708157. Maximum Number of Toys
Maximum Number of Toys
Summary
The problem "Maximum Number of Toys" requires you to find the maximum number of toys that can be purchased given a set of prices and a budget. This is an optimization problem, as you need to maximize the number of toys while staying within the budget.
Detailed Explanation
To solve this problem, we will use dynamic programming. The idea is to create a table where each cell represents the maximum number of toys that can be purchased with a certain amount of money. We start by initializing the first row and column of the table, then fill in the rest of the table based on the previous rows.
Here's the step-by-step breakdown:
- Initialize a 2D array
dpwith dimensions(budget + 1) x (num_toys + 1)wheredp[i][j]represents the maximum number of toys that can be purchased withiamount of money and considering the firstjtoys. - Fill in the first row and column of the table:
dp[0][j] = 0for allj, as you cannot purchase any toys with a budget of 0, anddp[i][0] = i / price[0]for alli, as you can always purchase at least one toy per unit of money. - Fill in the rest of the table:
dp[i][j] = max(dp[i - 1][k] + 1) for k from 0 to j. This means that if you have enough money to purchase at least one more toy, then do so and update the maximum number of toys. - The maximum number of toys is stored in
dp[budget][num_toys].
Here's an ASCII art diagram illustrating the dynamic programming approach:
+--------+--------+--------+
| 0 | 1 | 2 |
+--------+--------+--------+
| 10 | 5 | 3 |
| 20 | 4 | 2 |
| 30 | 3 | 1 |
| ... | ... | ... |
+--------+--------+--------+
Time complexity: O(budget * num_toys) Space complexity: O(budget * num_toys)
Optimized Solutions
Java
public int maxToys(int[] prices, int budget) {
int[][] dp = new int[budget + 1][prices.length + 1];
for (int i = 0; i <= budget; i++) {
dp[i][0] = i / prices[0];
}
for (int j = 0; j <= prices.length; j++) {
dp[0][j] = 0;
}
for (int i = 1; i <= budget; i++) {
for (int j = 1; j <= prices.length; j++) {
dp[i][j] = Math.max(dp[i - 1][k] + 1) for k from 0 to j);
}
}
return dp[budget][prices.length];
}