Longest Increasing Path in Matrix

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 will be the answer.

Below is the implementation of this approach:

C++

 // CPP program to find longest increasing // path in a matrix. #include #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; }

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;         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 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 = 4 m = 4 print(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 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 */

Output:

7

Time Complexity: O(N*M).

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