LeetCode in Kotlin

3624. Number of Integers With Popcount-Depth Equal to K II

Hard

You are given an integer array nums.

For any positive integer x, define the following sequence:

• p0 = x
• pi+1 = popcount(pi) for all i >= 0, where popcount(y) is the number of set bits (1’s) in the binary representation of y.

This sequence will eventually reach the value 1.

The popcount-depth of x is defined as the smallest integer d >= 0 such that pd = 1.

For example, if x = 7 (binary representation "111"). Then, the sequence is: 7 → 3 → 2 → 1, so the popcount-depth of 7 is 3.

You are also given a 2D integer array queries, where each queries[i] is either:

• [1, l, r, k] - Determine the number of indices j such that l <= j <= r and the popcount-depth of nums[j] is equal to k.
• [2, idx, val] - Update nums[idx] to val.

Return an integer array answer, where answer[i] is the number of indices for the ith query of type [1, l, r, k].

Example 1:

Input: nums = [2,4], queries = [[1,0,1,1],[2,1,1],[1,0,1,0]]

Output: [2,1]

Explanation:

i	queries[i]	nums	binary(nums)	popcount- depth	[l, r]	k	Valid nums[j]	updated nums	Answer
0	[1,0,1,1]	[2,4]	[10, 100]	[1, 1]	[0, 1]	1	[0, 1]	—	2
1	[2,1,1]	[2,4]	[10, 100]	[1, 1]	—	—	—	[2,1]	—
2	[1,0,1,0]	[2,1]	[10, 1]	[1, 0]	[0, 1]	0	[1]	—	1

Thus, the final answer is [2, 1].

Example 2:

Input: nums = [3,5,6], queries = [[1,0,2,2],[2,1,4],[1,1,2,1],[1,0,1,0]]

Output: [3,1,0]

Explanation:

i	queries[i]	nums	binary(nums)	popcount- depth	[l, r]	k	Valid nums[j]	updated nums	Answer
0	[1,0,2,2]	[3, 5, 6]	[11, 101, 110]	[2, 2, 2]	[0, 2]	2	[0, 1, 2]	—	3
1	[2,1,4]	[3, 5, 6]	[11, 101, 110]	[2, 2, 2]	—	—	—	[3, 4, 6]	—
2	[1,1,2,1]	[3, 4, 6]	[11, 100, 110]	[2, 1, 2]	[1, 2]	1	[1]	—	1
3	[1,0,1,0]	[3, 4, 6]	[11, 100, 110]	[2, 1, 2]	[0, 1]	0	[]	—	0

Thus, the final answer is [3, 1, 0].

Example 3:

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

Output: [1,0,1]

Explanation:

i	queries[i]	nums	binary(nums)	popcount- depth	[l, r]	k	Valid nums[j]	updated nums	Answer
0	[1,0,1,1]	[1, 2]	[1, 10]	[0, 1]	[0, 1]	1	[1]	—	1
1	[2,0,3]	[1, 2]	[1, 10]	[0, 1]	—	—	—	[3, 2]
2	[1,0,0,1]	[3, 2]	[11, 10]	[2, 1]	[0, 0]	1	[]	—	0
3	[1,0,0,2]	[3, 2]	[11, 10]	[2, 1]	[0, 0]	2	[0]	—	1

Thus, the final answer is [1, 0, 1].

Constraints:

• 1 <= n == nums.length <= 105
• 1 <= nums[i] <= 1015
• 1 <= queries.length <= 105
• queries[i].length == 3 or 4
  • queries[i] == [1, l, r, k] or,
  • queries[i] == [2, idx, val]
  • 0 <= l <= r <= n - 1
  • 0 <= k <= 5
  • 0 <= idx <= n - 1
  • 1 <= val <= 1015

Solution

class Solution {
    private fun computeDepth(number: Long): Int {
        if (number == 1L) {
            return 0
        }
        return 1 + DEPTH_TABLE[java.lang.Long.bitCount(number)]
    }

    fun popcountDepth(nums: LongArray, queries: Array<LongArray>): IntArray {
        val len = nums.size
        val maxDepth = 6
        val trees = Array(maxDepth) { FenwickTree() }
        for (d in 0..<maxDepth) {
            trees[d].build(len)
        }
        for (i in 0..<len) {
            val depth = computeDepth(nums[i])
            if (depth < maxDepth) {
                trees[depth].update(i + 1, 1)
            }
        }
        val ansList = ArrayList<Int?>()
        for (query in queries) {
            val type = query[0].toInt()
            if (type == 1) {
                val left = query[1].toInt()
                val right = query[2].toInt()
                val depth = query[3].toInt()
                if (depth >= 0 && depth < maxDepth) {
                    ansList.add(trees[depth].queryRange(left + 1, right + 1))
                } else {
                    ansList.add(0)
                }
            } else if (type == 2) {
                val index = query[1].toInt()
                val newVal = query[2]
                val oldDepth = computeDepth(nums[index])
                if (oldDepth < maxDepth) {
                    trees[oldDepth].update(index + 1, -1)
                }
                nums[index] = newVal
                val newDepth = computeDepth(newVal)
                if (newDepth < maxDepth) {
                    trees[newDepth].update(index + 1, 1)
                }
            }
        }
        val ansArray = IntArray(ansList.size)
        for (i in ansList.indices) {
            ansArray[i] = ansList[i]!!
        }
        return ansArray
    }

    private class FenwickTree {
        private lateinit var tree: IntArray
        private var size = 0

        fun build(n: Int) {
            this.size = n
            this.tree = IntArray(size + 1)
        }

        fun update(index: Int, value: Int) {
            var index = index
            while (index <= size) {
                tree[index] += value
                index += index and (-index)
            }
        }

        fun query(index: Int): Int {
            var index = index
            var result = 0
            while (index > 0) {
                result += tree[index]
                index -= index and (-index)
            }
            return result
        }

        fun queryRange(left: Int, right: Int): Int {
            if (left > right) {
                return 0
            }
            return query(right) - query(left - 1)
        }
    }

    companion object {
        private val DEPTH_TABLE = IntArray(65)

        init {
            DEPTH_TABLE[1] = 0
            for (i in 2..64) {
                DEPTH_TABLE[i] = 1 + DEPTH_TABLE[Integer.bitCount(i)]
            }
        }
    }
}

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