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feat: add solutions to lc problems: No.3383,3384 (#3850)
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solution/1800-1899/1812.Determine Color of a Chessboard Square/README.md

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观察棋盘我们发现,颜色相同的两个格子 $(x_1, y_1)$ 和 $(x_2, y_2)$ 满足 $x_1 + y_1$ 和 $x_2 + y_2$ 均为奇数或偶数。
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因此,我们可以根据 `coordinates` 获取对应的坐标 $(x, y)$,如果 $x + y$ 为奇数,则格子为白色,返回 `true`,否则返回 `false`
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因此,我们可以根据 $\textit{coordinates}$ 获取对应的坐标 $(x, y)$,如果 $x + y$ 为奇数,则格子为白色,返回 $\textit{true}$,否则返回 $\textit{false}$
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时间复杂度 $O(1)$,空间复杂度 $O(1)$。
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* @return {boolean}
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*/
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var squareIsWhite = function (coordinates) {
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const x = coordinates.charAt(0).charCodeAt();
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const y = coordinates.charAt(1).charCodeAt();
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return (x + y) % 2 == 1;
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return (coordinates[0].charCodeAt() + coordinates[1].charCodeAt()) % 2 == 1;
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};
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```
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solution/1800-1899/1812.Determine Color of a Chessboard Square/README_EN.md

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<!-- solution:start -->
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### Solution 1: Find the Pattern
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### Solution 1: Pattern Recognition
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By observing the chessboard, we find that two squares $(x_1, y_1)$ and $(x_2, y_2)$ with the same color satisfy that both $x_1 + y_1$ and $x_2 + y_2$ are either odd or even.
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Observing the chessboard, we find that two squares $(x_1, y_1)$ and $(x_2, y_2)$ with the same color satisfy that both $x_1 + y_1$ and $x_2 + y_2$ are either odd or even.
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Therefore, we can get the corresponding coordinates $(x, y)$ from `coordinates`. If $x + y$ is odd, then the square is white, return `true`, otherwise return `false`.
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Therefore, we can get the corresponding coordinates $(x, y)$ from $\textit{coordinates}$. If $x + y$ is odd, the square is white, and we return $\textit{true}$; otherwise, we return $\textit{false}$.
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The time complexity is $O(1)$, and the space complexity is $O(1)$.
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* @return {boolean}
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*/
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var squareIsWhite = function (coordinates) {
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const x = coordinates.charAt(0).charCodeAt();
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const y = coordinates.charAt(1).charCodeAt();
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return (x + y) % 2 == 1;
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return (coordinates[0].charCodeAt() + coordinates[1].charCodeAt()) % 2 == 1;
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};
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```
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solution/1800-1899/1812.Determine Color of a Chessboard Square/Solution.js

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* @return {boolean}
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*/
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var squareIsWhite = function (coordinates) {
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const x = coordinates.charAt(0).charCodeAt();
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const y = coordinates.charAt(1).charCodeAt();
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return (x + y) % 2 == 1;
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return (coordinates[0].charCodeAt() + coordinates[1].charCodeAt()) % 2 == 1;
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};

solution/1800-1899/1891.Cutting Ribbons/README_EN.md

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</ul>
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<p>Your task is to determine the <strong>maximum</strong> length of ribbon, <code>x</code>, that allows you to cut <em>exactly</em> <code>k</code> ribbons, each of length <code>x</code>. You can discard any leftover ribbon from the cuts. If it is <strong>impossible</strong> to cut <code>k</code> ribbons of the same length, return 0.</p>
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<p>Your task is to determine the <strong>maximum</strong> length of ribbon, <code>x</code>, that allows you to cut <em>at least</em> <code>k</code> ribbons, each of length <code>x</code>. You can discard any leftover ribbon from the cuts. If it is <strong>impossible</strong> to cut <code>k</code> ribbons of the same length, return 0.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>

solution/2500-2599/2557.Maximum Number of Integers to Choose From a Range II/README_EN.md

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<strong>Input:</strong> banned = [4,3,5,6], n = 7, maxSum = 18
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> You can choose the integers 1, 2, and 7.
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All these integers are in the range [1, 7], all do not appear in banned, and their sum is 18, which does not exceed maxSum.
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All these integers are in the range [1, 7], all do not appear in banned, and their sum is 10, which does not exceed maxSum.
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</pre>
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<p>&nbsp;</p>

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