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Submission Week Nine #9

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27 changes: 27 additions & 0 deletions week09/course_schedule.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,27 @@
// Time complexity: O(N), N = Number of courses.
// Space complexity: O(N), N = Number of courses.

bool hasCycle(vector<int> &col, vector<vector<int>> &graph, int u){
col[u] = 1;
bool status = false;
for(auto v: graph[u]){
if(col[v] == 1) status = true;
else if(col[v] == 0){
status |= hasCycle(col, graph, v);
}
}
col[u] = 2;
return status;
}

bool canFinish(int numCourses, vector<vector<int>> &prerequisites){
vector<vector<int>> graph(numCourses);
vector<int> col(numCourses, 0);
for(auto pre: prerequisites){
graph[pre[1]].push_back(pre[0]);
}
for(int i = 0; i < numCourses; i++){
if(hasCycle(col, graph, i)) return false;
}
return true;
}
7 changes: 7 additions & 0 deletions week09/maximum_depth_of_a_binary_tree.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,7 @@
// Time complexity: O(N), N = Number of nodes.
// Space complexity: O(1)
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Actually the space complexity is O(H) where H = depth of the recursion.


int maxDepth(TreeNode* root, int depth = 0){
if(!root) return depth;
return max(maxDepth(root->left, depth + 1), maxDepth(root->right, depth + 1));
}
31 changes: 31 additions & 0 deletions week09/number_of_islands.cpp
Original file line number Diff line number Diff line change
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// Time complexity: O(M * N), M = Number of rows, N = Number of columns of the grid.
// Space complexity: O(1)
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Same thing as above.


int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};

void dfs(vector<vector<char>> &grid, int row, int col){
int r = grid.size();
int c = grid[0].size();
if(row == r || col == c || row < 0 || col < 0) return;
if(grid[row][col] == '0' || grid[row][col] == '2') return;
grid[row][col] = '2';
for(int i = 0; i < 4; i++){
dfs(grid, row + dr[i], col + dc[i]);
}
}

int numIslands(vector<vector<char>> &grid){
int row = grid.size();
int col = grid[0].size();
int count = 0;
for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
if(grid[i][j] == '1'){
count++;
dfs(grid, i, j);
}
}
}
return count;
}