| Isogram |
diff --git a/src/cpp/RottenOranges.cpp b/src/cpp/RottenOranges.cpp
new file mode 100644
index 00000000..4c639476
--- /dev/null
+++ b/src/cpp/RottenOranges.cpp
@@ -0,0 +1,103 @@
+#include
+
+using namespace std;
+
+/*
+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.
+*/
+
+// Approach: BFS
+
+bool isValid(int nx, int ny, int m, int n)
+{
+ return (0 <= nx && nx < m) &&
+ (0 <= ny && ny < n);
+}
+
+// Time: O(mn) x 4
+// Space: O(mn)
+int orangesRotting(vector> &grid)
+{
+ int m = grid.size();
+ int n = grid[0].size();
+
+ int total = 0; // fresh + rotten oranges
+ int count = 0; // rotten oranges
+ int mins = 0; // minutes elapsed
+
+ queue> rotten; // {i, j}: position of rotten orange
+
+ // 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
+ }
+ }
+
+ int dx[] = {0, 0, -1, 1}; // 4-directions (x-axis)
+ int dy[] = {-1, 1, 0, 0}; // 4-directions (y-axis)
+
+ while (!rotten.empty())
+ {
+ int size = rotten.size(); // rotten oranges in current minute
+ count += size; // add to total rotten oranges
+
+ 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();
+
+ // 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];
+
+ // 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
+ }
+ }
+ }
+
+ if (!rotten.empty()) // if there are more rotten oranges
+ mins++;
+ }
+
+ if (total != count) // fresh oranges left
+ return -1;
+
+ return mins;
+}
+
+int main()
+{
+ vector> grid = {{2, 1, 1},
+ {1, 1, 0},
+ {0, 1, 1}};
+
+ cout << orangesRotting(grid) << endl; // 4
+
+ return 0;
+}
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+
+ 
+
+