|
| 1 | +""" |
| 2 | +Problem Link: https://practice.geeksforgeeks.org/problems/path-in-matrix3805/1 |
| 3 | + |
| 4 | +Given a NxN matrix of positive integers. There are only three possible moves from a cell Matrix[r][c]. |
| 5 | +Matrix [r+1] [c] |
| 6 | +Matrix [r+1] [c-1] |
| 7 | +Matrix [r+1] [c+1] |
| 8 | +Starting from any column in row 0, return the largest sum of any of the paths up to row N-1. |
| 9 | + |
| 10 | +Example 1: |
| 11 | +Input: N = 2 |
| 12 | +Matrix = {{348, 391}, |
| 13 | + {618, 193}} |
| 14 | +Output: 1009 |
| 15 | +Explaination: The best path is 391 -> 618. |
| 16 | +It gives the sum = 1009. |
| 17 | + |
| 18 | +Example 2: |
| 19 | +Input: N = 2 |
| 20 | +Matrix = {{2, 2}, |
| 21 | + {2, 2}} |
| 22 | +Output: 4 |
| 23 | +Explaination: No matter which path is |
| 24 | +chosen, the output is 4. |
| 25 | + |
| 26 | +Your Task: |
| 27 | +You do not need to read input or print anything. Your task is to complete the function maximumPath() which takes the size N |
| 28 | +and the Matrix as input parameters and returns the highest maximum path sum. |
| 29 | + |
| 30 | +Expected Time Complexity: O(N*N) |
| 31 | +Expected Auxiliary Space: O(N*N) |
| 32 | + |
| 33 | +Constraints: |
| 34 | +1 ≤ N ≤ 100 |
| 35 | +1 ≤ Matrix[i][j] ≤ 1000 |
| 36 | +""" |
| 37 | +class Solution: |
| 38 | + def maximumPath(self, N, Matrix): |
| 39 | + dp = [[0] * N for _ in range(N)] |
| 40 | + |
| 41 | + directions = [[1, 1], [1, -1], [1, 0]] |
| 42 | + |
| 43 | + max_path_sum = 0 |
| 44 | + |
| 45 | + for row in range(N-1, -1, -1): |
| 46 | + for col in range(N): |
| 47 | + max_val = 0 |
| 48 | + for r, c in directions: |
| 49 | + new_row = r + row |
| 50 | + new_col = c + col |
| 51 | + if new_row >= 0 and new_col >= 0 and new_row < N and new_col < N: |
| 52 | + max_val = max(max_val, dp[new_row][new_col]) |
| 53 | + |
| 54 | + dp[row][col] = Matrix[row][col] + max_val |
| 55 | + max_path_sum = max(max_path_sum, dp[row][col]) |
| 56 | + |
| 57 | + return max_path_sum |
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