[pull] master from wisdompeak:master #341

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		[2031.Count-Subarrays-With-More-Ones-Than-Zeros](https://github.com/wisdompeak/LeetCode/tree/master/Segment_Tree/2031.Count-Subarrays-With-More-Ones-Than-Zeros) (H)
		[2179.Count-Good-Triplets-in-an-Array](https://github.com/wisdompeak/LeetCode/tree/master/Segment_Tree/2179.Count-Good-Triplets-in-an-Array) (H)
		[2659.Make-Array-Empty](https://github.com/wisdompeak/LeetCode/tree/master/Segment_Tree/2659.Make-Array-Empty) (H)
	[3624.Number-of-Integers-With-Popcount-Depth-Equal-to-K-II](https://github.com/wisdompeak/LeetCode/tree/master/Segment_Tree/3624.Number-of-Integers-With-Popcount-Depth-Equal-to-K-II) (H-)

		#### [Design](https://github.com/wisdompeak/LeetCode/tree/master/Design)
		[380.Insert-Delete-GetRandom-O(1)](https://github.com/wisdompeak/LeetCode/tree/master/Design/380.Insert-Delete-GetRandom-O-1/) (M+)
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	using ll = long long;

	class BIT{
	public:
	int N;
	vector<long long>bitArr; // Note: all arrays are 1-index
	vector<long long>nums;
	long long M = 1e9+7;

	void init(int N)
	{
	this->N = N;
	bitArr.resize(N+1);
	nums.resize(N+1);
	}

	// increase nums[i] by delta
	void updateDelta(int i, long long delta) {
	int idx = i;
	while (idx <= N)
	{
	bitArr[idx]+=delta;
	// bitArr[idx] %= M;
	idx+=idx&(-idx);
	}
	}

	// sum of a range nums[1:j] inclusively
	long long queryPreSum(int idx){
	long long result = 0;
	while (idx){
	result += bitArr[idx];
	// result %= M;
	idx-=idx&(-idx);
	}
	return result;
	}

	// sum of a range nums[i:j] inclusively
	long long sumRange(int i, int j) {
	return queryPreSum(j)-queryPreSum(i-1);
	}
	};

	class Solution {
	public:
	int getDepth(ll x) {
	int count = 0;
	while (x!=1ll) {
	x = __builtin_popcountll(x);
	count++;
	}
	return count;
	}

	vector<int> popcountDepth(vector<long long>& nums, vector<vector<long long>>& queries) {
	int N = nums.size()+1;
	vector<BIT> bit_arr (5);
	for (int i=0; i<=4; i++)
	bit_arr[i].init(N);

	for (int i=0; i<nums.size(); i++) {
	int depth = getDepth(nums[i]);
	bit_arr[depth].updateDelta(i+1, 1);
	}

	vector<int>rets;
	for (auto&q: queries) {
	if (q[0]==2) {
	ll idx = q[1], val = q[2];
	int depth0 = getDepth(nums[idx]);
	int depth1 = getDepth(val);
	bit_arr[depth0].updateDelta(idx+1, -1);
	bit_arr[depth1].updateDelta(idx+1, 1);
	nums[idx] = val;
	} else {
	int l = q[1], r = q[2], k = q[3];
	if (k>=5)
	rets.push_back(0);
	else
	rets.push_back(bit_arr[k].sumRange(l+1,r+1));
	}
	}

	return rets;

	}
	};

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

	首先我们要意识到,任何一个长整型的popcount depth不可能很大。

	首先对于<=64的数,我们可以使用穷举的方法,发现最坏情况就是63,它的路径就是63->6->2->1。而剩下任意大于64的长整型(共有64个bit),它的第一步变化之后必然落入[1,64]之间。由之前的结论,其路径最多再走3步就会到1。所以结论是:任意一个长整型,其popcount depth不超过4.

	既然处理任何一个数只需要4步就能求出深度,那我们可以将所有nums都先处理一遍,将所有nums可以按照depth分类。于是对于每一种深度,我们可以知道有哪些数对应,将其index标记为1(否则标记为0),这样用树状数组就可以高效(log的时间)地算出任意区间内有多少数对应该深度(即求区间和),满足了第一类query的要求。同时树状数组支持对单点的动态修改,支持了第二类query的操作。

	因为深度的种类只有5种,我们只需要开5个这样的树状数组,因此空间复杂度也是符合要求的。

2 changes: 2 additions & 0 deletions Template/Bit_manipulation/Count_bit_1_numbers.cpp

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