Suppose we have a 2D matrix, we have to find the length of the longest strictly increasing path. To traverse the path we can move up, down, left, or right nor diagonally.
So, if the input is like
then the output will be 6, as The longest path is [1, 2, 4, 6, 7, 9]
To solve this, we will follow these steps −
n := row count of matrix , m := column count of matrix moves := a list of pairs to move up, down, left and right [[1, 0], [-1, 0], [0, 1], [0, -1]] Define a function dp() . This will take y, x if x and y are in range of matrix, then return 0 currVal := matrix[y, x] res := 0 for each d in moves, do (dy, dx) := d (newY, newX) := (y + dy, x + dx) if newY and newX are in range of matrix and matrix[newY, newX] > currVal, then res := maximum of res and dp(newY, newX) return res + 1 From the main method do the following: result := 0 for i in range 0 to n - 1, do for j in range 0 to m - 1, do result := maximum of result and dp(i, j) return result
Let us see the following implementation to get better understanding −
class Solution: def solve(self, matrix): n, m = len(matrix), len(matrix) moves = [[1, 0], [-1, 0], [0, 1], [0, -1]] def dp(y, x): if y < 0 or y >= n or x < 0 or x >= m: return 0 currVal = matrix[y][x] res = 0 for d in moves: dy, dx = d newY, newX = y + dy, x + dx if (newY >= 0 and newY < n and newX >= 0 and newX < m and matrix[newY][newX] > currVal): res = max(res, dp(newY, newX)) return res + 1 result = 0 for i in range(n): for j in range(m): result = max(result, dp(i, j)) return result ob = Solution() matrix = [ [2, 4, 6], [1, 5, 7], [3, 3, 9] ] print(ob.solve(matrix))
[ [2, 4, 6], [1, 5, 7], [3, 3, 9] ]