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| 1 | +#include <bits/stdc++.h> |
| 2 | + |
| 3 | +using namespace std; |
| 4 | + |
| 5 | +typedef long long int ll; |
| 6 | +typedef pair<int, int> ii; |
| 7 | +typedef vector<int> vi; |
| 8 | +typedef vector<string> vs; |
| 9 | + |
| 10 | +#define len(container) int((container).size()) |
| 11 | +#define all(c) (c).begin(), (c).end() |
| 12 | + |
| 13 | +template <typename ITER> |
| 14 | +void show(ITER begin, ITER end) { |
| 15 | + for (int i = 1; begin != end; i++) { |
| 16 | + printf("%d ", *(begin++)); |
| 17 | + if (i % 20 == 0 or begin == end) printf("\n"); |
| 18 | + } |
| 19 | +}; |
| 20 | + |
| 21 | +inline bool isValid(int x, int y, int R, int C) { |
| 22 | + return x >= 0 && x < R && y >= 0 && y < C; |
| 23 | +} |
| 24 | + |
| 25 | +struct TreeNode { |
| 26 | + int val; |
| 27 | + TreeNode* left; |
| 28 | + TreeNode* right; |
| 29 | + TreeNode(int x) : val(x), left(NULL), right(NULL) {} |
| 30 | +}; |
| 31 | + |
| 32 | +int maxSubsetSumNoAdjacent(vector<int> array) { |
| 33 | + const int N = array.size(); |
| 34 | + if (N == 0) { |
| 35 | + return 0; |
| 36 | + } |
| 37 | + if (N == 1) { |
| 38 | + return array[0]; |
| 39 | + } |
| 40 | + |
| 41 | + int tmp; |
| 42 | + int prev = 0, curr = array[0]; |
| 43 | + |
| 44 | + for (int i = 1; i < N; ++i) { |
| 45 | + tmp = prev; |
| 46 | + prev = curr; |
| 47 | + curr = max(tmp + array[i], curr); |
| 48 | + } |
| 49 | + |
| 50 | + return curr; |
| 51 | +} |
| 52 | + |
| 53 | +int numberOfWaysToMakeChange(int n, vector<int> denoms) { |
| 54 | + if (n == 0) { |
| 55 | + return 1; |
| 56 | + } |
| 57 | + const int D = denoms.size(); |
| 58 | + vector<int> ways(n + 1, 0); |
| 59 | + ways[0] = 1; |
| 60 | + for (int i = 0; i < D; ++i) { |
| 61 | + int d = denoms[i]; |
| 62 | + for (int j = n; j >= 0; --j) { |
| 63 | + int k = j / d; |
| 64 | + for (int l = 1; l <= k; ++l) { |
| 65 | + ways[j] += ways[j - l * d]; |
| 66 | + } |
| 67 | + } |
| 68 | + } |
| 69 | + return ways[n]; |
| 70 | +} |
| 71 | +// Instead of iterating in reverse way, we can just iterate normal way with |
| 72 | +// adding the denom value. Time: O(nd) | Space: O(n) |
| 73 | + |
| 74 | +int minNumberOfCoinsForChange(int n, vector<int> denoms) { |
| 75 | + vector<int> ways(n + 1, INT_MAX); |
| 76 | + ways[0] = 0; |
| 77 | + for (auto& denom : denoms) { |
| 78 | + for (int i = denom; i < n + 1; ++i) { |
| 79 | + if (ways[i - denom] != INT_MAX) |
| 80 | + ways[i] = min(ways[i], ways[i - denom] + 1); |
| 81 | + } |
| 82 | + } |
| 83 | + return ways[n] == INT_MAX ? -1 : ways[n]; |
| 84 | +} |
| 85 | + |
| 86 | +vector<vector<int>> maxSumIncreasingSubsequence(vector<int> array) { |
| 87 | + const int N = array.size(); |
| 88 | + vector<int> parents(N, -1); |
| 89 | + vector<int> lis(N, 0); // lis = array |
| 90 | + lis[0] = array[0]; |
| 91 | + for (int i = 1; i < N; ++i) { |
| 92 | + lis[i] = array[i]; |
| 93 | + for (int j = 0; j < i; ++j) { |
| 94 | + if (array[j] < array[i] && lis[i] < lis[j] + array[i]) { |
| 95 | + lis[i] = lis[j] + array[i]; |
| 96 | + parents[i] = j; |
| 97 | + } |
| 98 | + } |
| 99 | + } |
| 100 | + |
| 101 | + auto it = max_element(all(lis)); |
| 102 | + int st = it - lis.begin(); |
| 103 | + vi ans; |
| 104 | + |
| 105 | + while (st >= 0) { |
| 106 | + ans.push_back(array[st]); |
| 107 | + st = parents[st]; |
| 108 | + } |
| 109 | + |
| 110 | + reverse(all(ans)); |
| 111 | + |
| 112 | + return { |
| 113 | + {*it}, // Sum of sequence. |
| 114 | + ans, // Elements of the sequence. |
| 115 | + }; |
| 116 | +} |
| 117 | + |
| 118 | +vector<char> longestCommonSubsequence(string str1, string str2) { |
| 119 | + int l1 = str1.length(), l2 = str2.length(); |
| 120 | + vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1, 0)); |
| 121 | + int im = 0, jm = 0; |
| 122 | + for (int i = 1; i <= l1; ++i) { |
| 123 | + for (int j = 1; j <= l2; ++j) { |
| 124 | + if (str1[i - 1] == str2[j - 1]) { |
| 125 | + dp[i][j] = 1 + dp[i - 1][j - 1]; |
| 126 | + } else |
| 127 | + dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]); |
| 128 | + if (dp[i][j] > dp[im][jm]) { |
| 129 | + im = i; |
| 130 | + jm = j; |
| 131 | + } |
| 132 | + } |
| 133 | + } |
| 134 | + |
| 135 | + if (im == 0 && jm == 0) { |
| 136 | + return {}; |
| 137 | + } |
| 138 | + |
| 139 | + vector<char> ans; |
| 140 | + while (im >= 1 && jm >= 1) { // OR im != 0 && jm != 0 |
| 141 | + if (str1[im - 1] == str2[jm - 1]) { |
| 142 | + ans.push_back(str1[im - 1]); |
| 143 | + im--; |
| 144 | + jm--; |
| 145 | + } else if (dp[im][jm] == dp[im][jm - 1]) { |
| 146 | + jm--; |
| 147 | + } else |
| 148 | + im--; |
| 149 | + } |
| 150 | + reverse(ans.begin(), ans.end()); |
| 151 | + return ans; |
| 152 | +} |
| 153 | +// Time: O(nm) | Space: O(nm) |
| 154 | + |
| 155 | +int minNumberOfJumps(vector<int> array) { |
| 156 | + const int N = array.size(); |
| 157 | + vector<int> jumps(N, INT_MAX); |
| 158 | + jumps[0] = 0; |
| 159 | + for (int i = 0; i < N; ++i) { |
| 160 | + for (int j = i + 1; j <= min(N - 1, i + array[i]); ++j) { |
| 161 | + jumps[j] = min(jumps[j], 1 + jumps[i]); |
| 162 | + } |
| 163 | + } |
| 164 | + return jumps[N - 1]; |
| 165 | +} |
| 166 | +// Time: O(n^2) | Space: O(n) |
| 167 | +// Advance solution: O(n) using maxReach. |
| 168 | +// maxReach = array[0] |
| 169 | +// i = 1-> end: maxReach = max(maxReach, i + array[i]) |
| 170 | +// step-- |
| 171 | +// if (step==0) jump++; steps = maxReach - i |
| 172 | + |
| 173 | +// https://www.algoexpert.io/questions/Square%20of%20Zeroes |
| 174 | +// Time: O(n^3) | Space: O(n^2) |
| 175 | +bool squareOfZeroes(vector<vector<int>> matrix) { |
| 176 | + const int N = matrix.size(); |
| 177 | + vector<vector<int>> upper(N, vector<int>(N, 0)); |
| 178 | + vector<vector<int>> down(N, vector<int>(N, 0)); |
| 179 | + vector<vector<int>> left(N, vector<int>(N, 0)); |
| 180 | + vector<vector<int>> right(N, vector<int>(N, 0)); |
| 181 | + for (int i = 0; i < N; ++i) { |
| 182 | + for (int j = 1; j < N; ++j) { |
| 183 | + if (matrix[i][j] == 0) { |
| 184 | + left[i][j] = 1 + left[i][j - 1]; |
| 185 | + } |
| 186 | + } |
| 187 | + |
| 188 | + for (int j = N - 2; j >= 0; --j) { |
| 189 | + if (matrix[i][j] == 0) { |
| 190 | + right[i][j] = 1 + right[i][j + 1]; |
| 191 | + } |
| 192 | + } |
| 193 | + } |
| 194 | + |
| 195 | + for (int j = 0; j < N; ++j) { |
| 196 | + for (int i = 1; i < N; ++i) { |
| 197 | + if (matrix[i][j] == 0) { |
| 198 | + up[i][j] = 1 + up[i][j - 1]; |
| 199 | + } |
| 200 | + } |
| 201 | + |
| 202 | + for (int i = N - 2; i >= 0; ++i) { |
| 203 | + if (matrix[i][j] == 0) { |
| 204 | + down[i][j] = 1 + down[i][j + 1]; |
| 205 | + } |
| 206 | + } |
| 207 | + } |
| 208 | + // calc by check for each coordinate |
| 209 | + // the number of up and left, then check the down and right on the diagonal. |
| 210 | + return false; |
| 211 | +} |
| 212 | + |
| 213 | +int kadanesAlgorithm(vector<int> array) { |
| 214 | + int curr = 0; |
| 215 | + int maxSoFar = INT_MIN; |
| 216 | + for (int i = 0; i < array.size(); ++i) { |
| 217 | + curr = max(array[i], array[i] + curr); |
| 218 | + maxSoFar = max(maxSoFar, curr); |
| 219 | + } |
| 220 | + return maxSoFar; |
| 221 | +} |
| 222 | + |
| 223 | +// https://www.algoexpert.io/questions/Topological%20Sort |
| 224 | +unordered_map<int, vi> adjList; |
| 225 | +unordered_map<int, int> vis; |
| 226 | + |
| 227 | +bool dfs(int j, vi& ans) { |
| 228 | + if (vis[j] == 1) { |
| 229 | + return false; |
| 230 | + } else if (vis[j] == 2) { |
| 231 | + cout << j << endl; |
| 232 | + return true; |
| 233 | + } |
| 234 | + vis[j] = 2; |
| 235 | + for (auto neighbor : adjList[j]) { |
| 236 | + if (dfs(neighbor, ans)) { |
| 237 | + return true; |
| 238 | + } |
| 239 | + } |
| 240 | + vis[j] = 1; |
| 241 | + ans.push_back(j); |
| 242 | + |
| 243 | + return false; |
| 244 | +} |
| 245 | + |
| 246 | +vector<int> topologicalSort(vector<int> jobs, vector<vector<int>> deps) { |
| 247 | + vi ans; |
| 248 | + |
| 249 | + for (auto& dep : deps) { |
| 250 | + adjList[dep[0]].push_back(dep[1]); |
| 251 | + } |
| 252 | + |
| 253 | + for (auto& job : jobs) { |
| 254 | + if (dfs(job, ans)) { |
| 255 | + return {}; |
| 256 | + } |
| 257 | + } |
| 258 | + reverse(ans.begin(), ans.end()); |
| 259 | + return ans; |
| 260 | +} |
| 261 | + |
| 262 | +int main(int argc, char* argv[]) { |
| 263 | + // [1] |
| 264 | + // 75, 105, 120, 75, 90, 135 |
| 265 | + // [1, 2, 3, 4, 5, 6, 7, 8] |
| 266 | + // [[3, 1], [8, 1], [8, 7], [5, 7], [5, 2], [1, 4], [1, 6], [1, 2], [7, 6]] |
| 267 | + vector<int> denoms = {1, 5}; |
| 268 | + cout << numberOfWaysToMakeChange(6, denoms); |
| 269 | + return 0; |
| 270 | +} |
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