Given a matrix of N rows and M columns. From m[i][j], we can move to m[i+1][j], if m[i+1][j] > m[i][j], or can move to m[i][j+1] if m[i][j+1] > m[i][j]. The task is print longest path length if we start from (0, 0).
Examples:
Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
C++
// CPP program to find longest increasing // path in a matrix. #include <bits/stdc++.h> #define MAX 10 using namespace std; // Return the length of LIP in 2D matrix int LIP(int dp[][MAX], int mat[][MAX], int n, int m, int x, int y) { // If value not calculated yet. if (dp[x][y] < 0) { int result = 0; // If reach bottom left cell, return 1. if (x == n - 1 && y == m - 1) return dp[x][y] = 1; // If reach the corner of the matrix. if (x == n - 1 || y == m - 1) result = 1; // If value greater than below cell. if (mat[x][y] < mat[x + 1][y]) result = 1 + LIP(dp, mat, n, m, x + 1, y); // If value greater than left cell. if (mat[x][y] < mat[x][y + 1]) result = max(result, 1 + LIP(dp, mat, n, m, x, y + 1)); dp[x][y] = result; } return dp[x][y]; } // Wrapper function int wrapper(int mat[][MAX], int n, int m) { int dp[MAX][MAX]; memset(dp, -1, sizeof dp); return LIP(dp, mat, n, m, 0, 0); } // Driven Program int main() { int mat[][MAX] = { { 1, 2, 3, 4 }, { 2, 2, 3, 4 }, { 3, 2, 3, 4 }, { 4, 5, 6, 7 }, }; int n = 4, m = 4; cout << wrapper(mat, n, m) << endl; return 0; } |
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Java
// Java program to find longest increasing // path in a matrix. import java.util.*; class GFG { // Return the length of LIP in 2D matrix static int LIP(int dp[][], int mat[][], int n, int m, int x, int y) { // If value not calculated yet. if (dp[x][y] < 0) { int result = 0; // If reach bottom left cell, return 1. if (x == n - 1 && y == m - 1) return dp[x][y] = 1; // If reach the corner of the matrix. if (x == n - 1 || y == m - 1) result = 1; // If value greater than below cell. if (x + 1 < n && mat[x][y] < mat[x + 1][y]) result = 1 + LIP(dp, mat, n, m, x + 1, y); // If value greater than left cell. if (y + 1 < m && mat[x][y] < mat[x][y + 1]) result = Math.max(result, 1 + LIP(dp, mat, n, m, x, y + 1)); dp[x][y] = result; } return dp[x][y]; } // Wrapper function static int wrapper(int mat[][], int n, int m) { int dp[][] = new int[10][10]; for (int i = 0; i < 10; i++) Arrays.fill(dp[i], -1); return LIP(dp, mat, n, m, 0, 0); } /* Driver program to test above function */ public static void main(String[] args) { int mat[][] = { { 1, 2, 3, 4 }, { 2, 2, 3, 4 }, { 3, 2, 3, 4 }, { 4, 5, 6, 7 }, }; int n = 4, m = 4; System.out.println(wrapper(mat, n, m)); } } // This code is contributed by Arnav Kr. Mandal. |
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Python3
# Python3 program to find longest # increasing path in a matrix. MAX = 20 # Return the length of # LIP in 2D matrix def LIP(dp, mat, n, m, x, y): # If value not calculated yet. if (dp[x][y] < 0): result = 0 # If reach bottom left cell, # return 1. if (x == n - 1 and y == m - 1): dp[x][y] = 1 return dp[x][y] # If reach the corner # of the matrix. if (x == n - 1 or y == m - 1): result = 1 # If value greater than below cell. if (x + 1 < n and mat[x][y] < mat[x + 1][y]): result = 1 + LIP(dp, mat, n, m, x + 1, y) # If value greater than left cell. if (y + 1 < m and mat[x][y] < mat[x][y + 1]): result = max(result, 1 + LIP(dp, mat, n, m, x, y + 1)) dp[x][y] = result return dp[x][y] # Wrapper function def wrapper(mat, n, m): dp = [[-1 for i in range(MAX)] for i in range(MAX)] return LIP(dp, mat, n, m, 0, 0) # Driver Code mat = [[1, 2, 3, 4 ], [2, 2, 3, 4 ], [3, 2, 3, 4 ], [4, 5, 6, 7 ]] n = 4m = 4print(wrapper(mat, n, m)) # This code is contributed # by Sahil Shelangia |
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C#
// C# program to find longest increasing // path in a matrix. using System; public class GFG { // Return the length of LIP in 2D matrix static int LIP(int[, ] dp, int[, ] mat, int n, int m, int x, int y) { // If value not calculated yet. if (dp[x, y] < 0) { int result = 0; // If reach bottom left cell, return 1. if (x == n - 1 && y == m - 1) return dp[x, y] = 1; // If reach the corner of the matrix. if (x == n - 1 || y == m - 1) result = 1; // If value greater than below cell. if (x + 1 < n && mat[x, y] < mat[x + 1, y]) result = 1 + LIP(dp, mat, n, m, x + 1, y); // If value greater than left cell. if (y + 1 < m && mat[x, y] < mat[x, y + 1]) result = Math.Max(result, 1 + LIP(dp, mat, n, m, x, y + 1)); dp[x, y] = result; } return dp[x, y]; } // Wrapper function static int wrapper(int[, ] mat, int n, int m) { int[, ] dp = new int[10, 10]; for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { dp[i, j] = -1; } } return LIP(dp, mat, n, m, 0, 0); } /* Driver code */ public static void Main() { int[, ] mat = { { 1, 2, 3, 4 }, { 2, 2, 3, 4 }, { 3, 2, 3, 4 }, { 4, 5, 6, 7 }, }; int n = 4, m = 4; Console.WriteLine(wrapper(mat, n, m)); } } /* This code contributed by PrinciRaj1992 */ |
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Output:
7
Time Complexity: O(N*M).
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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