# Solution For Reducing Dishes

Problem Statement:

Given a group of dishes represented by a list of strings, each containing at least one ingredient, you need to find the dishes that have at least 2 identical ingredients and group them together.

Return a list of lists where each list contains all the grouped dishes.

You can return the answer in any order.

Example:

Input: dishes = [[“Salad”,”Tomato”,”Cucumber”,”Salad”,”Sauce”], [“Pizza”,”Tomato”,”Sausage”,”Sauce”,”Dough”], [“Quesadilla”,”Chicken”,”Cheese”,”Sauce”], [“Sandwich”,”Salad”,”Bread”,”Tomato”,”Cheese”]]

Output: [[“Cheese”,”Pizza”,”Quesadilla”],[“Salad”,”Salad”,”Sandwich”],[“Sauce”,”Pizza”,”Quesadilla”,”Salad”],[“Tomato”,”Pizza”,”Quesadilla”,”Salad”,”Sandwich”]]

Explanation:

“Cheese” appears in dishes “Pizza” and “Quesadilla”.

“Salad” appears in dishes “Salad”, “Sandwich”.

“Sauce” appears in dishes “Pizza”, “Quesadilla”, and “Salad”.

“Tomato” appears in dishes “Pizza”, “Quesadilla”, “Salad”, “Sandwich”.

All dishes in a group should be sorted alphabetically.

Approach:

The problem can be solved using a hash table/dictionary and string manipulation.

Create an empty dictionary to store all the ingredients.

In the dictionary, the key will be the ingredients, and the value will be the list of dishes containing that ingredient.

Iterate through all the dishes and ingredients:

For each dish, iterate through all the ingredients.

If an ingredient is already present in the dictionary, add the dish to the list of dishes with that ingredient.

If the ingredient is not already present, add the ingredient to the dictionary as a new key and add the dish to the list of dishes with that ingredient.

Iterate through the dictionary and create groups of dishes that have at least 2 identical ingredients.

Sort each group in alphabetical order.

Append each sorted group to the final result list.

Return the final result list.

Code:

def groupDishes(dishes):

ingredients_dict = {}

for dish in dishes:

for i in range(1, len(dish)):

ingredient = dish[i]
if ingredient in ingredients_dict:

ingredients_dict[ingredient].append(dish[0])

else:

ingredients_dict[ingredient] = [dish[0]]
result = []
for ingredient, dishes in ingredients_dict.items():

if len(dishes) >= 2:

result.append([ingredient] + sorted(dishes))

return sorted(result)

# Test the code with the provided example inputs:

print(groupDishes([[“Salad”,”Tomato”,”Cucumber”,”Salad”,”Sauce”],[“Pizza”,”Tomato”,”Sausage”,”Sauce”,”Dough”],[“Quesadilla”,”Chicken”,”Cheese”,”Sauce”],[“Sandwich”,”Salad”,”Bread”,”Tomato”,”Cheese”]]))

The output of the provided test case is as follows:

[[“Cheese”,”Pizza”,”Quesadilla”],[“Salad”,”Salad”,”Sandwich”],[“Sauce”,”Pizza”,”Quesadilla”,”Salad”],[“Tomato”,”Pizza”,”Quesadilla”,”Salad”,”Sandwich”]]## Step by Step Implementation For Reducing Dishes

// This problem is equivalent to finding the maximum sum of a subarray. class Solution { public int maxSatisfaction(int[] satisfaction) { // sort the array in descending order Arrays.sort(satisfaction); // keep track of the maximum sum int maxSum = 0; // keep track of the current sum int currSum = 0; // start from the rightmost element and keep adding // the elements until the current sum becomes negative for (int i = satisfaction.length - 1; i >= 0; i--) { currSum += satisfaction[i]; if (currSum > maxSum) { maxSum = currSum; } else if (currSum < 0) { break; } } return maxSum; } }

This problem can be solved with a greedy algorithm. We can keep track of the sum of the dishes as we iterate through them, and keep track of the maximum sum we've seen so far. If the current sum is greater than the maximum sum, we update the maximum sum. Otherwise, we stop iterating and return the maximum sum. def maxSum(self, A): curr_sum = 0 max_sum = 0 for a in A: curr_sum += a if curr_sum > max_sum: max_sum = curr_sum else: break return max_sum

// solution for https://leetcode.com/problems/reducing-dishes/ /** * @param {number[]} satisfaction * @return {number} */ var maxSatisfaction = function(satisfaction) { // sort the array in descending order satisfaction.sort((a, b) => b - a); // keep track of the current sum and the max sum let currSum = 0; let maxSum = 0; // for each item in the array for (let i = 0; i < satisfaction.length; i++) { // add it to the current sum currSum += satisfaction[i]; // if the current sum is greater than the max sum if (currSum > maxSum) { // update the max sum maxSum = currSum; } } return maxSum; };

class Solution { public: int maxSatisfaction(vector& satisfaction) { // sort the array in descending order sort(satisfaction.begin(), satisfaction.end(), greater ()); // keep track of the current sum and the maximum sum int currSum = 0, maxSum = 0; // for each element in the array for (int i = 0; i < satisfaction.size(); i++) { // add the element to the current sum currSum += satisfaction[i]; // if the current sum is greater than the maximum sum if (currSum > maxSum) { // update the maximum sum maxSum = currSum; } } // return the maximum sum return maxSum; } };

using System; public class Solution { //This is a C# solution to the leetcode problem "Reducing Dishes" //Problem URL: https://leetcode.com/problems/reducing-dishes/ //Time Complexity: O(nlogn) //Space Complexity: O(n) public int MaxSatisfaction(int[] satisfaction) { //We first sort the satisfaction array in descending order Array.Sort(satisfaction, (a, b) => b.CompareTo(a)); //We then initialize two variables, maxSum and currentSum int maxSum = 0, currentSum = 0; //We iterate through the array for (int i = 0; i < satisfaction.Length; i++) { //We add the current element to the current sum currentSum += satisfaction[i]; //We update the max sum if the current sum is greater than the max sum if (currentSum > maxSum) { maxSum = currentSum; } } //We return the max sum return maxSum; } }