Optimize Code Time and Space | Dataford Interview Questions

Your question is Optimize Code Time and Space. Take a moment with it on the right.

Talk me through your thinking if you like. When you're confident, submit your answer and I'll grade it like a real screen (7/10 or better passes).

Problem

Lowe's runs seasonal bundle promotions where two SKUs whose prices sum exactly to a promo total get auto-linked as a suggested bundle. A merchandising analyst wrote the function below to find those matching price pairs from a store's price list, but on a full department price list (several thousand SKUs) it takes minutes to run and burns memory the team can't account for.

def find_bundle_pairs(prices, target_total):
    all_pairs = []
    for i in range(len(prices)):
        for j in range(len(prices)):
            if i != j:
                all_pairs.append((i, j))

    matches = []
    for (i, j) in all_pairs:
        if prices[i] + prices[j] == target_total:
            matches.append((i, j))

    unique_matches = []
    for pair in matches:
        if pair not in unique_matches and (pair[1], pair[0]) not in unique_matches:
            unique_matches.append(pair)

    return unique_matches

Explain the time and space complexity problems in this function and describe exactly how you would rewrite it so it scales to a full department price list.

Your question is Optimize Code Time and Space. Take a moment with it on the right.

Talk me through your thinking if you like. When you're confident, submit your answer and I'll grade it like a real screen (7/10 or better passes).

def find_bundle_pairs(prices, target_total):
    all_pairs = []
    for i in range(len(prices)):
        for j in range(len(prices)):
            if i != j:
                all_pairs.append((i, j))

    matches = []
    for (i, j) in all_pairs:
        if prices[i] + prices[j] == target_total:
            matches.append((i, j))

    unique_matches = []
    for pair in matches:
        if pair not in unique_matches and (pair[1], pair[0]) not in unique_matches:
            unique_matches.append(pair)

    return unique_matches

Explain the time and space complexity problems in this function and describe exactly how you would rewrite it so it scales to a full department price list.