Longest Increasing Path in Matrix

22 Jun, 2022
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++
Java
Python3
C#
Javascript
// 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 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 program to find longest# increasing path in a matrix.MAX = 20 # Return the length of# LIP in 2D matrixdef 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 functiondef 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 Codemat = [[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# 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 */
<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)
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