3.10.2023
1512. Number of Good Pairs easy
blog post
Join me on Telegram
https://t.me/leetcode_daily_unstoppable/358
Problem TLDR
Count equal pairs
Intuition
The naive N^2 solution will work.
Another idea is to store the number frequency
so far and add it to the current result.
Approach
Let’s use Kotlin’s API:
with
fold
Complexity
Time complexity:
O(n)Space complexity:
O(n)
Code
fun numIdenticalPairs(nums: IntArray) = with(IntArray(101)) {
nums.fold(0) { r, t -> r + this[t].also { this[t]++ } }
}