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‎leetcode/0085-maximal-rectangle/0085-maximal-rectangle.cpp

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class Solution {
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public:
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int largestRectangleArea(vector<int>& histo) {
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stack<int> st;
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int maxA = 0, n = histo.size();
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for (int i = 0; i <= n; i++) {
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while (st.size() && (i == n || histo[st.top()] >= histo[i])) {
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int height = histo[st.top()]; st.pop();
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int width = st.empty() ? i: i - st.top() - 1;
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maxA = max(maxA, width * height);
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}
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st.push(i);
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}
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return maxA;
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}
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int maximalRectangle(vector<vector<char>>& matrix) {
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int maxiArea = 0, n = matrix.size(), m = matrix[0].size();
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vector<int> height(m, 0);
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++)
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if (matrix[i][j] == '1') height[j]++; else height[j] = 0;
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int area = largestRectangleArea(height);
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maxiArea = max(maxiArea, area);
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}
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return maxiArea;
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}
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};

‎leetcode/0085-maximal-rectangle/NOTES.md

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‎leetcode/0085-maximal-rectangle/README.md

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<h2><a href="https://leetcode.com/problems/maximal-rectangle/">85. Maximal Rectangle</a></h2><h3>Hard</h3><hr><div><p>Given a <code>rows x cols</code>&nbsp;binary <code>matrix</code> filled with <code>0</code>'s and <code>1</code>'s, find the largest rectangle containing only <code>1</code>'s and return <em>its area</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/maximal.jpg" style="width: 402px; height: 322px;">
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<pre><strong>Input:</strong> matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> The maximal rectangle is shown in the above picture.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> matrix = [["0"]]
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<strong>Output:</strong> 0
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre><strong>Input:</strong> matrix = [["1"]]
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<strong>Output:</strong> 1
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>rows == matrix.length</code></li>
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<li><code>cols == matrix[i].length</code></li>
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<li><code>1 &lt;= row, cols &lt;= 200</code></li>
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<li><code>matrix[i][j]</code> is <code>'0'</code> or <code>'1'</code>.</li>
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</ul>
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</div>

‎leetcode/0210-course-schedule-ii/0210-course-schedule-ii.cpp

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class Solution {
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public:
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vector<int> findOrder(int n, vector<vector<int>>& prereq) {
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vector<vector<int>> g(n);
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vector<int> indegree(n);
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for (auto e: prereq) {
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auto take = e[0], pre = e[1];
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g[pre].push_back(take),
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indegree[take]++;
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}
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queue<int> q;
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for (auto i = 0; i < n; i++) if (!indegree[i]) q.push(i);
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vector<int> res;
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while (q.size()) {
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auto course = q.front(); q.pop();
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res.push_back(course);
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for (auto to: g[course]) if (--indegree[to] == 0) q.push(to);
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}
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return res.size() == n ? res: vector<int>{};
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}
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};

‎leetcode/0210-course-schedule-ii/NOTES.md

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‎leetcode/0210-course-schedule-ii/README.md

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<h2><a href="https://leetcode.com/problems/course-schedule-ii/">210. Course Schedule II</a></h2><h3>Medium</h3><hr><div><p>There are a total of <code>numCourses</code> courses you have to take, labeled from <code>0</code> to <code>numCourses - 1</code>. You are given an array <code>prerequisites</code> where <code>prerequisites[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that you <strong>must</strong> take course <code>b<sub>i</sub></code> first if you want to take course <code>a<sub>i</sub></code>.</p>
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<ul>
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<li>For example, the pair <code>[0, 1]</code>, indicates that to take course <code>0</code> you have to first take course <code>1</code>.</li>
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</ul>
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<p>Return <em>the ordering of courses you should take to finish all courses</em>. If there are many valid answers, return <strong>any</strong> of them. If it is impossible to finish all courses, return <strong>an empty array</strong>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre><strong>Input:</strong> numCourses = 2, prerequisites = [[1,0]]
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<strong>Output:</strong> [0,1]
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<strong>Explanation:</strong> There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
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<strong>Output:</strong> [0,2,1,3]
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<strong>Explanation:</strong> There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
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So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre><strong>Input:</strong> numCourses = 1, prerequisites = []
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<strong>Output:</strong> [0]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= numCourses &lt;= 2000</code></li>
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<li><code>0 &lt;= prerequisites.length &lt;= numCourses * (numCourses - 1)</code></li>
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<li><code>prerequisites[i].length == 2</code></li>
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<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; numCourses</code></li>
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<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
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<li>All the pairs <code>[a<sub>i</sub>, b<sub>i</sub>]</code> are <strong>distinct</strong>.</li>
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</ul>
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</div>

‎leetcode/0258-add-digits/0258-add-digits.cpp

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class Solution {
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public:
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int addDigits(int num) {
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auto get_digit_sum = [](int n, int sum = 0) {
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while (n) sum += n % 10, n /= 10;
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return sum;
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};
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do num = get_digit_sum(num);
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while (num > 9 && num != 2);
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return num;
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}
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};

‎leetcode/0258-add-digits/NOTES.md

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‎leetcode/0258-add-digits/README.md

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<h2><a href="https://leetcode.com/problems/add-digits/">258. Add Digits</a></h2><h3>Easy</h3><hr><div><p>Given an integer <code>num</code>, repeatedly add all its digits until the result has only one digit, and return it.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre><strong>Input:</strong> num = 38
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> The process is
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38 --&gt; 3 + 8 --&gt; 11
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11 --&gt; 1 + 1 --&gt; 2
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Since 2 has only one digit, return it.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> num = 0
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<strong>Output:</strong> 0
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>0 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Follow up:</strong> Could you do it without any loop/recursion in <code>O(1)</code> runtime?</p>
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</div>

‎leetcode/0319-bulb-switcher/0319-bulb-switcher.cpp

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class Solution {
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public:
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int bulbSwitch(int n) {
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return sqrt(n);
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}
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};

‎leetcode/0319-bulb-switcher/README.md

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<h2><a href="https://leetcode.com/problems/bulb-switcher/">319. Bulb Switcher</a></h2><h3>Medium</h3><hr><div><p>There are <code>n</code> bulbs that are initially off. You first turn on all the bulbs, then&nbsp;you turn off every second bulb.</p>
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<p>On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the <code>i<sup>th</sup></code> round, you toggle every <code>i</code> bulb. For the <code>n<sup>th</sup></code> round, you only toggle the last bulb.</p>
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<p>Return <em>the number of bulbs that are on after <code>n</code> rounds</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/bulb.jpg" style="width: 421px; height: 321px;">
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<pre><strong>Input:</strong> n = 3
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> At first, the three bulbs are [off, off, off].
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After the first round, the three bulbs are [on, on, on].
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After the second round, the three bulbs are [on, off, on].
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After the third round, the three bulbs are [on, off, off].
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So you should return 1 because there is only one bulb is on.</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> n = 0
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<strong>Output:</strong> 0
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre><strong>Input:</strong> n = 1
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<strong>Output:</strong> 1
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
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</ul>
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</div>

‎leetcode/0427-construct-quad-tree copy/0427-construct-quad-tree.cpp

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/*
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// Definition for a QuadTree node.
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class Node {
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public:
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bool val;
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bool isLeaf;
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Node* topLeft;
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Node* topRight;
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Node* bottomLeft;
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Node* bottomRight;
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Node() {
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val = false;
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isLeaf = false;
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topLeft = NULL;
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topRight = NULL;
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bottomLeft = NULL;
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bottomRight = NULL;
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}
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Node(bool _val, bool _isLeaf) {
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val = _val;
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isLeaf = _isLeaf;
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topLeft = NULL;
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topRight = NULL;
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bottomLeft = NULL;
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bottomRight = NULL;
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}
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Node(bool _val, bool _isLeaf, Node* _topLeft, Node* _topRight, Node* _bottomLeft, Node* _bottomRight) {
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val = _val;
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isLeaf = _isLeaf;
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topLeft = _topLeft;
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topRight = _topRight;
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bottomLeft = _bottomLeft;
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bottomRight = _bottomRight;
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}
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};
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*/
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class Solution {
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public:
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Node* construct(vector<vector<int>>& grid) {
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function<Node*(int, int, int)> go = [&](auto x, auto y, auto len) {
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auto val = grid[x][y];
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auto isDiff = [&]() {
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for (int i = x; i < x + len; i++)
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for (int j = y; j < y + len; j++)
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if (grid[i][j] != val) return true;
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return false;
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};
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if (isDiff()) {
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auto l = len >> 1;
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auto tl = go(x, y, l), tr = go(x , y + l, l);
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auto bl = go(x + l, y, l), br = go(x + l, y + l, l);
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return new Node(val, false, tl, tr, bl, br);
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}
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return new Node(val, true);
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};
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return go(0, 0, grid.size());
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}
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};

‎leetcode/0427-construct-quad-tree copy/NOTES.md

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‎leetcode/0427-construct-quad-tree copy/README.md

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<h2><a href="https://leetcode.com/problems/construct-quad-tree/">427. Construct Quad Tree</a></h2><h3>Medium</h3><hr><div><p>Given a <code>n * n</code> matrix <code>grid</code> of <code>0's</code> and <code>1's</code> only. We want to represent <code>grid</code> with a Quad-Tree.</p>
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<p>Return <em>the root of the Quad-Tree representing </em><code>grid</code>.</p>
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<p>A Quad-Tree is a tree data structure in which each internal node has exactly four children. Besides, each node has two attributes:</p>
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<ul>
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<li><code>val</code>: True if the node represents a grid of 1's or False if the node represents a grid of 0's. Notice that you can assign the <code>val</code> to True or False when <code>isLeaf</code> is False, and both are accepted in the answer.</li>
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<li><code>isLeaf</code>: True if the node is a leaf node on the tree or False if the node has four children.</li>
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</ul>
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<pre>class Node {
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public boolean val;
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public boolean isLeaf;
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public Node topLeft;
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public Node topRight;
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public Node bottomLeft;
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public Node bottomRight;
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}</pre>
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<p>We can construct a Quad-Tree from a two-dimensional area using the following steps:</p>
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<ol>
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<li>If the current grid has the same value (i.e all <code>1's</code> or all <code>0's</code>) set <code>isLeaf</code> True and set <code>val</code> to the value of the grid and set the four children to Null and stop.</li>
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<li>If the current grid has different values, set <code>isLeaf</code> to False and set <code>val</code> to any value and divide the current grid into four sub-grids as shown in the photo.</li>
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<li>Recurse for each of the children with the proper sub-grid.</li>
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</ol>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/02/11/new_top.png" style="width: 777px; height: 181px;">
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<p>If you want to know more about the Quad-Tree, you can refer to the <a href="https://en.wikipedia.org/wiki/Quadtree">wiki</a>.</p>
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<p><strong>Quad-Tree format:</strong></p>
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<p>You don't need to read this section for solving the problem. This is only if you want to understand the output format here. The output represents the serialized format of a Quad-Tree using level order traversal, where <code>null</code> signifies a path terminator where no node exists below.</p>
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<p>It is very similar to the serialization of the binary tree. The only difference is that the node is represented as a list <code>[isLeaf, val]</code>.</p>
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<p>If the value of <code>isLeaf</code> or <code>val</code> is True we represent it as <strong>1</strong> in the list <code>[isLeaf, val]</code> and if the value of <code>isLeaf</code> or <code>val</code> is False we represent it as <strong>0</strong>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/02/11/grid1.png" style="width: 777px; height: 99px;">
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<pre><strong>Input:</strong> grid = [[0,1],[1,0]]
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<strong>Output:</strong> [[0,1],[1,0],[1,1],[1,1],[1,0]]
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<strong>Explanation:</strong> The explanation of this example is shown below:
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Notice that 0 represnts False and 1 represents True in the photo representing the Quad-Tree.
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<img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/e1tree.png" style="width: 777px; height: 186px;">
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<p><img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/e2mat.png" style="width: 777px; height: 343px;"></p>
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<pre><strong>Input:</strong> grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
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<strong>Output:</strong> [[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
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<strong>Explanation:</strong> All values in the grid are not the same. We divide the grid into four sub-grids.
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The topLeft, bottomLeft and bottomRight each has the same value.
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The topRight have different values so we divide it into 4 sub-grids where each has the same value.
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Explanation is shown in the photo below:
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<img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/e2tree.png" style="width: 777px; height: 328px;">
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == grid.length == grid[i].length</code></li>
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<li><code>n == 2<sup>x</sup></code> where <code>0 &lt;= x &lt;= 6</code></li>
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</ul>
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</div>

‎leetcode/0502-ipo copy/0502-ipo.cpp

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class Solution {
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public:
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int findMaximizedCapital(int k,
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int w,
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vector<int>& pfs,
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vector<int>& cap) {
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int n = pfs.size();
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vector<pair<int, int>> p;
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for (int i = 0; i < n; i++) p.emplace_back(cap[i], pfs[i]);
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sort(p.begin(), p.end());
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priority_queue<int> pq;
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for (int i = 0, curr = 0; i < k; i++) {
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while (curr < n && p[curr].first <= w) pq.push(p[curr++].second);
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if (pq.empty()) break;
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w += pq.top();
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pq.pop();
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}
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return w;
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}
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};

‎leetcode/0502-ipo copy/NOTES.md

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