Package g3601_3700.s3624_number_of_integers_with_popcount_depth_equal_to_k_ii

Class Solution

java.lang.Object

g3601_3700.s3624_number_of_integers_with_popcount_depth_equal_to_k_ii.Solution

public class Solution
extends Object

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-<br>depth	[l, r]	k	Valid<br>nums[j]	updated<br>nums	Answer
0	[1,0,1,1]	[2,4]	[10, 100]	[1, 1]	[0, 1]	1	[0, 1]	\u2014	2
1	[2,1,1]	[2,4]	[10, 100]	[1, 1]	\u2014	\u2014	\u2014	[2,1]	\u2014
2	[1,0,1,0]	[2,1]	[10, 1]	[1, 0]	[0, 1]	0	[1]	\u2014	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-<br>depth	[l, r]	k	Valid<br>nums[j]	updated<br>nums	Answer
0	[1,0,2,2]	[3, 5, 6]	[11, 101, 110]	[2, 2, 2]	[0, 2]	2	[0, 1, 2]	\u2014	3
1	[2,1,4]	[3, 5, 6]	[11, 101, 110]	[2, 2, 2]	\u2014	\u2014	\u2014	[3, 4, 6]	\u2014
2	[1,1,2,1]	[3, 4, 6]	[11, 100, 110]	[2, 1, 2]	[1, 2]	1	[1]	\u2014	1
3	[1,0,1,0]	[3, 4, 6]	[11, 100, 110]	[2, 1, 2]	[0, 1]	0	[]	\u2014	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-<br>depth	[l, r]	k	Valid<br>nums[j]	updated<br>nums	Answer
0	[1,0,1,1]	[1, 2]	[1, 10]	[0, 1]	[0, 1]	1	[1]	\u2014	1
1	[2,0,3]	[1, 2]	[1, 10]	[0, 1]	\u2014	\u2014	\u2014	[3, 2]
2	[1,0,0,1]	[3, 2]	[11, 10]	[2, 1]	[0, 0]	1	[]	\u2014	0
3	[1,0,0,2]	[3, 2]	[11, 10]	[2, 1]	[0, 0]	2	[0]	\u2014	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

Constructor Summary

Constructors

Constructor	Description
Solution()

Method Summary

Modifier and Type	Method	Description
int[]	popcountDepth(long[] nums, long[][] queries)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

Solution

public Solution()

Method Details

popcountDepth

public int[] popcountDepth(long[] nums,
 long[][] queries)