What is the time complexity of Count Good Meals?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Good Meals?

This problem uses Array, Hash Table. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Good Meals?

Insight 1: We need to efficiently find pairs whose sum is a power of 2. Since individual deliciousness values are at most 2^20, the maximum possible sum is 2 * 2^20 = 2^21. Therefore, we only need to check powers of 2 up to 2^21. Insight 2: A hash table (or frequency map) is an efficient data structure to count the occurrences of each deliciousness value. This allows us to quickly determine the number of complements needed to reach a power of 2.

Which companies ask LeetCode 1711. Count Good Meals in interviews?

This problem is frequently asked at Robinhood, Swiggy.

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Count Good Meals - LeetCode 1711 Solution

Count Good Meals - Complete Solution Guide

Count Good Meals is LeetCode problem 1711, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two. You can pick any two different foods to make a good meal. Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i th item of food, return the number of different good meals you can make from this list modulo 10 9 + 7 . Note that items with different indices are considered different even if they have the same deliciousness

Detailed Explanation

The problem asks us to find the number of 'good meals' we can create from a given list of food item deliciousness values. A 'good meal' consists of exactly two *different* food items, and their combined deliciousness must be a power of 2 (e.g., 2, 4, 8, 16, ...). The input is an array `deliciousness` of integers representing the deliciousness of each food item. We need to return the total number of good meals modulo 10^9 + 7. The constraints specify that the array size is up to 10^5 and individual deliciousness values can be up to 2^20.

Solution Approach

The solution iterates through the `deliciousness` array. For each deliciousness value, it checks all possible powers of 2 (from 2^0 to 2^21) to determine the 'complement' needed to reach that power of 2. Using a hash table, it efficiently looks up the number of occurrences of that complement. The number of complements is added to a running count. Finally, the result is taken modulo 10^9 + 7 before being returned.

Step-by-Step Algorithm

Step 1: Initialize a hash table (or frequency map) to store the count of each deliciousness value. Initialize a counter variable `count` to 0 and the modulo constant `MOD` to 10^9 + 7.
Step 2: Iterate through the `deliciousness` array.
Step 3: For each deliciousness value `d`, iterate through the possible powers of 2 (from 2^0 to 2^21).
Step 4: Calculate the 'complement' needed to reach the current power of 2: `complement = power_of_two - d`.
Step 5: Look up the number of occurrences of the `complement` in the hash table. Add this count to the running `count`.
Step 6: Increment the count of the current deliciousness value `d` in the hash table.
Step 7: After iterating through all deliciousness values, return the `count` modulo `MOD`.

Key Insights

Insight 1: We need to efficiently find pairs whose sum is a power of 2. Since individual deliciousness values are at most 2^20, the maximum possible sum is 2 * 2^20 = 2^21. Therefore, we only need to check powers of 2 up to 2^21.
Insight 2: A hash table (or frequency map) is an efficient data structure to count the occurrences of each deliciousness value. This allows us to quickly determine the number of complements needed to reach a power of 2.
Insight 3: We must ensure to return the result modulo 10^9 + 7 to prevent integer overflow as the number of good meals can be very large.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Robinhood, Swiggy.

Count Good Meals | LeetCode 1711

Medium

Array Hash Table

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int countPairs(int[] deliciousness) {
        int MOD = (int)1e9 + 7;
        
        Map<Integer, Integer> freq = new HashMap<>();
        int count = 0;

        for (int d : deliciousness) {
            for (int i = 0; i < 22; i++) {
                int power_of_two = 1 << i;
                int complement = power_of_two - d;
                count += freq.getOrDefault(complement, 0);
            }
            
            freq.put(d, freq.getOrDefault(d, 0) + 1);
        }
        
        return count % MOD;
    }
}

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Problem Info

Description

Problem 1711: Count Good Meals - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

RobinhoodSwiggy

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1711

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a hash table (or frequency map) to store the count of each deliciousness value. Initialize a counter variable `count` to 0 and the modulo constant `MOD` to 10^9 + 7.
2
Step 2: Iterate through the `deliciousness` array.
3
Step 3: For each deliciousness value `d`, iterate through the possible powers of 2 (from 2^0 to 2^21).
4
Step 4: Calculate the 'complement' needed to reach the current power of 2: `complement = power_of_two - d`.
5
Step 5: Look up the number of occurrences of the `complement` in the hash table. Add this count to the running `count`.
6
Step 6: Increment the count of the current deliciousness value `d` in the hash table.
7
Step 7: After iterating through all deliciousness values, return the `count` modulo `MOD`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the single loop through the input deliciousness array. Inside this loop, a fixed number of computations are performed (22 power of 2 checks) for each element, giving a linear relationship with the input size. The space complexity is determined by the hash map which stores the frequencies of different deliciousness values, growing linearly with the number of distinct values in the input array, up to a maximum of n, where n is the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting and Two Pointers

Sort the `deliciousness` array. Then, for each element, use two pointers to search for a complement that results in a power of 2. Binary search could also be used for faster complement lookup.

Time: O(n log n) due to sorting.Space: O(1) if sorting in-place, or O(n) depending on the sorting algorithm.

Trade-offs: This approach avoids using a hash table, which could be beneficial in situations with extremely limited memory. However, the sorting step makes it slower than the hash table approach.

✨

Key Insights

1Insight 1: We need to efficiently find pairs whose sum is a power of 2. Since individual deliciousness values are at most 2^20, the maximum possible sum is 2 * 2^20 = 2^21. Therefore, we only need to check powers of 2 up to 2^21.
2Insight 2: A hash table (or frequency map) is an efficient data structure to count the occurrences of each deliciousness value. This allows us to quickly determine the number of complements needed to reach a power of 2.
3Insight 3: We must ensure to return the result modulo 10^9 + 7 to prevent integer overflow as the number of good meals can be very large.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Duplicate deliciousness values. The hash table correctly handles duplicate values by incrementing their counts, which ensures that all possible pairs are considered.
Edge case 2: Large number of good meals. The modulo operation (10^9 + 7) prevents integer overflow when calculating the total count.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the modulo operation. The number of good meals can be very large, leading to integer overflow if the result is not taken modulo 10^9 + 7.

✗Common mistake 2: Incorrectly iterating through the powers of 2. It's crucial to calculate powers of 2 until 2^21 because the maximum deliciousness sum is 2 * 2^20 = 2^21.

✗Common mistake 3: Counting the same pair twice (e.g., (a, b) and (b, a)). Since the problem specifies *different* food items, we need to ensure that the same combination is not counted in reverse order. The problem statement guarantees that items at different indices are considered different, which is addressed by the algorithm.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Recommendation systems. Identifying items that complement each other based on some scoring function (e.g., recommending movies that are liked together). In this case, good meals analogy refers to recommending the combinations that have a high chance of being selected together.

🏦Real-world scenario 2: Financial analysis. Finding pairs of assets whose combined value meets a certain target (e.g., finding stocks that when combined, have a predicted growth rate meeting a specific investment goal).

🎮Real-world scenario 3: Network analysis. Identifying pairs of nodes in a network that, when combined, reach a certain network performance threshold (e.g., finding two servers that when combined their throughput performance hits a target).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Good Meals"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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