Add a solution to Shortest Path to Get Food

Author: soapyigu

BFS/ShortestPathGetFood.swift

@@ -0,0 +1,61 @@
+/**
+ * Question Link: https://leetcode.com/problems/shortest-path-to-get-food/
+ * Primary idea: BFS to go over all possible word paths until the cell is food, avoid hitting the obstacle.
+ * 
+ * Time Complexity: O(nm), Space Complexity: O(nm)
+ */
+
+class ShortestPathGetFood {
+    func getFood(_ grid: [[Character]]) -> Int {
+        let m = grid.count, n = grid[0].count
+        var isVisited = Array(repeating: Array(repeating: false, count: n), count: m)
+        
+        let start = findStart(grid)
+        
+        isVisited[start.0][start.1] = true
+        var queue = [Point(i: start.0, j: start.1, len: 0)]
+        
+        while !queue.isEmpty {
+            let point = queue.removeFirst()
+            
+            if grid[point.i][point.j] == "#" {
+                return point.len
+            }
+            
+            for dir in [(0, 1), (0, -1), (1, 0), (-1, 0)] {
+                let (x, y) = (point.i + dir.0, point.j + dir.1)
+                
+                guard x >= 0 && x < m && y >= 0 && y < n && !isVisited[x][y] else {
+                    continue
+                }
+                
+                if grid[x][y] == "X" {
+                    continue
+                }
+                
+                isVisited[x][y] = true
+                queue.append(Point(i: x, j: y, len: point.len + 1))
+            }
+        }
+        
+        return -1
+    }
+    
+    private func findStart(_ grid: [[Character]]) -> (Int, Int) {
+        for i in 0..<grid.count {
+            for j in 0..<grid[0].count {
+                if grid[i][j] == "*" {
+                    return (i, j)
+                }
+            }
+        }
+        
+        return (-1, -1)
+    }
+    
+    struct Point {
+        var i: Int
+        var j: Int
+        var len: Int
+    }
+}