Is Finding Pairs With a Certain Sum easy or hard?

The problem is rated Medium because it combines array processing with data structure design. The core logic is straightforward once you recognize that maintaining a frequency map allows efficient pair counting across repeated queries.

Finding Pairs With a Certain Sum Python/Java solution

In Python, use a dictionary or collections.Counter to store frequencies of nums2. In Java, use a HashMap . Both implementations update frequencies during add operations and iterate through nums1 during count queries.

How to solve Finding Pairs With a Certain Sum in O(n)?

Each count query can be handled in O(n1) time using a frequency hash map. Iterate through nums1, compute the complement total - x, and retrieve how many times it appears in nums2 from the map. The add operation updates the map in constant time.

What is the best approach for Finding Pairs With a Certain Sum?

The HashMap frequency approach is the most efficient design. Store counts of every value in nums2, then for each element x in nums1 check how many times total - x appears using a hash lookup. Add operations update the frequency map in O(1), while count queries run in O(n1) time.

Is Finding Pairs With a Certain Sum asked at Google/Amazon/Meta?

Design-style problems that combine pair sums with dynamic updates appear frequently in interviews at companies like Amazon, Google, and Meta. They test knowledge of hash maps, frequency counting, and designing efficient query operations.

What data structure is used in Finding Pairs With a Certain Sum?

A hash table (HashMap or unordered_map) is the key data structure. It stores the frequency of each value in nums2, enabling constant-time lookups when searching for complements during pair counting.

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

Using the optimal HashMap design, the add operation runs in O(1) time and the count operation runs in O(n1), where n1 is the size of nums1. Space complexity is O(n2) to store frequencies of values from nums2.

Description Hints 2 Solution Video 19:47

1865. Finding Pairs With a Certain Sum

Medium61.6% acceptance8 companies

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 add(int index, int val) Adds val to nums2[index], i.e., apply nums2[index] += val.
int count(int tot) Returns the number of pairs (i, j) such that nums1[i] + nums2[j] == tot.

Example 1:

Input
["FindSumPairs", "count", "add", "count", "count", "add", "add", "count"]
[[[1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]], [7], [3, 2], [8], [4], [0, 1], [1, 1], [7]]
Output
[null, 8, null, 2, 1, null, null, 11]

Explanation
FindSumPairs findSumPairs = new FindSumPairs([1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]);
findSumPairs.count(7);  // return 8; pairs (2,2), (3,2), (4,2), (2,4), (3,4), (4,4) make 2 + 5 and pairs (5,1), (5,5) make 3 + 4
findSumPairs.add(3, 2); // now nums2 = [1,4,5,4,5,4]
findSumPairs.count(8);  // return 2; pairs (5,2), (5,4) make 3 + 5
findSumPairs.count(4);  // return 1; pair (5,0) makes 3 + 1
findSumPairs.add(0, 1); // now nums2 = [2,4,5,4,5,4]
findSumPairs.add(1, 1); // now nums2 = [2,5,5,4,5,4]
findSumPairs.count(7);  // return 11; pairs (2,1), (2,2), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,4) make 2 + 5 and pairs (5,3), (5,5) make 3 + 4

Constraints: