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 the length of the longest increasing sequence for submatrix starting from the ith row and jth 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 the base case with the length of longest increasing subsequence 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 10using namespace std;// Return the length of LIP in 2D matrixint 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 right 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 != n-1) // x reaches last row 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(y != m-1) // y reaches last column 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 functionint 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 Programint 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;} |
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 right 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. |
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 right 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 |
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 right 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 */ |
Javascript
<script> // Javascript program to find longest increasing path in a matrix. // Return the length of LIP in 2D matrix function LIP(dp, mat, n, m, x, y) { // If value not calculated yet. if (dp[x][y] < 0) { let result = 0; // If reach bottom right 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 function wrapper(mat, n, m) { let dp = new Array(10); for (let i = 0; i < 10; i++) { dp[i] = new Array(10); for (let j = 0; j < 10; j++) { dp[i][j] = -1; } } return LIP(dp, mat, n, m, 0, 0); } let mat = [ [ 1, 2, 3, 4 ], [ 2, 2, 3, 4 ], [ 3, 2, 3, 4 ], [ 4, 5, 6, 7 ], ]; let n = 4, m = 4; document.write(wrapper(mat, n, m)); </script> |
Output
7
Time Complexity: O(N*M).
Space Complexity: O(N*M)
This article is contributed by Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
