Fashion Trends

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Unit 4 Session 1 Advanced (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Q: What is the structure of the input?

  • A: The input is a list of dictionaries, where each dictionary represents a brand with a "name" key and a "materials" key containing a list of strings.

• Q: What is the output?

  • A: The output is a list of materials or practices (strings) that appear more than once across all brands.

• Q: What should the function return if no materials are trending?

  • A: The function should return an empty list.

• Q: Are there any constraints on the input, such as the presence of the "materials" key in each dictionary?

  • A: It is assumed that each dictionary in the list will have both a "name" key and a "materials" key with corresponding values.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Count the occurrences of each material using a dictionary, then iterate through the dictionary to identify and return the materials that appear more than once.

1) Initialize an empty dictionary called `material_count`.
2) For each `brand` in `brands`:
   a) For each `material` in `brand["materials"]`:
      i) If the `material` is already in `material_count`, increment its count.
      ii) If the `material` is not in `material_count`, add it with a count of 1.
3) Initialize an empty list called `trending_materials`.
4) For each `material, count` in `material_count.items()`:
   a) If `count` > 1, append the `material` to `trending_materials`.
5) Return the `trending_materials` list.

**⚠️ Common Mistakes**

- Forgetting to correctly initialize the material count when encountering a material for the first time.
- Not checking that a material appears more than once before marking it as trending.

I-mplement

def find_trending_materials(brands):
    material_count = {}
    trending_materials = []

    for brand in brands:
        for material in brand["materials"]:
            if material in material_count:
                material_count[material] += 1
            else:
                material_count[material] = 1

    for material, count in material_count.items():
        if count > 1:
            trending_materials.append(material)

    return trending_materials