LeetCode Entry
1512. Number of Good Pairs
Count equal pairs
1512. Number of Good Pairs easy
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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]++ } }
}