Sell Diminishing Valued Colored Balls - Leetcode Solution

Problem Statement:

You have an inventory of colored balls, and you want to sell each ball. There are k types of colored balls in the inventory, numbered from 1 to k. You want to sell all the balls in such a way that:

• Each day you can sell exactly one ball of any color.
• After you sell a ball, you cannot sell another ball of that color on the same day.
• The kth ball you sell is colored with a number x such that 1 <= x <= k. In other words, the last ball you sell must be colored with one of the k colors.
• You want to maximize the number of days you can sell the balls.

Write a function to return the maximum number of days you can sell the balls to meet the conditions.

Approach:

To maximize the number of days you can sell the balls, you need to sell the balls with diminishing value of colors. In other words, you should sell the balls in decreasing order of their count in the inventory. Since you can sell only one ball of each color each day, you can sell the balls in decreasing order of their count until you run out of different colors. After that, you can sell the remaining balls of the color with the highest count until you have sold all the balls. The color of the last ball should be x, where x is between 1 and k.

Algorithm:

1. Count the number of balls of each color in the inventory.
2. Sort the colors by decreasing count.
3. Add the count of the first k-1 colors to determine the maximum number of days you can sell the balls without selling the last kth color.
4. Subtract the minimum count of the first k-1 colors from the count of the kth color to determine how many more days you can sell the balls if you sell the kth color.
5. Add the result of step 4 to the result of step 3 to determine the maximum number of days you can sell the balls.
6. Return the result computed in step 5.

Code:

int maxDays(vector<int>& inventory, int orders) { sort(inventory.begin(), inventory.end(), greater<int>()); int k = inventory.size(); long long count = 0, ans = 0; for (int i = 0; i < k && inventory[i] > 0; ++i) { long long diff = i == 0 ? inventory[i] : inventory[i] - inventory[i - 1]; long long days = (k - i) * diff; if (days <= orders) { orders -= days; count += diff; ans += (k - i) * (2 * count - diff + 1) * diff / 2; } else { long long q = orders / (k - i), r = orders % (k - i); count += q; ans += (k - i) * (2 * count - q + 1) * q / 2; ans += r * (count + count - r + 1) / 2; orders = 0; } } ans += (long long)orders * (2 * count - k + 1 + orders) / 2; return ans % 1000000007; }

Time Complexity:

The time complexity of the algorithm is O(k log k), where k is the number of colors in the inventory. This is because we need to sort the colors by decreasing count, which takes O(k log k) time.

Space Complexity:

The space complexity of the algorithm is O(k), where k is the number of colors in the inventory. This is because we need to store the count of each color in the inventory.

1