What is the time complexity of 3Sum With Multiplicity?

The optimal solution has a time complexity of O(1).

What data structures are used to solve 3Sum With Multiplicity?

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

What are the key insights for solving 3Sum With Multiplicity?

Insight 1: The problem can be solved efficiently by using a counting approach. Instead of iterating through all possible triplets, we can count the occurrences of each number in the array and then iterate through possible values for the triplet, calculating the number of ways to form the target sum based on these counts. Insight 2: The constraint that arr[i] is between 0 and 100 is crucial because it allows us to use an array of size 101 to store the counts of each number.

Which companies ask LeetCode 923. 3Sum With Multiplicity in interviews?

This problem is frequently asked at Quora.

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3Sum With Multiplicity - LeetCode 923 Solution

Q: How to solve 3Sum With Multiplicity?

The solution uses a counting approach to efficiently determine the number of triplets that sum to the target. First, it counts the occurrences of each number in the input array `arr` using an array `counts` of size 101. Then, it iterates through all possible pairs of numbers (x, y) from 0 to 100. For each pair, it calculates the required third number z as `target - x - y`. If z is within the valid range [0, 100] and y <= z, it checks for different cases: (1) x == y == z, (2) x == y, (3) y == z, and (4) all three are distinct. Based on the case, it applies the appropriate combination formula using the `counts` array to calculate the number of ways to form the triplet. Finally, the total count of triplets is returned modulo 10^9 + 7.

3Sum With Multiplicity - Complete Solution Guide

3Sum With Multiplicity is LeetCode problem 923, 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

Given an integer array arr , and an integer target , return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target . As the answer can be very large, return it modulo 10 9 + 7 . Example 1: Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8 Output: 20 Explanation: Enumerating by the values (arr[i], arr[j], arr[k]): (1, 2, 5) occurs 8 times; (1, 3, 4) occurs 8 times; (2, 2, 4) occurs 2 times; (2, 3, 3) occurs 2 times. Example 2: Input: arr = [1,1,2,2,2,2], target = 5 O

Detailed Explanation

The problem asks us to find the number of triplets (i, j, k) in a given array `arr` such that `i < j < k` and `arr[i] + arr[j] + arr[k] == target`. Since the array may contain duplicate numbers, we need to count the multiplicity of each triplet. The answer can be very large, so we need to return it modulo 10^9 + 7. The array elements are integers in the range [0, 100], and the target is in the range [0, 300].

Solution Approach

The solution uses a counting approach to efficiently determine the number of triplets that sum to the target. First, it counts the occurrences of each number in the input array `arr` using an array `counts` of size 101. Then, it iterates through all possible pairs of numbers (x, y) from 0 to 100. For each pair, it calculates the required third number z as `target - x - y`. If z is within the valid range [0, 100] and y <= z, it checks for different cases: (1) x == y == z, (2) x == y, (3) y == z, and (4) all three are distinct. Based on the case, it applies the appropriate combination formula using the `counts` array to calculate the number of ways to form the triplet. Finally, the total count of triplets is returned modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize an array `counts` of size 101 to store the frequency of each number in `arr`.
Step 2: Iterate through `arr` and update `counts` by incrementing `counts[num]` for each number `num` in `arr`.
Step 3: Initialize the answer `ans` to 0.
Step 4: Iterate through all possible values of `x` from 0 to 100.
Step 5: Iterate through all possible values of `y` from `x` to 100.
Step 6: Calculate `z = target - x - y`.
Step 7: Check if `y <= z <= 100`. If not, continue to the next iteration.
Step 8: If `x == y == z`, then `ans += counts[x] * (counts[x] - 1) * (counts[x] - 2) // 6` if `counts[x] >= 3`.
Step 9: Else if `x == y`, then `ans += (counts[x] * (counts[x] - 1) // 2) * counts[z]` if `counts[x] >= 2`.
Step 10: Else if `y == z`, then `ans += counts[x] * (counts[y] * (counts[y] - 1) // 2)` if `counts[y] >= 2`.
Step 11: Else, `ans += counts[x] * counts[y] * counts[z]`.
Step 12: Take the modulo of `ans` by 10^9 + 7 after each addition to prevent overflow. In the provided code the modulo operation is done at the end of the calculation.
Step 13: Return `ans % (10**9 + 7)`.

Key Insights

Insight 1: The problem can be solved efficiently by using a counting approach. Instead of iterating through all possible triplets, we can count the occurrences of each number in the array and then iterate through possible values for the triplet, calculating the number of ways to form the target sum based on these counts.
Insight 2: The constraint that arr[i] is between 0 and 100 is crucial because it allows us to use an array of size 101 to store the counts of each number.
Insight 3: Handling different cases for x, y, and z (x == y == z, x == y, y == z, x != y != z) is essential for correct counting. We need to apply appropriate combinations formulas for each scenario to avoid overcounting or undercounting.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Two Pointers, Sorting, Counting.

Companies

Asked at: Quora.

3Sum With Multiplicity | LeetCode 923

Medium

Array Hash Table Two Pointers Sorting Counting

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public int threeSumMulti(int[] arr, int target) {
        int MOD = 1000000007;
        long[] counts = new long[101];
        for (int num : arr) {
            counts[num]++;
        }
        
        long ans = 0;
        
        for (int x = 0; x <= 100; x++) {
            for (int y = x; y <= 100; y++) {
                int z = target - x - y;
                
                if (y <= z && z <= 100) {
                    if (x == y && y == z) {
                        if (counts[x] >= 3) {
                            ans += counts[x] * (counts[x] - 1) * (counts[x] - 2) / 6;
                        }
                    } else if (x == y) {
                        if (counts[x] >= 2) {
                            ans += (counts[x] * (counts[x] - 1) / 2) * counts[z];
                        }
                    } else if (y == z) {
                        if (counts[y] >= 2) {
                            ans += counts[x] * (counts[y] * (counts[y] - 1) / 2);
                        }
                    } else {
                        ans += counts[x] * counts[y] * counts[z];
                    }
                }
            }
        }
        
        return (int)(ans % MOD);
    }
}

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Problem Info

Description

Problem 923: 3Sum With Multiplicity - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers Sorting Counting

Companies

Quora

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:923

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

13 steps

1
Step 1: Initialize an array `counts` of size 101 to store the frequency of each number in `arr`.
2
Step 2: Iterate through `arr` and update `counts` by incrementing `counts[num]` for each number `num` in `arr`.
3
Step 3: Initialize the answer `ans` to 0.
4
Step 4: Iterate through all possible values of `x` from 0 to 100.
5
Step 5: Iterate through all possible values of `y` from `x` to 100.
6
Step 6: Calculate `z = target - x - y`.
7
Step 7: Check if `y <= z <= 100`. If not, continue to the next iteration.
8
Step 8: If `x == y == z`, then `ans += counts[x] * (counts[x] - 1) * (counts[x] - 2) // 6` if `counts[x] >= 3`.
9
Step 9: Else if `x == y`, then `ans += (counts[x] * (counts[x] - 1) // 2) * counts[z]` if `counts[x] >= 2`.
10
Step 10: Else if `y == z`, then `ans += counts[x] * (counts[y] * (counts[y] - 1) // 2)` if `counts[y] >= 2`.
11
Step 11: Else, `ans += counts[x] * counts[y] * counts[z]`.
12
Step 12: Take the modulo of `ans` by 10^9 + 7 after each addition to prevent overflow. In the provided code the modulo operation is done at the end of the calculation.
13
Step 13: Return `ans % (10**9 + 7)`.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(1) since the loops iterate through a fixed range of numbers (0 to 100) independent of the input array size. The space complexity is also O(1) because the counts array has a fixed size, again independent of the input array size.

📖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

Two Pointers

Sort the array first and then use a two-pointer approach. Iterate through the array with index `i`, and for each `i`, use two pointers to find pairs `j` and `k` such that `arr[i] + arr[j] + arr[k] == target` and `i < j < k`. This needs careful handling of duplicates.

Time: O(N^2), where N is the length of the array. Sorting takes O(N log N), and the two-pointer approach takes O(N^2).Space: O(log N) or O(N), depending on the sorting algorithm used.

Trade-offs: This approach could be useful if the numbers in the array were not restricted to the range 0-100. However, it is generally less efficient than the counting approach when the range of numbers is small because of the sorting overhead and duplicate handling complexities.

✨

Key Insights

1Insight 1: The problem can be solved efficiently by using a counting approach. Instead of iterating through all possible triplets, we can count the occurrences of each number in the array and then iterate through possible values for the triplet, calculating the number of ways to form the target sum based on these counts.
2Insight 2: The constraint that arr[i] is between 0 and 100 is crucial because it allows us to use an array of size 101 to store the counts of each number.
3Insight 3: Handling different cases for x, y, and z (x == y == z, x == y, y == z, x != y != z) is essential for correct counting. We need to apply appropriate combinations formulas for each scenario to avoid overcounting or undercounting.

📚

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!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

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

Edge Cases

Edge case 1: When `x == y == z`, we need to ensure we have at least three occurrences of that number in the array. The condition `counts[x] >= 3` handles this.
Edge case 2: When `x == y` or `y == z`, we need to ensure we have at least two occurrences of the repeated number. The conditions `counts[x] >= 2` and `counts[y] >= 2` handle these respectively.
Edge case 3: The `target` value might be small or large. We need to ensure that `z` remains within the valid range of [0, 100] to avoid accessing out-of-bounds memory in the `counts` array. The condition `y <= z && z <= 100` handles this.
Edge case 4: The result can be very large, so it is necessary to perform the modulo operation at the end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the modulo operation correctly. The intermediate sums can become very large, so it's important to apply the modulo operation (`% (10**9 + 7)`) to prevent integer overflow.

✗Common mistake 2: Overcounting or undercounting triplets due to duplicates. Failing to distinguish between the cases where all three numbers are the same, two numbers are the same, or all three numbers are different can lead to incorrect results.

✗Common mistake 3: Not considering the constraints of the problem. Forgetting that the numbers are within the range 0-100 can lead to inefficient solutions.

✗Common mistake 4: Incorrectly calculating the number of combinations when elements are repeated.

✅

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: Finding combinations of items that meet a specific price target in a retail setting. Given the price of items, the problem aims to find the number of ways to buy three items that add up to the specified budget.

🏦Real-world scenario 2: Identifying patterns in network traffic. Suppose you have network packets of different sizes and you want to find the number of combinations of three packets that reach a specific size, for security analysis or optimizing network performance.

💡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.

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: June 24, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 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 "3Sum With Multiplicity"?

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.

🤖 Need help with this problem?