LeetCode in Kotlin

2426. Number of Pairs Satisfying Inequality

Hard

You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that:

• 0 <= i < j <= n - 1 and
• nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

Example 1:

Input: nums1 = [3,2,5], nums2 = [2,2,1], diff = 1

Output: 3

Explanation:

There are 3 pairs that satisfy the conditions:

1. i = 0, j = 1: 3 - 2 <= 2 - 2 + 1. Since i < j and 1 <= 1, this pair satisfies the conditions.

2. i = 0, j = 2: 3 - 5 <= 2 - 1 + 1. Since i < j and -2 <= 2, this pair satisfies the conditions.

3. i = 1, j = 2: 2 - 5 <= 2 - 1 + 1. Since i < j and -3 <= 2, this pair satisfies the conditions.

Therefore, we return 3.

Example 2:

Input: nums1 = [3,-1], nums2 = [-2,2], diff = -1

Output: 0

Explanation:

Since there does not exist any pair that satisfies the conditions, we return 0.

Constraints:

• n == nums1.length == nums2.length
• 2 <= n <= 105
• -104 <= nums1[i], nums2[i] <= 104
• -104 <= diff <= 104

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    private lateinit var cnt: LongArray

    private fun find(x: Int, n: Int): Long {
        var x = x
        var res: Long = 0
        x = x.coerceAtMost(n)
        while (x > 0) {
            res += cnt[x]
            x -= x and -x
        }
        return res
    }

    private fun update(x: Int, n: Int) {
        var x = x
        while (x <= n) {
            cnt[x] = cnt[x] + 1
            x += x and -x
        }
    }

    fun numberOfPairs(nums1: IntArray, nums2: IntArray, diff: Int): Long {
        val n = nums1.size
        val nums = IntArray(n)
        var min = Int.MAX_VALUE
        var max = Int.MIN_VALUE
        for (i in 0 until n) {
            nums[i] = nums1[i] - nums2[i]
            min = min.coerceAtMost(nums[i])
            max = max.coerceAtLeast(nums[i])
        }
        max = max - min + 2
        var ans: Long = 0
        cnt = LongArray(50000)
        for (i in 0 until n) {
            nums[i] = nums[i] - min + 1
            ans += find(nums[i] + diff, max)
            update(nums[i], max)
        }
        return ans
    }
}

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