LeetCode in Kotlin
2364. Count Number of Bad Pairs
Medium
You are given a 0-indexed integer array nums. A pair of indices (i, j) is a bad pair if i < j and j - i != nums[j] - nums[i].
Return the total number of bad pairs in nums.
Example 1:
Input: nums = [4,1,3,3]
Output: 5
Explanation:
The pair (0, 1) is a bad pair since 1 - 0 != 1 - 4.
The pair (0, 2) is a bad pair since 2 - 0 != 3 - 4, 2 != -1.
The pair (0, 3) is a bad pair since 3 - 0 != 3 - 4, 3 != -1.
The pair (1, 2) is a bad pair since 2 - 1 != 3 - 1, 1 != 2.
The pair (2, 3) is a bad pair since 3 - 2 != 3 - 3, 1 != 0.
There are a total of 5 bad pairs, so we return 5.
Example 2:
Input: nums = [1,2,3,4,5]
Output: 0
Explanation: There are no bad pairs.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
class Solution {
fun countBadPairs(nums: IntArray): Long {
val seen = HashMap<Int, Int>()
var count: Long = 0
for (i in nums.indices) {
val diff = i - nums[i]
count += if (seen.containsKey(diff)) {
(i - seen[diff]!!).toLong()
} else {
i.toLong()
}
seen[diff] = seen.getOrDefault(diff, 0) + 1
}
return count
}
}