Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Greedy Approach with Sorting - Complex approach
- Dynamic Programming - Complex approach
Approach 1: Greedy Approach with Sorting
The key insight for solving this problem is to sort the satisfaction levels in ascending order and then decide whether to include each dish from the end (highest satisfaction) to the beginning (lowest satisfaction).
After sorting, we can greedily include dishes with the highest satisfaction levels first, as they will contribute more to the total like-time coefficient.
For each dish, we need to decide whether including it will increase the total like-time coefficient. If a dish has a negative satisfaction level, it might still be worth including if it allows us to increase the time coefficient of dishes with higher satisfaction levels.
Let's define total as the sum of satisfaction levels of all dishes we've included so far, and result as the maximum sum of like-time coefficients.
For each dish, we update total by adding its satisfaction level, and update result by adding total. If result decreases after including a dish, we stop and return the previous result.
Algorithm:
- Sort the satisfaction array in ascending order.
- Initialize total = 0 (sum of satisfaction levels) and result = 0 (maximum sum of like-time coefficients).
- Iterate through the sorted array from right to left (highest to lowest satisfaction):
- a. Add the current satisfaction level to total.
- b. Add total to result.
- c. If result decreases (i.e., result < previous result), break and return the previous result.
- Return result.
Time Complexity:
Where n is the number of dishes. We need to sort the array, which takes O(n log n) time, and then iterate through it once, which takes O(n) time.
Space Complexity:
We only use a constant amount of extra space regardless of the input size.
Approach 2: Dynamic Programming
We can also solve this problem using dynamic programming. The key is to define a state that represents whether we've included a dish or not.
Let's define dp[i][j] as the maximum sum of like-time coefficients we can get by considering the first i dishes and including j dishes.
For each dish, we have two options:
- Include the dish:
dp[i][j] = dp[i-1][j-1] + satisfaction[i-1] * j - Exclude the dish:
dp[i][j] = dp[i-1][j]
We take the maximum of these two options.
The final answer is the maximum value in dp[n][j] for all j from 0 to n.
Algorithm:
- Sort the satisfaction array in ascending order.
- Initialize a 2D DP array dp of size (n+1) x (n+1), where dp[i][j] represents the maximum sum of like-time coefficients we can get by considering the first i dishes and including j dishes.
- Set dp[0][0] = 0 (base case: no dishes, no like-time coefficient).
- For each dish i from 1 to n:
- a. For each number of dishes j from 1 to i:
- i. Include the dish: dp[i][j] = dp[i-1][j-1] + satisfaction[i-1] * j
- ii. Exclude the dish: dp[i][j] = dp[i-1][j]
- iii. dp[i][j] = max(include, exclude)
- Return the maximum value in dp[n][j] for all j from 0 to n.
Time Complexity:
Where n is the number of dishes. We need to fill a 2D DP array of size (n+1) x (n+1).
Space Complexity:
We use a 2D DP array of size (n+1) x (n+1).
Pseudocode
123456789101112131415161718192021function maxSatisfaction(satisfaction): // Sort the satisfaction array in ascending order sort(satisfaction) total = 0 // Sum of satisfaction levels result = 0 // Maximum sum of like-time coefficients // Iterate through the sorted array from right to left for i from n-1 to 0: // Add the current satisfaction level to total total += satisfaction[i] // Add total to result result += total // If result decreases, break and return the previous result if total < 0: result -= total break return resultString Problems
Learning Objective
Understand different approaches to solve the reducing dishes problem and analyze their efficiency.
Solution Approaches
Approach 1: Greedy Approach with Sorting
The key insight for solving this problem is to sort the satisfaction levels in ascending order and then decide whether to include each dish from the end (highest satisfaction) to the beginning (lowest satisfaction).
After sorting, we can greedily include dishes with the highest satisfaction levels first, as they will contribute more to the total like-time coefficient.
For each dish, we need to decide whether including it will increase the total like-time coefficient. If a dish has a negative satisfaction level, it might still be worth including if it allows us to increase the time coefficient of dishes with higher satisfaction levels.
Let's define total as the sum of satisfaction levels of all dishes we've included so far, and result as the maximum sum of like-time coefficients.
For each dish, we update total by adding its satisfaction level, and update result by adding total. If result decreases after including a dish, we stop and return the previous result.
Algorithm:
- Sort the satisfaction array in ascending order.
- Initialize total = 0 (sum of satisfaction levels) and result = 0 (maximum sum of like-time coefficients).
- Iterate through the sorted array from right to left (highest to lowest satisfaction):
- a. Add the current satisfaction level to total.
- b. Add total to result.
- c. If result decreases (i.e., result < previous result), break and return the previous result.
- Return result.
Approach 2: Dynamic Programming
We can also solve this problem using dynamic programming. The key is to define a state that represents whether we've included a dish or not.
Let's define dp[i][j] as the maximum sum of like-time coefficients we can get by considering the first i dishes and including j dishes.
For each dish, we have two options:
- Include the dish:
dp[i][j] = dp[i-1][j-1] + satisfaction[i-1] * j - Exclude the dish:
dp[i][j] = dp[i-1][j]
We take the maximum of these two options.
The final answer is the maximum value in dp[n][j] for all j from 0 to n.
Algorithm:
- Sort the satisfaction array in ascending order.
- Initialize a 2D DP array dp of size (n+1) x (n+1), where dp[i][j] represents the maximum sum of like-time coefficients we can get by considering the first i dishes and including j dishes.
- Set dp[0][0] = 0 (base case: no dishes, no like-time coefficient).
- For each dish i from 1 to n:
- a. For each number of dishes j from 1 to i:
- i. Include the dish: dp[i][j] = dp[i-1][j-1] + satisfaction[i-1] * j
- ii. Exclude the dish: dp[i][j] = dp[i-1][j]
- iii. dp[i][j] = max(include, exclude)
- Return the maximum value in dp[n][j] for all j from 0 to n.
Complexity Analysis
Greedy Approach with Sorting Approach
Time Complexity
Where n is the number of dishes. We need to sort the array, which takes O(n log n) time, and then iterate through it once, which takes O(n) time.
Space Complexity
We only use a constant amount of extra space regardless of the input size.
Dynamic Programming Approach
Time Complexity
Where n is the number of dishes. We need to fill a 2D DP array of size (n+1) x (n+1).
Space Complexity
We use a 2D DP array of size (n+1) x (n+1).
Pseudocode
Greedy Approach with Sorting Approach
123456789101112131415161718192021function maxSatisfaction(satisfaction): // Sort the satisfaction array in ascending order sort(satisfaction) total = 0 // Sum of satisfaction levels result = 0 // Maximum sum of like-time coefficients // Iterate through the sorted array from right to left for i from n-1 to 0: // Add the current satisfaction level to total total += satisfaction[i] // Add total to result result += total // If result decreases, break and return the previous result if total < 0: result -= total break return resultDynamic Programming Approach
123456789101112131415161718192021222324252627function maxSatisfaction(satisfaction): // Sort the satisfaction array in ascending order sort(satisfaction) n = length of satisfaction // Initialize DP array dp = new 2D array of size (n+1) x (n+1), initialized with 0 // Fill DP array for i from 1 to n: for j from 1 to i: // Include the dish include = dp[i-1][j-1] + satisfaction[i-1] * j // Exclude the dish exclude = dp[i-1][j] // Take the maximum dp[i][j] = max(include, exclude) // Find the maximum value in dp[n][j] for all j from 0 to n result = 0 for j from 0 to n: result = max(result, dp[n][j]) return result