[pull] master from wisdompeak:master #361

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		假设n的二进制长度是maxLen,那么对于任何1<=L<maxLen的二进制回文数自然都符合"小于n"的条件。这些二进制回文数与它的"前半段"有一一对应的关系。也就是说,我们遍历所有的"前半段",就能构造出所有长度为L的二进制回文数。举个例子,如果L=8,它一半的长度h=4,那么我们遍历1000\~1111,将它们各自翻转拼接,就对应了所有长度为8的二进制回文数。注意,如果L=7,它一半的长度h也是4,我们同样遍历1000~1111,将它们各自翻转拼接(但是有一个中轴位),它们依然能对应所有长度为7的二进制回文数。

	接下来考虑L=maxLen,且"小于等于n"的二进制回文数。比较简单的处理就是用二分搜值。计算`h=(L+1)/2`,然后"前半段"有下限`mn=1<<(h-1)`,上限是`mx=(1<<h)-1`,我们二分搜值找到最大的、小于等于n的数m,那么`m-mn+1`就是答案。但是注意,存在并没有解的可能。比如h=1时,就有`mn>mx`的情况。
	接下来考虑L=maxLen,且"小于等于n"的二进制回文数。比较简单的处理就是用二分搜值。计算`h=(L+1)/2`,然后"前半段"有下限`mn=1<<(h-1)`,上限是`mx=(1<<h)-1`,我们二分搜值找到最大的、小于等于n的数m,那么`m-mn+1`就是答案。但是注意,存在并没有解的可能。比如h=1时,就有`mn>mx`的情况,所以需要对收敛的解做二次验证。

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	class Solution(object):
	def makeTheIntegerZero(self,num1,num2):
	x = num1
	y = num2
	k = 1

	while True:
	x = x - y
	if x < k:
	return -1

	if bin(x).count('1') <= k:
	return k

	k = k + 1

	# I hope this can help anyone whos resolving this huge problem on python ;)

1 change: 1 addition & 0 deletions Readme.md

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		[2434.Using-a-Robot-to-Print-the-Lexicographically-Smallest-String](https://github.com/wisdompeak/LeetCode/tree/master/Stack/2434.Using-a-Robot-to-Print-the-Lexicographically-Smallest-String) (H-)
		[2454.Next-Greater-Element-IV](https://github.com/wisdompeak/LeetCode/tree/master/Stack/2454.Next-Greater-Element-IV) (H-)
		[3113.Find-the-Number-of-Subarrays-Where-Boundary-Elements-Are-Maximum](https://github.com/wisdompeak/LeetCode/tree/master/Stack/3113.Find-the-Number-of-Subarrays-Where-Boundary-Elements-Are-Maximum) (M)
	[3676.Count-Bowl-Subarrays](https://github.com/wisdompeak/LeetCode/tree/master/Stack/3676.Count-Bowl-Subarrays) (M+)
		* ``monotonic stack: other usages``
		[084.Largest-Rectangle-in-Histogram](https://github.com/wisdompeak/LeetCode/tree/master/Stack/084.Largest-Rectangle-in-Histogram) (H)
		[2334.Subarray-With-Elements-Greater-Than-Varying-Threshold](https://github.com/wisdompeak/LeetCode/tree/master/Stack/2334.Subarray-With-Elements-Greater-Than-Varying-Threshold) (M+)
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	class Solution {
	public:
	long long bowlSubarrays(vector<int>& nums) {
	int n = nums.size();
	vector<int>nextGreater(n, -1);
	vector<int>prevGreater(n, -1);
	vector<int>st;
	for (int i=0; i<n; i++) {
	while (!st.empty() && nums[st.back()]<nums[i]) st.pop_back();
	if (!st.empty()) prevGreater[i] = st.back();
	st.push_back(i);
	}
	st.clear();

	for (int i=n-1; i>=0; i--) {
	while (!st.empty() && nums[st.back()]<nums[i]) st.pop_back();
	if (!st.empty()) nextGreater[i] = st.back();
	st.push_back(i);
	}

	long long ret = 0;
	for (int i=0; i<n; i++) {
	if (prevGreater[i]!=-1 && i-prevGreater[i]>=2) ret++;
	if (nextGreater[i]!=-1 && nextGreater[i]-i>=2) ret++;
	}
	return ret;
	}
	};

7 changes: 7 additions & 0 deletions Stack/3676.Count-Bowl-Subarrays/Readme.md

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	### 3676.Count-Bowl-Subarrays

	非常有趣的题目。

	我们很容易想到,对于每个nums[i]考虑其作为一个端点时,另一个端点应该在哪些位置。我们不妨认为nums[i]是左边且较低的端点,那么我们很容易找到对应的next greater element,比如说j,这是下一个可以作为右端点的位置,而这恰恰也是它所对应的唯一的右端点。如果再往右寻找右端点,那么j处在bowl内部就高于了左端点,不符合条件。

	于是我们就可以得出结论,对于每个nums[i],只有唯一的next greater element可以配对为右边且更高的端点。同理,对于每个nums[i],只有唯一的prev greater element可以配对为左边且更高的端点。这样我们就枚举除了所有的bowl的形状。

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