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C++用 -1 mask法達到branchless解Leetcode1513 Number of Substrings With Only 1...

C++用 -1 mask法達到branchless解Leetcode1513  Number of Substrings With Only 1s|Py3 1-liner

其實很容易，想練功，試試一行解，或是把if分岔拿掉

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Py3 C++ 2D sweep解Leetcode 2536 Increment Submatrices by One

Py3 C++ 2D sweep解Leetcode 2536  Increment Submatrices by One

使用2D掃描，方法為line sweep的2D推廣，方便好用。

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C++ Py3 Euclidean Algorithm解Leetcode 2169 Count Operations to Obtain Zero

C++ Py3 Euclidean Algorithm解Leetcode 2169  Count Operations to Obtain Zero

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Py3 C++導出遞迴公式解難題Leetcode1611 Minimum One Bit Operations to Make Integer...

Py3 C++導出遞迴公式解難題Leetcode1611  Minimum One Bit Operations to Make Integers Zero從導出公式到寫出code，一次完成解Leetcode難題

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C++ py3 C Rust速解Leetcode 1526 Minimum Number of Increments on Subarrays ...

C++ py3 C Rust adjacent difference與loop速解Leetcode 1526 Minimum Number of Increments on Subarrays to Form a Target Array

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python requests讀取台積電股票json資訊爬蟲練習

url="https://www.twse.com.tw/exchangeReport/STOCK_DAY?response=json&date=20220330&stockNo=2330"

py3 C++ Line sweep解Leetcode 3346 Maximum Frequency of an Element After ...

py3 cpp Line sweep解Leetcode 3346  Maximum Frequency of an Element After Performing Operations I

使用 line sweep不用排序，可得線性解

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Py3 C++計數bitmask與2 pointers速解Leetcode2273 Find Resultant Array After Re...

Py3 C++計數bitmask與2 pointers速解Leetcode2273  Find Resultant Array After Removing Anagrams

字串長度最長也只有10，四個bits一組，用bitmask解題

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The maximum length of a string is only 10, and four bits are grouped together. Use bitmask to solve the problem.

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C++ Py3 計數排序與partial sum解Leetcode 2300 Successful Pairs of Spells and ...

C++ Py3  計數排序與partial sum解Leetcode 2300  Successful Pairs of Spells and Potions

Portions的極大值小於等於10萬是可進行記數排序的關鍵，既然可以採用記數排序，後面的二元搜尋也可以透過partial sum的手法加速。

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