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Longest Increasing Path in a Matrix in Python
Suppose we have one matrix; we have to find the length of the longest increasing path. From each cell, we can either move to four directions − left, right, up or down. We cannot move diagonally or move outside of the boundary.
So, if the input is like
| 9 | 9 | 4 |
| 6 | 6 | 8 |
| 2 | 1 | 1 |
then the output will be 4 as the longest increasing path is [3, 4, 5, 6].
To solve this, we will follow these steps −
Define a function solve(). This will take i,j,matrix
if dp[i,j] is non-zero, then
return dp[i, j]
dp[i, j] := 1
temp := 0
for r in range i-1 to i+2, do
for c in range j-1 to j+2, do
if r is same as i and c is same as j or(|r-i| is same as 1 and |c-j| is same as 1) , then
go for the next iteration
if c>=0 and r>=0 and r< row count of matrix and c < col size of matrix[0] and matrix[r, c]>matrix[i, j], then
temp := maximum of temp, solve(r, c, matrix)
dp[i, j] := dp[i, j] + temp
return dp[i, j]
From the main method do the following −
if not matrix is non-zero, then
return 0
dp := a matrix of size same as given matrix and fill with 0
ans := 0
for i in range 0 to size of matrix, do
for j in range 0 to size of matrix[0], do
if dp[i, j] is same as 0, then
solve(i, j, matrix)
return ans
Example
Let us see the following implementation to get better understanding −
class Solution(object): def solve(self,i,j,matrix): if self.dp[i][j]: return self.dp[i][j] self.dp[i][j] = 1 temp = 0 for r in range(i-1,i+2): for c in range(j-1,j+2): if r==i and c==j or (abs(r-i)==1 and abs(c-j)==1): continue if c>=0 and r>=0 and r<len(matrix) and c<len(matrix[0]) and matrix[r][c]>matrix[i][j]: temp = max(temp,self.solve(r,c,matrix)) self.dp[i][j]+=temp return self.dp[i][j] def longestIncreasingPath(self, matrix): if not matrix: return 0 self.dp = [ [0 for i in range(len(matrix[0]))] for j in range(len(matrix))] self.ans = 0 for i in range(len(matrix)): for j in range(len(matrix[0])): if self.dp[i][j]==0: self.solve(i,j,matrix) self.ans = max(self.ans,self.dp[i][j]) return self.ans ob = Solution() print(ob.longestIncreasingPath([[9,9,4],[6,6,8],[2,1,1]]))
Input
[[9,9,4],[6,6,8],[2,1,1]]
Output
4
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