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Merge pull request #324 from Hardvan/rotten-oranges
Add Rotten Oranges Graph Problem using BFS
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,103 @@ | ||
| #include <bits/stdc++.h> | ||
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| using namespace std; | ||
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| /* | ||
| Problem Statement: | ||
| Given an m x n grid, where each cell has the following values: | ||
| 2 – Rotten orange | ||
| 1 – Fresh orange | ||
| 0 – Empty cell | ||
| Every minute, if a Fresh orange is adjacent to a Rotten Orange in 4-direction | ||
| (upward, downwards, right, and left) it becomes Rotten. | ||
| Return the minimum number of minutes required such that | ||
| none of the cells has a Fresh Orange. | ||
| If it's not possible, return -1. | ||
| */ | ||
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| // Approach: BFS | ||
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| bool isValid(int nx, int ny, int m, int n) | ||
| { | ||
| return (0 <= nx && nx < m) && | ||
| (0 <= ny && ny < n); | ||
| } | ||
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| // Time: O(mn) x 4 | ||
| // Space: O(mn) | ||
| int orangesRotting(vector<vector<int>> &grid) | ||
| { | ||
| int m = grid.size(); | ||
| int n = grid[0].size(); | ||
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| int total = 0; // fresh + rotten oranges | ||
| int count = 0; // rotten oranges | ||
| int mins = 0; // minutes elapsed | ||
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| queue<pair<int, int>> rotten; // {i, j}: position of rotten orange | ||
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| // Count the fresh & rotten oranges, push rotten oranges into queue | ||
| for (int i = 0; i < m; i++) | ||
| { | ||
| for (int j = 0; j < n; j++) | ||
| { | ||
| if (grid[i][j] != 0) // Rotten or Fresh orange | ||
| total++; | ||
| if (grid[i][j] == 2) // Rotten | ||
| rotten.push({i, j}); // Push position of rotten orange | ||
| } | ||
| } | ||
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| int dx[] = {0, 0, -1, 1}; // 4-directions (x-axis) | ||
| int dy[] = {-1, 1, 0, 0}; // 4-directions (y-axis) | ||
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| while (!rotten.empty()) | ||
| { | ||
| int size = rotten.size(); // rotten oranges in current minute | ||
| count += size; // add to total rotten oranges | ||
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| while (size--) // Each rotten orange in current minute | ||
| { | ||
| // Pop the front rotten orange | ||
| int x = rotten.front().first; | ||
| int y = rotten.front().second; | ||
| rotten.pop(); | ||
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| // Check for fresh oranges in 4-directions | ||
| for (int i = 0; i < 4; i++) | ||
| { | ||
| // New coordinates | ||
| int nx = x + dx[i]; | ||
| int ny = y + dy[i]; | ||
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| // Valid, fresh orange | ||
| if (isValid(nx, ny, m, n) && grid[nx][ny] == 1) | ||
| { | ||
| grid[nx][ny] = 2; // make it rotten | ||
| rotten.push({nx, ny}); // push it into queue | ||
| } | ||
| } | ||
| } | ||
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| if (!rotten.empty()) // if there are more rotten oranges | ||
| mins++; | ||
| } | ||
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| if (total != count) // fresh oranges left | ||
| return -1; | ||
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| return mins; | ||
| } | ||
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| int main() | ||
| { | ||
| vector<vector<int>> grid = {{2, 1, 1}, | ||
| {1, 1, 0}, | ||
| {0, 1, 1}}; | ||
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| cout << orangesRotting(grid) << endl; // 4 | ||
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| return 0; | ||
| } |