:
        return 1
    
    if memo[i][money] != -1:
        return memo[i][money]
    
    count = 0
    
    for j in range(i, len(prices)):
        count += dfs(prices, j, money - prices[j], memo)
    
    memo[i][money] = count
    return count
    
def numberOfWaysToBuy(prices, money):
    memo = [[-1] * (money+1) for _ in range(len(prices))]
    return dfs(prices, 0, money, memo)

Time Complexity: O(n*m), where n is the length of the prices array and m is the money to spend on pens and pencils.

Space Complexity: O(n*m), for memoization table.

1