{ "data": { "question": { "questionId": "3941", "questionFrontendId": "3624", "categoryTitle": "Algorithms", "boundTopicId": 3727791, "title": "Number of Integers With Popcount-Depth Equal to K II", "titleSlug": "number-of-integers-with-popcount-depth-equal-to-k-ii", "content": "
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.
\n\nThe 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]
.
\n
Example 1:
\n\nInput: nums = [2,4], queries = [[1,0,1,1],[2,1,1],[1,0,1,0]]
\n\nOutput: [2,1]
\n\nExplanation:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\tValid \n\t\t\t nums[j] | \n\t\t\tupdated \n\t\t\t nums | \n\t\t\tAnswer | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,1,1] | \n\t\t\t[2,4] | \n\t\t\t[10, 100] | \n\t\t\t[1, 1] | \n\t\t\t[0, 1] | \n\t\t\t1 | \n\t\t\t[0, 1] | \n\t\t\t— | \n\t\t\t2 | \n\t\t
1 | \n\t\t\t[2,1,1] | \n\t\t\t[2,4] | \n\t\t\t[10, 100] | \n\t\t\t[1, 1] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[2,1] | \n\t\t\t— | \n\t\t
2 | \n\t\t\t[1,0,1,0] | \n\t\t\t[2,1] | \n\t\t\t[10, 1] | \n\t\t\t[1, 0] | \n\t\t\t[0, 1] | \n\t\t\t0 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
Thus, the final answer
is [2, 1]
.
Example 2:
\n\nInput: nums = [3,5,6], queries = [[1,0,2,2],[2,1,4],[1,1,2,1],[1,0,1,0]]
\n\nOutput: [3,1,0]
\n\nExplanation:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\tValid \n\t\t\t nums[j] | \n\t\t\tupdated \n\t\t\t nums | \n\t\t\tAnswer | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,2,2] | \n\t\t\t[3, 5, 6] | \n\t\t\t[11, 101, 110] | \n\t\t\t[2, 2, 2] | \n\t\t\t[0, 2] | \n\t\t\t2 | \n\t\t\t[0, 1, 2] | \n\t\t\t— | \n\t\t\t3 | \n\t\t
1 | \n\t\t\t[2,1,4] | \n\t\t\t[3, 5, 6] | \n\t\t\t[11, 101, 110] | \n\t\t\t[2, 2, 2] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[3, 4, 6] | \n\t\t\t— | \n\t\t
2 | \n\t\t\t[1,1,2,1] | \n\t\t\t[3, 4, 6] | \n\t\t\t[11, 100, 110] | \n\t\t\t[2, 1, 2] | \n\t\t\t[1, 2] | \n\t\t\t1 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
3 | \n\t\t\t[1,0,1,0] | \n\t\t\t[3, 4, 6] | \n\t\t\t[11, 100, 110] | \n\t\t\t[2, 1, 2] | \n\t\t\t[0, 1] | \n\t\t\t0 | \n\t\t\t[] | \n\t\t\t— | \n\t\t\t0 | \n\t\t
Thus, the final answer
is [3, 1, 0]
.
Example 3:
\n\nInput: nums = [1,2], queries = [[1,0,1,1],[2,0,3],[1,0,0,1],[1,0,0,2]]
\n\nOutput: [1,0,1]
\n\nExplanation:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\tValid \n\t\t\t nums[j] | \n\t\t\tupdated \n\t\t\t nums | \n\t\t\tAnswer | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,1,1] | \n\t\t\t[1, 2] | \n\t\t\t[1, 10] | \n\t\t\t[0, 1] | \n\t\t\t[0, 1] | \n\t\t\t1 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
1 | \n\t\t\t[2,0,3] | \n\t\t\t[1, 2] | \n\t\t\t[1, 10] | \n\t\t\t[0, 1] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[3, 2] | \n\t\t\t\n\t\t |
2 | \n\t\t\t[1,0,0,1] | \n\t\t\t[3, 2] | \n\t\t\t[11, 10] | \n\t\t\t[2, 1] | \n\t\t\t[0, 0] | \n\t\t\t1 | \n\t\t\t[] | \n\t\t\t— | \n\t\t\t0 | \n\t\t
3 | \n\t\t\t[1,0,0,2] | \n\t\t\t[3, 2] | \n\t\t\t[11, 10] | \n\t\t\t[2, 1] | \n\t\t\t[0, 0] | \n\t\t\t2 | \n\t\t\t[0] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
Thus, the final answer
is [1, 0, 1]
.
\n
Constraints:
\n\n1 <= n == nums.length <= 105
1 <= nums[i] <= 1015
1 <= queries.length <= 105
queries[i].length == 3
or 4
\n\tqueries[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
给你一个整数数组 nums
。
对于任意正整数 x
,定义以下序列:
p0 = x
pi+1 = popcount(pi)
,对于所有 i >= 0
,其中 popcount(y)
表示整数 y
的二进制表示中 1 的个数。这个序列最终会收敛到值 1。
\n\npopcount-depth(位计数深度)定义为满足 pd = 1
的最小整数 d >= 0
。
例如,当 x = 7
(二进制表示为 \"111\"
)时,该序列为:7 → 3 → 2 → 1
,因此 7 的 popcount-depth 为 3。
此外,给定一个二维整数数组 queries
,其中每个 queries[i]
可以是以下两种类型之一:
[1, l, r, k]
- 计算在区间 [l, r]
中,满足 nums[j]
的 popcount-depth 等于 k
的索引 j
的数量。[2, idx, val]
- 将 nums[idx]
更新为 val
。返回一个整数数组 answer
,其中 answer[i]
表示第 i
个类型为 [1, l, r, k]
的查询的结果。
\n\n
示例 1:
\n\n输入: nums = [2,4], queries = [[1,0,1,1],[2,1,1],[1,0,1,0]]
\n\n输出: [2,1]
\n\n解释:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\t有效 \n\t\t\t nums[j] | \n\t\t\t更新后的 \n\t\t\t nums | \n\t\t\t答案 | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,1,1] | \n\t\t\t[2,4] | \n\t\t\t[10, 100] | \n\t\t\t[1, 1] | \n\t\t\t[0, 1] | \n\t\t\t1 | \n\t\t\t[0, 1] | \n\t\t\t— | \n\t\t\t2 | \n\t\t
1 | \n\t\t\t[2,1,1] | \n\t\t\t[2,4] | \n\t\t\t[10, 100] | \n\t\t\t[1, 1] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[2,1] | \n\t\t\t— | \n\t\t
2 | \n\t\t\t[1,0,1,0] | \n\t\t\t[2,1] | \n\t\t\t[10, 1] | \n\t\t\t[1, 0] | \n\t\t\t[0, 1] | \n\t\t\t0 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
因此,最终 answer
为 [2, 1]
。
示例 2:
\n\n输入:nums = [3,5,6], queries = [[1,0,2,2],[2,1,4],[1,1,2,1],[1,0,1,0]]
\n\n输出:[3,1,0]
\n\n解释:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\t有效 \n\t\t\t nums[j] | \n\t\t\t更新后的 \n\t\t\t nums | \n\t\t\t答案 | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,2,2] | \n\t\t\t[3, 5, 6] | \n\t\t\t[11, 101, 110] | \n\t\t\t[2, 2, 2] | \n\t\t\t[0, 2] | \n\t\t\t2 | \n\t\t\t[0, 1, 2] | \n\t\t\t— | \n\t\t\t3 | \n\t\t
1 | \n\t\t\t[2,1,4] | \n\t\t\t[3, 5, 6] | \n\t\t\t[11, 101, 110] | \n\t\t\t[2, 2, 2] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[3, 4, 6] | \n\t\t\t— | \n\t\t
2 | \n\t\t\t[1,1,2,1] | \n\t\t\t[3, 4, 6] | \n\t\t\t[11, 100, 110] | \n\t\t\t[2, 1, 2] | \n\t\t\t[1, 2] | \n\t\t\t1 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
3 | \n\t\t\t[1,0,1,0] | \n\t\t\t[3, 4, 6] | \n\t\t\t[11, 100, 110] | \n\t\t\t[2, 1, 2] | \n\t\t\t[0, 1] | \n\t\t\t0 | \n\t\t\t[] | \n\t\t\t— | \n\t\t\t0 | \n\t\t
因此,最终 answer
为 [3, 1, 0]
。
示例 3:
\n\n输入:nums = [1,2], queries = [[1,0,1,1],[2,0,3],[1,0,0,1],[1,0,0,2]]
\n\n输出:[1,0,1]
\n\n解释:
\n\ni | \n\t\t\tqueries[i] | \n\t\t\tnums | \n\t\t\tbinary(nums ) | \n\t\t\tpopcount- \n\t\t\tdepth | \n\t\t\t[l, r] | \n\t\t\tk | \n\t\t\t有效 \n\t\t\t nums[j] | \n\t\t\t更新后的 \n\t\t\t nums | \n\t\t\t答案 | \n\t\t
---|---|---|---|---|---|---|---|---|---|
0 | \n\t\t\t[1,0,1,1] | \n\t\t\t[1, 2] | \n\t\t\t[1, 10] | \n\t\t\t[0, 1] | \n\t\t\t[0, 1] | \n\t\t\t1 | \n\t\t\t[1] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
1 | \n\t\t\t[2,0,3] | \n\t\t\t[1, 2] | \n\t\t\t[1, 10] | \n\t\t\t[0, 1] | \n\t\t\t— | \n\t\t\t— | \n\t\t\t— | \n\t\t\t[3, 2] | \n\t\t\t\n\t\t |
2 | \n\t\t\t[1,0,0,1] | \n\t\t\t[3, 2] | \n\t\t\t[11, 10] | \n\t\t\t[2, 1] | \n\t\t\t[0, 0] | \n\t\t\t1 | \n\t\t\t[] | \n\t\t\t— | \n\t\t\t0 | \n\t\t
3 | \n\t\t\t[1,0,0,2] | \n\t\t\t[3, 2] | \n\t\t\t[11, 10] | \n\t\t\t[2, 1] | \n\t\t\t[0, 0] | \n\t\t\t2 | \n\t\t\t[0] | \n\t\t\t— | \n\t\t\t1 | \n\t\t
因此,最终 answer
为 [1, 0, 1]
。
\n\n
提示:
\n\n1 <= n == nums.length <= 105
1 <= nums[i] <= 1015
1 <= queries.length <= 105
queries[i].length == 3
或 4
\n\tqueries[i] == [1, l, r, k]
或queries[i] == [2, idx, val]
0 <= l <= r <= n - 1
0 <= k <= 5
0 <= idx <= n - 1
1 <= val <= 1015
depth[i]
for each nums[i]
by applying popcount until you reach 1.",
"Maintain six Fenwick trees fenw[0]
through fenw[5]
, where fenw[d]
stores a 1 at index i
iff depth[i] == d
.",
"For an update [2, idx, val]
, remove index idx from its old fenw[old_depth]
and insert into fenw[new_depth]
; for a query [1, l, r, k]
, return fenw[k].query(r) - fenw[k].query(l-1)
."
],
"solution": null,
"status": null,
"sampleTestCase": "[2,4]\n[[1,0,1,1],[2,1,1],[1,0,1,0]]",
"metaData": "{\n \"name\": \"popcountDepth\",\n \"params\": [\n {\n \"name\": \"nums\",\n \"type\": \"long[]\"\n },\n {\n \"type\": \"long[][]\",\n \"name\": \"queries\"\n }\n ],\n \"return\": {\n \"type\": \"integer[]\"\n }\n}",
"judgerAvailable": true,
"judgeType": "large",
"mysqlSchemas": [],
"enableRunCode": true,
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