|
| 1 | +""" |
| 2 | +Problem Link: https://practice.geeksforgeeks.org/problems/gold-mine-problem2608/1 |
| 3 | + |
| 4 | +Given a gold mine called M of (n x m) dimensions. Each field in this mine contains a positive integer which is the amount |
| 5 | +of gold in tons. Initially the miner can start from any row in the first column. From a given cell, the miner can move |
| 6 | +to the cell diagonally up towards the right |
| 7 | +to the right |
| 8 | +to the cell diagonally down towards the right |
| 9 | +Find out maximum amount of gold which he can collect. |
| 10 | + |
| 11 | +Example 1: |
| 12 | +Input: n = 3, m = 3 |
| 13 | +M = {{1, 3, 3}, |
| 14 | + {2, 1, 4}, |
| 15 | + {0, 6, 4}}; |
| 16 | +Output: 12 |
| 17 | + |
| 18 | +Explaination: |
| 19 | +The path is {(1,0) -> (2,1) -> (2,2)}. |
| 20 | + |
| 21 | +Example 2: |
| 22 | +Input: n = 4, m = 4 |
| 23 | +M = {{1, 3, 1, 5}, |
| 24 | + {2, 2, 4, 1}, |
| 25 | + {5, 0, 2, 3}, |
| 26 | + {0, 6, 1, 2}}; |
| 27 | +Output: 16 |
| 28 | +Explaination: |
| 29 | +The path is {(2,0) -> (3,1) -> (2,2) |
| 30 | +-> (2,3)} or {(2,0) -> (1,1) -> (1,2) |
| 31 | +-> (0,3)}. |
| 32 | + |
| 33 | +Your Task: |
| 34 | +You do not need to read input or print anything. Your task is to complete the function maxGold() which takes the |
| 35 | +values n, m and the mine M as input parameters and returns the maximum amount of gold that can be collected. |
| 36 | + |
| 37 | +Expected Time Complexity: O(n*m) |
| 38 | +Expected Auxiliary Space: O(n*m) |
| 39 | + |
| 40 | +Constraints: |
| 41 | +1 ≤ n, m ≤ 50 |
| 42 | +1 ≤ M[i][j] ≤ 100 |
| 43 | +""" |
| 44 | +class Solution: |
| 45 | + def maxGold(self, n, m, M): |
| 46 | + cache = [[0] * m for _ in range(n)] |
| 47 | + |
| 48 | + directions = [[0, 1], [-1, 1], [1, 1]] |
| 49 | + |
| 50 | + max_gold = 0 |
| 51 | + for col in range(m-1, -1, -1): |
| 52 | + for row in range(n-1, -1, -1): |
| 53 | + max_val = 0 |
| 54 | + for direction in directions: |
| 55 | + new_row = row + direction[0] |
| 56 | + new_col = col + direction[1] |
| 57 | + |
| 58 | + if not (new_row >= 0 and new_row < n and new_col < m): |
| 59 | + continue |
| 60 | + |
| 61 | + max_val = max(max_val, cache[new_row][new_col]) |
| 62 | + |
| 63 | + cache[row][col] = M[row][col] + max_val |
| 64 | + |
| 65 | + if col == 0: |
| 66 | + max_gold = max(max_gold, cache[row][col]) |
| 67 | + |
| 68 | + return max_gold |
| 69 | + |
| 70 | + |
| 71 | +class Solution1: |
| 72 | + def maxGold(self, n, m, M): |
| 73 | + cache = [[0] * m for _ in range(n)] |
| 74 | + |
| 75 | + directions = [[0, 1], [-1, 1], [1, 1]] |
| 76 | + queue = [] |
| 77 | + for row in range(n): |
| 78 | + queue.append([row, 0, M[row][0]]) |
| 79 | + |
| 80 | + max_gold = 0 |
| 81 | + while queue: |
| 82 | + |
| 83 | + for _ in range(len(queue)): |
| 84 | + row, col, cur_gold = queue.pop(0) |
| 85 | + cache[row][col] = cur_gold |
| 86 | + max_gold = max(max_gold, cur_gold) |
| 87 | + |
| 88 | + for direction in directions: |
| 89 | + next_row = row + direction[0] |
| 90 | + next_col = col + direction[1] |
| 91 | + if not (next_row >= 0 and next_row < n and next_col < m and cache[next_row][next_col] < cur_gold + M[next_row][next_col]): |
| 92 | + continue |
| 93 | + |
| 94 | + queue.append([next_row, next_col, cur_gold + M[next_row][next_col]]) |
| 95 | + |
| 96 | + return max_gold |
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