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05- May/21- Shortest Bridge/21- Shortest Bridge (Lama Salah).cpp
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// Author: Lama Salah | ||
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/* | ||
-- THE IDEA -- | ||
Use a combination of depth-first search (DFS) and breadth-first search (BFS) algorithms. | ||
The DFS function is used to mark the first island as visited and push its cells into the queue. | ||
The BFS function then performs a breadth-first search to find the shortest distance from the first island to the second island. | ||
*/ | ||
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class Solution { | ||
int n; // Size of the grid. | ||
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// Direction arrays for exploring neighboring cells (right, left, down, up). | ||
int dx[4] = {0, 0, 1, -1}; | ||
int dy[4] = {1, -1, 0, 0}; | ||
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vector<vector<int>> grid; // Input grid | ||
queue<vector<int>> q; // Queue for BFS | ||
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public: | ||
// 'valid' function to check if coordinates (x, y) are within the grid bounds. | ||
bool valid(int x, int y) { | ||
return x >= 0 && y >= 0 && x < n && y < n; | ||
} | ||
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void dfs(int i, int j) { | ||
q.push({i, j}); // Push the current cell (i, j) into the queue. | ||
grid[i][j] = -1; // Mark the current cell as visited (-1). | ||
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// Explore all four neighboring cells. | ||
for (int k = 0; k < 4; k++) { | ||
// Calculate the new x-coordinate and y-coordinate of the neighboring cell. | ||
int nx = i + dx[k]; | ||
int ny = j + dy[k]; | ||
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// Check if the neighboring cell is within bounds and unvisited. | ||
if (valid(nx, ny) && grid[nx][ny] == 1) { | ||
dfs(nx, ny); // Recursively call DFS on the neighboring cell. | ||
} | ||
} | ||
} | ||
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int bfs(int i, int j) { | ||
// Initialize the distance between the two islands as 0. | ||
int distance = 0; | ||
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// Process cells in the queue until it becomes empty. | ||
while (!q.empty()) { | ||
int sz = q.size(); // Get the current number of cells in the queue. | ||
while (sz--) { | ||
// Get the front cell from the queue and pop it. | ||
auto curr = q.front(); | ||
q.pop(); | ||
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// Explore all four neighboring cells. | ||
for (int k = 0; k < 4; k++) { | ||
// Calculate the new x-coordinate and y-coordinate of the neighboring cell. | ||
int nx = curr[0] + dx[k]; | ||
int ny = curr[1] + dy[k]; | ||
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// Check if the neighboring cell is within bounds and unvisited. | ||
if (valid(nx, ny) && grid[nx][ny] != -1) { | ||
// Check if the neighboring cell is the second island (the second island is found). | ||
// If it is, return the current distance. | ||
if (grid[nx][ny] == 1) { | ||
return distance; | ||
} | ||
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// Mark the neighboring cell as visited (-1). | ||
grid[nx][ny] = -1; | ||
// Push the neighboring cell into the queue. | ||
q.push({nx, ny}); | ||
} | ||
} | ||
} | ||
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// Increment the distance after processing all cells in the current level. | ||
distance++; | ||
} | ||
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// Return the shortest distance between the two islands. | ||
return distance; | ||
} | ||
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int shortestBridge(vector<vector<int>>& g) { | ||
this->grid = g; // Store the input grid. | ||
this->n = grid.size(); // Get the size of the grid. | ||
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// Iterate through the grid to find the first island. | ||
for (int i = 0; i < n; i++) { | ||
for (int j = 0; j < n; j++) { | ||
// If a cell with value 1 is found (first island). | ||
if (g[i][j] == 1) { | ||
dfs(i, j); // Perform DFS to mark the first island as visited and push its cells into the queue. | ||
return bfs(i, j); // Call BFS to find the shortest bridge between the two islands. | ||
} | ||
} | ||
} | ||
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return 0; // If no islands are found, return 0. | ||
} | ||
}; |