create alphabetBoardPath.py

algorithms/strings/alphabetBoardPath.py

@@ -0,0 +1,66 @@
+"""
+On an alphabet board, we start at position (0, 0), corresponding to character board[0][0].
+
+Here, board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"], as shown in the diagram below.
+
+board = [
+"abcde", 
+"fghij", 
+"klmno", 
+"pqrst", 
+"uvwxy", 
+"z"]
+
+We may make the following moves:
+
+'U' moves our position up one row, if the position exists on the board;
+'D' moves our position down one row, if the position exists on the board;
+'L' moves our position left one column, if the position exists on the board;
+'R' moves our position right one column, if the position exists on the board;
+'!' adds the character board[r][c] at our current position (r, c) to the answer.
+(Here, the only positions that exist on the board are positions with letters on them.)
+
+Return a sequence of moves that makes our answer equal to target in the minimum number of moves.  You may return any path that does so.
+
+Example:
+Input: target = "code"
+Output: "RR!DDRR!UUL!R!"
+
+Time complexity: O(n)
+
+Space complextity: O(n)
+"""
+
+def alphabetBoardPath(self, target: str) -> str:
+
+    board = ["abcde", "fghij", "klmno", "pqrst", "uvwxy", "z"]
+    dct = {}
+    # mapping each letter to its position in the board
+    for i in range(len(board)):
+        for j in range(len(board[i])):
+            dct[board[i][j]] = (i, j)
+    curPos = (0, 0)
+    # iterate through every char in target, keeping track of current letter and destination
+    # add path from current letter to destination to path
+
+    path = ''
+    for letter in target:
+
+        if curPos == dct[letter]:
+            path += '!'
+
+        else:
+            ver = curPos[0] - dct[letter][0]
+            hor = curPos[1] - dct[letter][1]
+            yMove = 'U' * ver if ver > 0 else 'D' * (-1 * ver)
+            xMove = 'L' * hor if hor > 0 else 'R' * (-1 * hor)
+
+            # traversal if destination == 'z': horizontal -> vertical else vertical -> horizontal
+
+            if letter == 'z':
+                path  += xMove + yMove + '!'
+            else:
+                path += (yMove + xMove + '!')
+            curPos = dct[letter]
+
+    return path