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Commit 1a4aa1f

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‎july/11.cpp

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#include <bits/stdc++.h>
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using namespace std;
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typedef long long int ll;
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typedef pair<int, int> ii;
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typedef vector<int> vi;
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typedef vector<string> vs;
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#define len(container) int((container).size())
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#define all(c) (c).begin(), (c).end()
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const int M = 1e9 + 7;
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template <typename ITER>
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void show(ITER begin, ITER end) {
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for (int i = 1; begin != end; i++) {
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printf("%d ", *(begin++));
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if (i % 20 == 0 or begin == end) printf("\n");
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}
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};
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template <typename T>
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void addMod(T& a, T b) {
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a = (a + b) % M;
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}
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inline bool isValid(int x, int y, int R, int C) {
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return x >= 0 && x < R && y >= 0 && y < C;
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}
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struct TreeNode {
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int val;
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TreeNode* left;
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TreeNode* right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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class Solution {
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public:
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double myPow(double x, int n) {
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if (n < 0) {
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if (n == INT_MIN) return abs(x) == 1.0 ? 1 : 0;
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return 1 / myPow(x, -n);
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}
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if (n == 0) return 1;
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if (n == 1) {
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return x;
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}
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if (x == 1.0) return 1;
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if (n % 2) {
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auto c = myPow(x, (n - 1) / 2);
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return x * c * c;
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}
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double c = myPow(x, n / 2);
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return c * c;
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/*
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if (n == std::numeric_limits<int>::lowest()) {
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return myPow(1 / x, -(n + 1)) / x;
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}
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if (n < 0) {
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return myPow(1 / x, -n);
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}
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double ans = 1;
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while (n) {
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if (n & 1 == 1) ans *= x;
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x *= x;
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n >>= 1;
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}
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return ans;*/
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}
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// https://leetcode.com/problems/is-graph-bipartite/
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bool colorHelper(vector<vector<int>> g, vector<int>& colors, int v,
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int color) {
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if (colors[v] != -1 && colors[v] != color) return false;
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colors[v] = color;
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for (auto n : g[v]) {
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if (!colorHelper(g, colors, n, 1 - color)) return false;
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}
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return true;
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}
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bool isBipartite(vector<vector<int>>& graph) {
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const int N = graph.size();
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vector<int> colors(N, -1);
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for (int i = 0; i < N; ++i) {
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if (colors[i] == -1) {
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if (!colorHelper(graph, colors, i, 0)) return false;
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}
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}
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return true;
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}
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// https://leetcode.com/problems/custom-sort-string/
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string customSortString(string S, string T) {
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int count[26] = {0};
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int i = 0;
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for (auto& c : S) {
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count[c - 'a'] = i++;
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}
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auto cmp = [&count](const char& a, const char& b) {
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return count[a - 'a'] < count[b - 'a'];
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};
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std::sort(std::begin(T), std::end(T), cmp);
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return T;
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}
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// https://leetcode.com/problems/task-scheduler/
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int leastInterval(vector<char>& tasks, int n) {
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int ans = 0;
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const int T = tasks.size();
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if (n == 0) {
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return T;
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}
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int count[26] = {0};
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for (auto& c : tasks) {
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count[c - 'A']++;
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}
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vector<ii> sch;
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for (int i = 0; i < 26; ++i) {
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if (count[i]) sch.emplace_back(count[i], i);
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}
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std::sort(std::begin(sch), std::end(sch), std::greater<ii>());
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int lastSize = 0;
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while (!sch.empty()) {
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lastSize = sch.size();
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for (auto& s : sch) {
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s.second--;
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}
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ans += max(n, lastSize);
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for (int i = lastSize - 1; i >= 0; --i) {
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if (sch[i].second == 0) {
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sch.pop_back();
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} else
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break;
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}
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}
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return ans - lastSize;
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}
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// very hard greedy problem
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// int ans = (count-1)*(n+1);
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// for(auto e : mp) if(e.second == count) ans++;
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// return max((int)tasks.size(), ans);
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// https://leetcode.com/problems/kth-smallest-element-in-a-bst/
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int kthSmallest(TreeNode* root, int k) {
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if (root == nullptr) return 0;
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int v = getTreeSize(root->left);
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if (v >= k) {
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return kthSmallest(root->left, k);
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}
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if (v + 1 == k) {
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return root->val;
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}
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return kthSmallest(root->right, k - v - 1);
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}
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// Other solutions:
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// - using stack for inorder traversal
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// - using recursion for inorder travesal
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// - using binary search like above.
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// Follow up:
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// - If tree has frequent insert/delete, then we could cache the getTreeSize
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int getTreeSize(TreeNode* r) {
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if (r == NULL) {
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return 0;
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}
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return 1 + getTreeSize(r->left) + getTreeSize(r->right);
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}
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// https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
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string minRemoveToMakeValid(string s) {
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string ans;
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return ans;
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}
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};
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int main(int argc, char* argv[]) {
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// Solution s = Solution();
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return 0;
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}
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/*
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* @lc app=leetcode id=329 lang=cpp
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*
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* [329] Longest Increasing Path in a Matrix
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*
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* https://leetcode.com/problems/longest-increasing-path-in-a-matrix/description/
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*
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* algorithms
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* Hard (42.25%)
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* Likes: 1909
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* Dislikes: 36
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* Total Accepted: 146.4K
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* Total Submissions: 339.1K
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* Testcase Example: '[[9,9,4],[6,6,8],[2,1,1]]'
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*
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* Given an integer matrix, find the length of the longest increasing path.
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*
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* From each cell, you can either move to four directions: left, right, up or
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* down. You may NOT move diagonally or move outside of the boundary (i.e.
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* wrap-around is not allowed).
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*
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* Example 1:
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*
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*
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* Input: nums =
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* [
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* ⁠ [9,9,4],
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* ⁠ [6,6,8],
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* ⁠ [2,1,1]
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* ]
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* Output: 4
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* Explanation: The longest increasing path is [1, 2, 6, 9].
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*
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*
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* Example 2:
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*
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*
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* Input: nums =
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* [
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* ⁠ [3,4,5],
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* ⁠ [3,2,6],
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* ⁠ [2,2,1]
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* ]
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* Output: 4
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* Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally
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* is not allowed.
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*
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*
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*/
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#include <vector>
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using namespace std;
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// @lc code=start
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class Solution {
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public:
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int dirs[5] = {1, 0, -1, 0, 1};
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int R, C;
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int dfs(vector<vector<int>>& scores, vector<vector<int>>& matrix, int a,
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int b) {
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if (scores[a][b] != -1) {
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return scores[a][b];
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}
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int score = -1;
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for (int i = 0; i < 4; ++i) {
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int x = a + dirs[i];
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int y = b + dirs[i + 1];
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if (x >= 0 && x < R && y >= 0 && y < C) {
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if (matrix[x][y] > matrix[a][b]) {
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score = max(score, 1 + dfs(scores, matrix, x, y));
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}
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}
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}
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scores[a][b] = score;
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return scores[a][b];
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}
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int longestIncreasingPath(vector<vector<int>>& matrix) {
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R = matrix.size();
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if (R == 0) {
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return 0;
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}
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C = matrix[0].size();
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int ans = 0;
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vector<vector<int>> scores(R, vector<int>(C, -1));
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for (int i = 0; i < R; ++i) {
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for (int j = 0; j < C; ++j) {
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ans = max(ans, dfs(scores, matrix, i, j));
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}
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}
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return ans;
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode id=714 lang=cpp
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*
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* [714] Best Time to Buy and Sell Stock with Transaction Fee
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*
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* https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/
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*
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* algorithms
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* Medium (53.15%)
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* Likes: 1516
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* Dislikes: 46
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* Total Accepted: 67.9K
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* Total Submissions: 125.1K
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* Testcase Example: '[1,3,2,8,4,9]\n2'
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*
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* Your are given an array of integers prices, for which the i-th element is
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* the price of a given stock on day i; and a non-negative integer fee
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* representing a transaction fee.
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* You may complete as many transactions as you like, but you need to pay the
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* transaction fee for each transaction. You may not buy more than 1 share of
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* a stock at a time (ie. you must sell the stock share before you buy again.)
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* Return the maximum profit you can make.
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*
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* Example 1:
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*
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* Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
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* Output: 8
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* Explanation: The maximum profit can be achieved by:
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* Buying at prices[0] = 1Selling at prices[3] = 8Buying at prices[4] =
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* 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) =
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* 8.
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*
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*
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*
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* Note:
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* 0 < prices.length .
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* 0 < prices[i] < 50000.
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* 0 .
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*
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef vector<int> vi;
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typedef pair<int, int> ii;
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typedef long long int ll;
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#define len(container) int((container).size())
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#define all(c) (c).begin(), (c).end()
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template <int group = 20, typename ITER>
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void show(const char* note, ITER begin, ITER end) {
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cout << note;
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for (int i = 1; begin != end; i++) {
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std::cout << *(begin++) << ' ';
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if (i % group == 0 or begin == end) cout << endl;
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}
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};
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struct TreeNode {
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int val;
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TreeNode* left;
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TreeNode* right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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// @lc code=start
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class Solution {
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public:
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int maxProfit(vector<int>& prices, int fee) {
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const int N = prices.size();
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int k = N / 2;
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vector<vi> dp(k + 1, vector<int>(N, 0));
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int res = 0;
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// sell at position i
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for (int i = 1; i <= k; i++) {
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for (int j = 1; j < N; j++) {
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dp[i][j] = dp[i - 1][j];
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int msf = 0;
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for (int l = 0; l < j; ++l) {
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if (l) msf = max(msf, dp[i - 1][l - 1]);
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if (prices[l] <= prices[j]) {
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dp[i][j] = max(dp[i][j], msf + prices[j] - prices[l] - fee);
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}
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}
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res = max(res, dp[i][j]);
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}
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}
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return res;
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}
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};
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// @lc code=end
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