LeetCode in Kotlin

3649. Number of Perfect Pairs

Medium

You are given an integer array nums.

A pair of indices (i, j) is called perfect if the following conditions are satisfied:

• i < j
• Let a = nums[i], b = nums[j]. Then:
  • min(|a - b|, |a + b|) <= min(|a|, |b|)
  • max(|a - b|, |a + b|) >= max(|a|, |b|)

Return the number of distinct perfect pairs.

Note: The absolute value |x| refers to the non-negative value of x.

Example 1:

Input: nums = [0,1,2,3]

Output: 2

Explanation:

There are 2 perfect pairs:

(i, j)	(a, b)	min(|a − b|, |a + b|)	min(|a|, |b|)	max(|a − b|, |a + b|)	max(|a|, |b|)
(1, 2)	(1, 2)	min(|1 − 2|, |1 + 2|) = 1	1	max(|1 − 2|, |1 + 2|) = 3	2
(2, 3)	(2, 3)	min(|2 − 3|, |2 + 3|) = 1	2	max(|2 − 3|, |2 + 3|) = 5	3

Example 2:

Input: nums = [-3,2,-1,4]

Output: 4

Explanation:

There are 4 perfect pairs:

(i, j)	(a, b)	min(|a − b|, |a + b|)	min(|a|, |b|)	max(|a − b|, |a + b|)	max(|a|, |b|)
(0, 1)	(-3, 2)	min(|-3 - 2|, |-3 + 2|) = 1	2	max(|-3 - 2|, |-3 + 2|) = 5	3
(0, 3)	(-3, 4)	min(|-3 - 4|, |-3 + 4|) = 1	3	max(|-3 - 4|, |-3 + 4|) = 7	4
(1, 2)	(2, -1)	min(|2 - (-1)|, |2 + (-1)|) = 1	1	max(|2 - (-1)|, |2 + (-1)|) = 3	2
(1, 3)	(2, 4)	min(|2 - 4|, |2 + 4|) = 2	2	max(|2 - 4|, |2 + 4|) = 6	4

Example 3:

Input: nums = [1,10,100,1000]

Output: 0

Explanation:

There are no perfect pairs. Thus, the answer is 0.

Constraints:

• 2 <= nums.length <= 105
• -109 <= nums[i] <= 109

Solution

import kotlin.math.abs

class Solution {
    fun perfectPairs(nums: IntArray): Long {
        val n = nums.size
        val arr = LongArray(n)
        for (i in 0..<n) {
            arr[i] = abs(nums[i].toLong())
        }
        arr.sort()
        var cnt: Long = 0
        var r = 0
        for (i in 0..<n) {
            if (r < i) {
                r = i
            }
            while (r + 1 < n && arr[r + 1] <= 2 * arr[i]) {
                r++
            }
            cnt += (r - i).toLong()
        }
        return cnt
    }
}

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