What is the time complexity of Successful Pairs of Spells and Potions?

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

What data structures are used to solve Successful Pairs of Spells and Potions?

This problem uses Array, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Successful Pairs of Spells and Potions?

Insight 1: Sorting the `potions` array allows us to use binary search for efficient counting. Insight 2: For each spell, we need to find the minimum potion strength required to form a successful pair: `potion >= success / spell`. We must handle integer division carefully.

Which companies ask LeetCode 2300. Successful Pairs of Spells and Potions in interviews?

This problem is frequently asked at Goldman Sachs.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Successful Pairs of Spells and Potions - LeetCode 2300 Solution

Successful Pairs of Spells and Potions - Complete Solution Guide

Successful Pairs of Spells and Potions is LeetCode problem 2300, 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

You are given two positive integer arrays spells and potions , of length n and m respectively, where spells[i] represents the strength of the i th spell and potions[j] represents the strength of the j th potion. You are also given an integer success . A spell and potion pair is considered successful if the product of their strengths is at least success . Return an integer array pairs of length n where pairs[i] is the number of potions that will form a successful pair with the i th spell. Example

Detailed Explanation

The problem requires us to find, for each spell in the `spells` array, how many potions in the `potions` array can form a successful pair. A successful pair is defined as a spell and potion whose product is greater than or equal to a given `success` value. The goal is to return an array `pairs` where `pairs[i]` represents the number of potions that form a successful pair with the `i`-th spell.

Solution Approach

The provided solution uses a binary search approach after sorting the `potions` array. For each spell, it calculates the minimum potion strength required for a successful pair, then uses binary search (specifically, `bisect_left` or a similar implementation) to find the index of the first potion that meets this minimum strength. The number of potions to the right of this index (inclusive) represents the number of successful pairs for that spell.

Step-by-Step Algorithm

Step 1: Sort the `potions` array in ascending order. This enables efficient searching.
Step 2: Iterate through the `spells` array. For each `spell` value:
Step 3: Calculate the minimum required potion strength: `min_potion_strength = (success + spell - 1) // spell`. This ensures we get the ceiling of `success / spell` when using integer division, meaning we always round up to the nearest integer.
Step 4: Use binary search (e.g., `bisect_left` in Python, or a custom binary search function in Java/C/C++) to find the index of the first potion in the sorted `potions` array that is greater than or equal to `min_potion_strength`.
Step 5: Calculate the number of successful pairs for the current spell: `count = m - index`, where `m` is the length of the `potions` array and `index` is the index found in the previous step.
Step 6: Add the `count` to the `pairs` array.
Step 7: Return the `pairs` array.

Key Insights

Insight 1: Sorting the `potions` array allows us to use binary search for efficient counting.
Insight 2: For each spell, we need to find the minimum potion strength required to form a successful pair: `potion >= success / spell`. We must handle integer division carefully.
Insight 3: Using `bisect_left` (or equivalent binary search) helps find the index of the first potion that meets the minimum strength requirement. The number of potions from that index to the end of the `potions` array are the successful pairs.

Complexity Analysis

Time Complexity: O(n log m)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Goldman Sachs.

Successful Pairs of Spells and Potions | LeetCode 2300

Medium

Array Two Pointers Binary Search Sorting

Time: O(n log m)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        int m = potions.length;
        
        int[] pairs = new int[spells.length];
        for (int i = 0; i < spells.length; i++) {
            int spell = spells[i];
            // We need spell * potion >= success, which means potion >= success / spell.
            // To handle integer division correctly for the lower bound, we calculate the ceiling of success / spell.
            long minPotionStrength = (success + spell - 1) / spell;
            
            // Use binary search to find the index of the first potion that meets the minimum strength.
            // bisect_left gives the index of the first element >= target.
            int idx = binarySearch(potions, minPotionStrength);
            
            // The number of successful potions is the total number of potions minus this index.
            // All potions from this index to the end are valid.
            int count = m - idx;
            pairs[i] = count;
        }
        
        return pairs;
    }
    
    private int binarySearch(int[] potions, long target) {
        int left = 0;
        int right = potions.length - 1;
        int result = potions.length; // Default to length if no potion is strong enough

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (potions[mid] >= target) {
                result = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2300: Successful Pairs of Spells and Potions - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

Goldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2300

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: Sort the `potions` array in ascending order. This enables efficient searching.
2
Step 2: Iterate through the `spells` array. For each `spell` value:
3
Step 3: Calculate the minimum required potion strength: `min_potion_strength = (success + spell - 1) // spell`. This ensures we get the ceiling of `success / spell` when using integer division, meaning we always round up to the nearest integer.
4
Step 4: Use binary search (e.g., `bisect_left` in Python, or a custom binary search function in Java/C/C++) to find the index of the first potion in the sorted `potions` array that is greater than or equal to `min_potion_strength`.
5
Step 5: Calculate the number of successful pairs for the current spell: `count = m - index`, where `m` is the length of the `potions` array and `index` is the index found in the previous step.
6
Step 6: Add the `count` to the `pairs` array.
7
Step 7: Return the `pairs` array.

💡

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 log m)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting `potions` takes O(m log m). For each spell, the minimum potion strength is computed in O(1). The binary search takes O(log m) time. These steps are repeated for each of the n spells, resulting in O(n log m). The overall time complexity is O(m log m + n log m). If we assume m log m is less significant than n log m, then the complexity is O(n log m). The space complexity is O(1) because the algorithm operates in-place or with a fixed number of additional variables.

📖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

Brute Force

Iterate through each spell and then iterate through each potion to check if the product is greater than or equal to `success`. Count the successful pairs for each spell.

Time: O(n*m), where n is the length of the `spells` array and m is the length of the `potions` array.Space: O(1)

Trade-offs: This approach is simple to implement but inefficient for large input sizes. It should only be considered when n and m are very small. The sorted array with Binary Search is significantly better.

✨

Key Insights

1Insight 1: Sorting the `potions` array allows us to use binary search for efficient counting.
2Insight 2: For each spell, we need to find the minimum potion strength required to form a successful pair: `potion >= success / spell`. We must handle integer division carefully.
3Insight 3: Using `bisect_left` (or equivalent binary search) helps find the index of the first potion that meets the minimum strength requirement. The number of potions from that index to the end of the `potions` array are the successful pairs.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `success` is very large. The `min_potion_strength` calculation could potentially overflow if `success` is close to the maximum possible value. The code avoids overflow by carefully calculating the ceiling of `success / spell` using the expression `(success + spell - 1) // spell`.
Edge case 2: `potions` array is empty. The binary search will still function correctly, but the result will be `potions.length`, which is 0. The subsequent calculation `count = m - index` will result in `count = 0 - 0` which equals to zero, thus handling the case correctly. No successful pairs will be found.
Edge case 3: No potion is strong enough to form a successful pair with a given spell. The binary search will return `potions.length`, resulting in `count = 0` successful pairs.
Edge case 4: All potions are strong enough. The binary search will return index 0. `count` will be `potions.length`, the total number of potions.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating `spell * potion`. The problem description mentions `success` can be up to 10^10, so directly multiplying `spell` and `potion` without using `long` for intermediate calculations can cause overflow.

✗Common mistake 2: Incorrectly calculating the minimum potion strength required. Incorrectly dividing `success / spell` due to not ceiling the value.

✗Common mistake 3: Not sorting the `potions` array before performing binary search. This will lead to incorrect results.

✅

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: E-commerce: Matching products with customer preferences. `Spells` could represent customer preference scores, `potions` product features, and `success` a minimum compatibility score. This approach can efficiently find products that meet or exceed the customer's requirements.

🏦Real-world scenario 2: Resource allocation: Assigning tasks to workers. `Spells` represent the skill level of each worker, `potions` represent the difficulty of each task, and `success` represents a minimum performance threshold. The algorithm efficiently determines the tasks that each worker is qualified to handle.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: October 19, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Successful Pairs of Spells and Potions"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?