You have information about n different recipes. You are given a string array recipes and a 2D string array ingredients. The ith recipe has the name recipes[i], and you can create it if you have all the needed ingredients from ingredients[i]. A recipe can also be an ingredient for other recipes, i.e., ingredients[i] may contain a string that is in recipes.
You are also given a string array supplies containing all the ingredients that you initially have, and you have an infinite supply of all of them.
Return a list of all the recipes that you can create. You may return the answer in any order.
Note that two recipes may contain each other in their ingredients.
Example 1:
Input: recipes = ["bread"], ingredients = [["yeast","flour"]], supplies = ["yeast","flour","corn"] Output: ["bread"] Explanation: We can create "bread" since we have the ingredients "yeast" and "flour".
Example 2:
Input: recipes = ["bread","sandwich"], ingredients = [["yeast","flour"],["bread","meat"]], supplies = ["yeast","flour","meat"] Output: ["bread","sandwich"] Explanation: We can create "bread" since we have the ingredients "yeast" and "flour". We can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread".
Example 3:
Input: recipes = ["bread","sandwich","burger"], ingredients = [["yeast","flour"],["bread","meat"],["sandwich","meat","bread"]], supplies = ["yeast","flour","meat"] Output: ["bread","sandwich","burger"] Explanation: We can create "bread" since we have the ingredients "yeast" and "flour". We can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread". We can create "burger" since we have the ingredient "meat" and can create the ingredients "bread" and "sandwich".
Constraints:
n == recipes.length == ingredients.length1 <= n <= 1001 <= ingredients[i].length, supplies.length <= 1001 <= recipes[i].length, ingredients[i][j].length, supplies[k].length <= 10recipes[i], ingredients[i][j], and supplies[k] consist only of lowercase English letters.recipes and supplies combined are unique.ingredients[i] does not contain any duplicate values.class Solution {
public List<String> findAllRecipes(
String[] recipes,
List<List<String>> ingredients,
String[] supplies
) {
// Store available supplies
Set<String> availableSupplies = new HashSet<>();
// Map recipe to its index
Map<String, Integer> recipeToIndex = new HashMap<>();
// Map ingredient to recipes that need it
Map<String, List<String>> dependencyGraph = new HashMap<>();
// Initialize available supplies
for (String supply : supplies) {
availableSupplies.add(supply);
}
// Create recipe to index mapping
for (int idx = 0; idx < recipes.length; idx++) {
recipeToIndex.put(recipes[idx], idx);
}
// Count of non-supply ingredients needed for each recipe
int[] inDegree = new int[recipes.length];
// Build dependency graph
for (int recipeIdx = 0; recipeIdx < recipes.length; recipeIdx++) {
for (String ingredient : ingredients.get(recipeIdx)) {
if (!availableSupplies.contains(ingredient)) {
// Add edge: ingredient -> recipe
dependencyGraph.putIfAbsent(
ingredient,
new ArrayList<String>()
);
dependencyGraph.get(ingredient).add(recipes[recipeIdx]);
inDegree[recipeIdx]++;
}
}
}
// Start with recipes that only need supplies
Queue<Integer> queue = new LinkedList<>();
for (int recipeIdx = 0; recipeIdx < recipes.length; recipeIdx++) {
if (inDegree[recipeIdx] == 0) {
queue.add(recipeIdx);
}
}
// Process recipes in topological order
List<String> createdRecipes = new ArrayList<>();
while (!queue.isEmpty()) {
int recipeIdx = queue.poll();
String recipe = recipes[recipeIdx];
createdRecipes.add(recipe);
// Skip if no recipes depend on this one
if (!dependencyGraph.containsKey(recipe)) continue;
// Update recipes that depend on current recipe
for (String dependentRecipe : dependencyGraph.get(recipe)) {
if (--inDegree[recipeToIndex.get(dependentRecipe)] == 0) {
queue.add(recipeToIndex.get(dependentRecipe));
}
}
}
return createdRecipes;
}
}