What is the time complexity of Finding Pairs With a Certain Sum?

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

What data structures are used to solve Finding Pairs With a Certain Sum?

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

What are the key insights for solving Finding Pairs With a Certain Sum?

Insight 1: Storing frequencies of elements in `nums1` and `nums2` using hash maps allows for efficient counting of pairs that sum to the target. Insight 2: Using a hash map (or counter in Python) for frequencies enables O(1) lookups when counting pairs and updating frequencies after an `add` operation.

Which companies ask LeetCode 1865. Finding Pairs With a Certain Sum in interviews?

This problem is frequently asked at Databricks, Quora.

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Finding Pairs With a Certain Sum - LeetCode 1865 Solution

Q: How to solve Finding Pairs With a Certain Sum?

The provided solution uses hash maps (or Counters in Python) to store the frequencies of elements in `nums1` and `nums2`. During initialization, the frequencies of each element in both arrays are calculated and stored. The `add` operation updates the value at the specified index in `nums2`, and the corresponding frequency in the `freq2` hash map is adjusted. The `count` operation iterates through the frequency map of `nums1`. For each number in `nums1`, it calculates the required complement in `nums2` to reach the target sum `tot`. It then multiplies the frequency of that number in `nums1` by the frequency of its complement in `nums2` and adds the result to the total count.

Finding Pairs With a Certain Sum - Complete Solution Guide

Finding Pairs With a Certain Sum is LeetCode problem 1865, 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 integer arrays nums1 and nums2 . You are tasked to implement a data structure that supports queries of two types: Add a positive integer to an element of a given index in the array nums2 . Count the number of pairs (i, j) such that nums1[i] + nums2[j] equals a given value ( 0 <= i < nums1.length and 0 <= j < nums2.length ). Implement the FindSumPairs class: FindSumPairs(int[] nums1, int[] nums2) Initializes the FindSumPairs object with two integer arrays nums1 and nums2 . void

Detailed Explanation

The problem requires implementing a `FindSumPairs` class that can handle two types of queries: `add` and `count`. The class is initialized with two integer arrays, `nums1` and `nums2`. The `add(index, val)` operation increases the value of `nums2[index]` by `val`. The `count(tot)` operation returns the number of pairs (i, j) such that `nums1[i] + nums2[j] == tot`. In essence, we need an efficient way to update elements in `nums2` and quickly determine the number of pairs from `nums1` and `nums2` that sum up to a given target value `tot`.

Solution Approach

The provided solution uses hash maps (or Counters in Python) to store the frequencies of elements in `nums1` and `nums2`. During initialization, the frequencies of each element in both arrays are calculated and stored. The `add` operation updates the value at the specified index in `nums2`, and the corresponding frequency in the `freq2` hash map is adjusted. The `count` operation iterates through the frequency map of `nums1`. For each number in `nums1`, it calculates the required complement in `nums2` to reach the target sum `tot`. It then multiplies the frequency of that number in `nums1` by the frequency of its complement in `nums2` and adds the result to the total count.

Step-by-Step Algorithm

Step 1: Initialize the `FindSumPairs` class with `nums1` and `nums2`. Create frequency maps `freq1` and `freq2` to store the count of each element in `nums1` and `nums2`, respectively.
Step 2: Implement the `add(index, val)` method. Decrement the frequency of the old value at `nums2[index]` in `freq2`. If the frequency becomes 0, remove the element from `freq2`. Update the value at `nums2[index]` by adding `val`. Increment the frequency of the new value at `nums2[index]` in `freq2`.
Step 3: Implement the `count(tot)` method. Iterate through the `freq1` map. For each element `num` in `nums1` and its corresponding frequency `freq`, calculate the target `target = tot - num`. Check if the `target` exists in `freq2`. If it does, multiply `freq` with the frequency of `target` in `freq2` and add it to the running `result`. Return the `result`.

Key Insights

Insight 1: Storing frequencies of elements in `nums1` and `nums2` using hash maps allows for efficient counting of pairs that sum to the target.
Insight 2: Using a hash map (or counter in Python) for frequencies enables O(1) lookups when counting pairs and updating frequencies after an `add` operation.
Insight 3: The `add` operation only modifies `nums2`, so the frequencies of `nums1` remain constant. Only the frequencies of `nums2` need to be updated.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Design.

Companies

Asked at: Databricks, Quora.

Finding Pairs With a Certain Sum | LeetCode 1865

Medium

Array Hash Table Design

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

Solution Code

Available in:

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

class FindSumPairs {

    private Map<Integer, Integer> freq1;
    private int[] nums2;
    private Map<Integer, Integer> freq2;

    public FindSumPairs(int[] nums1, int[] nums2) {
        this.freq1 = new HashMap<>();
        for (int num : nums1) {
            freq1.put(num, freq1.getOrDefault(num, 0) + 1);
        }
        this.nums2 = nums2;
        this.freq2 = new HashMap<>();
        for (int num : nums2) {
            freq2.put(num, freq2.getOrDefault(num, 0) + 1);
        }
    }
    
    public void add(int index, int val) {
        int old_val = this.nums2[index];
        freq2.put(old_val, freq2.get(old_val) - 1);
        if (freq2.get(old_val) == 0) {
            freq2.remove(old_val);
        }
        
        this.nums2[index] += val;
        int new_val = this.nums2[index];
        freq2.put(new_val, freq2.getOrDefault(new_val, 0) + 1);
    }
    
    public int count(int tot) {
        int result = 0;
        for (Map.Entry<Integer, Integer> entry : freq1.entrySet()) {
            int num = entry.getKey();
            int freq = entry.getValue();
            int target = tot - num;
            result += freq * freq2.getOrDefault(target, 0);
        }
        return result;
    }
}

/**
 * Your FindSumPairs object will be instantiated and called as such:
 * FindSumPairs obj = new FindSumPairs(nums1, nums2);
 * obj.add(index,val);
 * int param_2 = obj.count(tot);
 */

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Problem Info

Description

Problem 1865: Finding Pairs With a Certain Sum - Detailed explanation and solution approaches.

Topics

Array Hash Table Design

Companies

DatabricksQuora

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1865

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

3 steps

1
Step 1: Initialize the `FindSumPairs` class with `nums1` and `nums2`. Create frequency maps `freq1` and `freq2` to store the count of each element in `nums1` and `nums2`, respectively.
2
Step 2: Implement the `add(index, val)` method. Decrement the frequency of the old value at `nums2[index]` in `freq2`. If the frequency becomes 0, remove the element from `freq2`. Update the value at `nums2[index]` by adding `val`. Increment the frequency of the new value at `nums2[index]` in `freq2`.
3
Step 3: Implement the `count(tot)` method. Iterate through the `freq1` map. For each element `num` in `nums1` and its corresponding frequency `freq`, calculate the target `target = tot - num`. Check if the `target` exists in `freq2`. If it does, multiply `freq` with the frequency of `target` in `freq2` and add it to the running `result`. Return the `result`.

💡

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 `count` function iterates through the `freq1` hash map, which can have at most `n` elements, and performs constant-time operations for each element. The `add` function involves constant-time hash map operations. The space complexity is determined by the space required to store the frequency maps, which depend on the number of distinct elements in the input arrays.

📖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 all pairs of elements in `nums1` and `nums2` and count the number of pairs that sum to `tot`. The add method would just directly modify the array element.

Time: O(m*k) for count, O(1) for add, where m and k are the lengths of `nums1` and `nums2` respectively.Space: O(1)

Trade-offs: The brute force approach has a much higher time complexity for the `count` operation, making it inefficient for large arrays and frequent `count` calls. However, it requires minimal extra space. This approach is simpler to implement but not suitable for the given constraints due to its poor time complexity.

✨

Key Insights

1Insight 1: Storing frequencies of elements in `nums1` and `nums2` using hash maps allows for efficient counting of pairs that sum to the target.
2Insight 2: Using a hash map (or counter in Python) for frequencies enables O(1) lookups when counting pairs and updating frequencies after an `add` operation.
3Insight 3: The `add` operation only modifies `nums2`, so the frequencies of `nums1` remain constant. Only the frequencies of `nums2` need to be updated.

📚

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: `nums1` or `nums2` is empty. The code handles this correctly because the loops won't execute, and the `count` method will return 0.
Edge case 2: The frequency of an element becomes zero after decrementing in the `add` function. The code explicitly removes the element from the `freq2` hash map to save space and prevent unnecessary lookups.
Edge case 3: Target is very large or very small such that calculating the complement results in integer overflow. The provided solutions do not specifically address overflow, but the constraint `1 <= tot <= 10^9` implies that if `nums1[i]` and `nums2[j]` are within their constraints, then there will be no overflow if they add up to at most `10^9`. This is handled as `if (target >= 0 && target < obj->maxVal)` condition in C code

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the frequency map `freq2` after the `add` operation. This will lead to incorrect counts in the `count` method.

✗Common mistake 2: Not handling the case where the frequency of an element becomes zero after the `add` operation. This can lead to memory leaks (in languages like C) and incorrect calculations if the zero frequency is still considered.

✗Common mistake 3: Not considering the constraints of the problem. A brute-force approach might seem simpler initially but will likely fail due to time limit exceeded errors for larger inputs.

✅

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 product recommendations. `nums1` represents user purchase history, `nums2` represents available items. The `count` operation finds pairs (user, item) with a sum (compatibility score) equal to a threshold to recommend relevant items.

🏦Real-world scenario 2: Database query optimization. The algorithm can be used to efficiently count the number of records from two tables that satisfy a join condition based on the sum of two attributes. The `add` operation can represent updates to one of the tables.

💡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: August 5, 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

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

@InterviewProJun 30

❓

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 "Finding Pairs With a Certain Sum"?

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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