Count Equal and Divisible Pairs in an Array | Brute Force | Optimal | Detailed Maths | Leetcode 2176
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Hi Everyone, this is the 80th video of our Playlist "Leetcode Easy".Now we will be solving an easy array problem - Count Equal and Divisible Pairs in an Array | Brute Force | Optimal | Detailed Maths | Leetcode 2176 | codestorywithMIK
We will first solve it from Brute Force and then improve it using maths. We will deep dive in Mathematical part so that we understand the reason behind everything.
Problem Name : Count Good Triplets | Simple Approach | Slight Improvements | Leetcode 1534 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Leetcode Easy/Count Equal and Divisible Pairs in an Array.cpp
Leetcode Link : https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array
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Video Summary :
The problem is to count pairs (i, j) where nums[i] == nums[j] and (i * j) % k == 0. We group indices by value to satisfy the first condition efficiently. For the second condition, we use number theory: for each index i, we compute gcd(i, k) and determine valid j values using precomputed divisors of k. A hashmap tracks how many indices satisfy the needed conditions, allowing for efficient pair counting.
Timelines
00:00 - Introduction
0:18 - Motivation
0:33 - Problem Explanation
1:25 - Brute Force
3:00 - Optimal - Detailed Math Behind it
11:35 - Dry Run
20:55 - Story Points
22:03 - Coding it up
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