How do you solve Finding Pairs With a Certain Sum on LeetCode?

Finding Pairs With a Certain Sum (LeetCode #1865) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(m) space complexity. Key insight: Using a HashMap allows for quick lookups, significantly reducing the time complexity.

What is the brute force solution for Finding Pairs With a Certain Sum?

The brute force approach for Finding Pairs With a Certain Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of elements from nums1 and nums2 to see if their sum equals the target value. While this is straightforward, it can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Finding Pairs With a Certain Sum?

Finding Pairs With a Certain Sum uses the following concepts: Array, Hash Table, Design. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Finding Pairs With a Certain Sum?

Always consider the data structure that can optimize your solution, such as HashMaps for counting. Think about how changes to the data (like adding values) affect your counting logic. Dry-run your solution with sample inputs to ensure correctness before coding.

What are common mistakes when solving Finding Pairs With a Certain Sum?

Not updating the frequency map correctly after adding values to nums2. Forgetting to handle cases where the complement does not exist in nums2.

#1865

Finding Pairs With a Certain Sum

Medium

ArrayHash TableDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(m)

💡

Intuition

Time O(n + m)Space O(m)

Instead of checking every pair, we can use a HashMap to count occurrences of numbers in nums2. This allows us to quickly find how many numbers in nums2 can pair with each number in nums1 to reach the target sum.

⚙️

Algorithm

5 steps

1Step 1: Create a HashMap to store the frequency of each number in nums2.
2Step 2: For the count method, iterate through nums1 and for each element, calculate the required complement (tot - num1).
3Step 3: Use the HashMap to get the frequency of the complement in nums2.
4Step 4: Sum these frequencies to get the total count of pairs.
5Step 5: For the add method, update the HashMap with the new value.

solution.py18 lines

1from collections import Counter
2
3class FindSumPairs:
4    def __init__(self, nums1, nums2):
5        self.nums1 = nums1
6        self.nums2 = nums2
7        self.counter = Counter(nums2)
8
9    def add(self, index, val):
10        self.counter[self.nums2[index]] -= 1
11        self.nums2[index] += val
12        self.counter[self.nums2[index]] += 1
13
14    def count(self, tot):
15        count = 0
16        for num1 in self.nums1:
17            count += self.counter[tot - num1]
18        return count

ℹ

Complexity note: The time complexity is O(n + m) because we build the frequency map in O(m) and then count pairs in O(n). The space complexity is O(m) for storing the frequency of elements in nums2.

1Using a HashMap allows for quick lookups, significantly reducing the time complexity.
2The problem can be viewed as a counting problem where we need to efficiently count complements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.