Find the longest path in a matrix with given constraints

Given a n*n matrix where all numbers are distinct, find the maximum length path (starting from any cell) such that all cells along the path are in increasing order with a difference of 1. 
We can move in 4 directions from a given cell (i, j), i.e., we can move to (i+1, j) or (i, j+1) or (i-1, j) or (i, j-1) with the condition that the adjacent cells have a difference of 1.

Example:

Input:  mat[][] = {{1, 2, 9}
                   {5, 3, 8}
                   {4, 6, 7}}
Output: 4
The longest path is 6-7-8-9. 

The idea is simple, we calculate longest path beginning with every cell. Once we have computed longest for all cells, we return maximum of all longest paths. One important observation in this approach is many overlapping sub-problems. Therefore this problem can be optimally solved using Dynamic Programming. 

Below is Dynamic Programming based implementation that uses a lookup table dp[][] to check if a problem is already solved or not.

C++

// C++ program to find the longest path in a matrix
// with given constraints
#include <bits/stdc++.h>
#define n 3
using namespace std;

// Returns length of the longest path beginning with
// mat[i][j]. This function mainly uses lookup table
// dp[n][n]
int findLongestFromACell(int i, int j, int mat[n][n],
                         int dp[n][n])
{
    if (i < 0 || i >= n || j < 0 || j >= n)
        return 0;

    // If this subproblem is already solved
    if (dp[i][j] != -1)
        return dp[i][j];

    // To store the path lengths in all the four directions
    int x = INT_MIN, y = INT_MIN, z = INT_MIN, w = INT_MIN;

    // Since all numbers are unique and in range from 1 to
    // n*n, there is atmost one possible direction from any
    // cell
    if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
        x = 1 + findLongestFromACell(i, j + 1, mat, dp);

    if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
        y = 1 + findLongestFromACell(i, j - 1, mat, dp);

    if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
        z = 1 + findLongestFromACell(i - 1, j, mat, dp);

    if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
        w = 1 + findLongestFromACell(i + 1, j, mat, dp);

    // If none of the adjacent fours is one greater we will
    // take 1 otherwise we will pick maximum from all the
    // four directions
    return dp[i][j] = max({x, y, z, w, 1});
}

// Returns length of the longest path beginning with any
// cell
int finLongestOverAll(int mat[n][n])
{
    int result = 1; // Initialize result

    // Create a lookup table and fill all entries in it as
    // -1
    int dp[n][n];
    memset(dp, -1, sizeof dp);

    // Compute longest path beginning from all cells
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (dp[i][j] == -1)
                findLongestFromACell(i, j, mat, dp);

            // Update result if needed
            result = max(result, dp[i][j]);
        }
    }

    return result;
}

// Driver program
int main()
{
    int mat[n][n]
        = { { 1, 2, 9 }, { 5, 3, 8 }, { 4, 6, 7 } };
    cout << "Length of the longest path is "
         << finLongestOverAll(mat);
    return 0;
}

Java

// Java program to find the longest path in a matrix
// with given constraints

class GFG {
    public static int n = 3;

    // Function that returns length of the longest path
    // beginning with mat[i][j]
    // This function mainly uses lookup table dp[n][n]
    static int findLongestFromACell(int i, int j,
                                    int mat[][], int dp[][])
    {
        // Base case
        if (i < 0 || i >= n || j < 0 || j >= n)
            return 0;

        // If this subproblem is already solved
        if (dp[i][j] != -1)
            return dp[i][j];

        // To store the path lengths in all the four
        // directions
        int x = Integer.MIN_VALUE, y = Integer.MIN_VALUE,
            z = Integer.MIN_VALUE, w = Integer.MIN_VALUE;
        // Since all numbers are unique and in range from 1
        // to n*n, there is atmost one possible direction
        // from any cell
        if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
            x = dp[i][j]
                = 1
                  + findLongestFromACell(i, j + 1, mat, dp);

        if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
            y = dp[i][j]
                = 1
                  + findLongestFromACell(i, j - 1, mat, dp);

        if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
            z = dp[i][j]
                = 1
                  + findLongestFromACell(i - 1, j, mat, dp);

        if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
            w = dp[i][j]
                = 1
                  + findLongestFromACell(i + 1, j, mat, dp);

        // If none of the adjacent fours is one greater we
        // will take 1 otherwise we will pick maximum from
        // all the four directions
        return dp[i][j]
            = Math.max(
                x,
                Math.max(y, Math.max(z, Math.max(w, 1))));
    }

    // Function that returns length of the longest path
    // beginning with any cell
    static int finLongestOverAll(int mat[][])
    {
        // Initialize result
        int result = 1;

        // Create a lookup table and fill all entries in it
        // as -1
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                dp[i][j] = -1;

        // Compute longest path beginning from all cells
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dp[i][j] == -1)
                    findLongestFromACell(i, j, mat, dp);

                // Update result if needed
                result = Math.max(result, dp[i][j]);
            }
        }

        return result;
    }

    // driver program
    public static void main(String[] args)
    {
        int mat[][]
            = { { 1, 2, 9 }, { 5, 3, 8 }, { 4, 6, 7 } };
        System.out.println("Length of the longest path is "
                           + finLongestOverAll(mat));
    }
}

// Contributed by Pramod Kumar

Python3

# Python3 program to find the longest path in a matrix
# with given constraints

n = 3
# Returns length of the longest path beginning with mat[i][j].
# This function mainly uses lookup table dp[n][n]


def findLongestFromACell(i, j, mat, dp):
    # Base case
    if (i < 0 or i >= n or j < 0 or j >= n):
        return 0

    # If this subproblem is already solved
    if (dp[i][j] != -1):
        return dp[i][j]

    # To store the path lengths in all the four directions
    x, y, z, w = -1, -1, -1, -1

    # Since all numbers are unique and in range from 1 to n * n,
    # there is atmost one possible direction from any cell
    if (j < n-1 and ((mat[i][j] + 1) == mat[i][j + 1])):
        x = 1 + findLongestFromACell(i, j + 1, mat, dp)

    if (j > 0 and (mat[i][j] + 1 == mat[i][j-1])):
        y = 1 + findLongestFromACell(i, j-1, mat, dp)

    if (i > 0 and (mat[i][j] + 1 == mat[i-1][j])):
        z = 1 + findLongestFromACell(i-1, j, mat, dp)

    if (i < n-1 and (mat[i][j] + 1 == mat[i + 1][j])):
        w = 1 + findLongestFromACell(i + 1, j, mat, dp)

    # If none of the adjacent fours is one greater we will take 1
    # otherwise we will pick maximum from all the four directions
    dp[i][j] = max(x, max(y, max(z, max(w, 1))))
    return dp[i][j]


# Returns length of the longest path beginning with any cell
def finLongestOverAll(mat):
    result = 1  # Initialize result

    # Create a lookup table and fill all entries in it as -1
    dp = [[-1 for i in range(n)]for i in range(n)]

    # Compute longest path beginning from all cells
    for i in range(n):
        for j in range(n):
            if (dp[i][j] == -1):
                findLongestFromACell(i, j, mat, dp)
            # Update result if needed
            result = max(result, dp[i][j])
    return result


# Driver program
mat = [[1, 2, 9],
       [5, 3, 8],
       [4, 6, 7]]
print("Length of the longest path is ", finLongestOverAll(mat))

# this code is improved by sahilshelangia

Javascript

<script>
// JavaScript program to find the longest path in a matrix
// with given constraints
let n = 3;

// Returns length of the longest path beginning with mat[i][j].
// This function mainly uses lookup table dp[n][n]
function findLongestFromACell( i, j, mat, dp){
    if (i < 0 || i >= n || j < 0 || j >= n)
        return 0;

    // If this subproblem is already solved
    if (dp[i][j] != -1)
        return dp[i][j];

    // To store the path lengths in all the four directions

    let x,y,z,w;
    x = -1;
    y = -1;
    z = -1
    w = -1;

    // Since all numbers are unique and in range from 1 to n*n,
    // there is atmost one possible direction from any cell
    if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
        x = 1 + findLongestFromACell(i, j + 1, mat, dp);

    if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
        y = 1 + findLongestFromACell(i, j - 1, mat, dp);

    if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
        z = 1 + findLongestFromACell(i - 1, j, mat, dp);

    if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
        w = 1 + findLongestFromACell(i + 1, j, mat, dp);

    // If none of the adjacent fours is one greater we will take 1
    // otherwise we will pick maximum from all the four directions
    dp[i][j] = Math.max(x, Math.max(y, Math.max(z, Math.max(w, 1))));
    return dp[i][j];
}

// Returns length of the longest path beginning with any cell
function finLongestOverAll( mat){
    let result = 1; // Initialize result

    // Create a lookup table and fill all entries in it as -1
    var dp = [];


    for( var y = 0; y < n; y++ ) {
    dp[ y ] = [];
    for( var x = 0; x < n; x++ ) {
        dp[ y ][ x ] = -1;
    }
}

    // Compute longest path beginning from all cells
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            if (dp[i][j] == -1)
                findLongestFromACell(i, j, mat, dp);

            // Update result if needed
            result = Math.max(result, dp[i][j]);
        }
    }

    return result;
}

// Driver program
let mat = [[ 1, 2, 9 ],
            [ 5, 3, 8 ],
            [ 4, 6, 7 ]];

document.write("Length of the longest path is ");
document.write( finLongestOverAll(mat));

</script>

C#

// C# program to find the longest path
// in a matrix with given constraints
using System;

class GFG {
    public static int n = 3;

    // Function that returns length of
    // the longest path beginning with mat[i][j]
    // This function mainly uses lookup
    // table dp[n][n]
    public static int findLongestFromACell(int i, int j,
                                           int[][] mat,
                                           int[][] dp)
    {
        // Base case
        if (i < 0 || i >= n || j < 0 || j >= n) {
            return 0;
        }

        // If this subproblem is
        // already solved
        if (dp[i][j] != -1) {
            return dp[i][j];
        }

        // To store the path lengths in all the four
        // directions
        int x = int.MinValue, y = int.MinValue,
            z = int.MinValue, w = int.MinValue;

        // Since all numbers are unique and
        // in range from 1 to n*n, there is
        // atmost one possible direction
        // from any cell
        if (j < n - 1
            && ((mat[i][j] + 1) == mat[i][j + 1])) {
            x = dp[i][j]
                = 1
                  + findLongestFromACell(i, j + 1, mat, dp);
        }

        if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1])) {
            y = dp[i][j]
                = 1
                  + findLongestFromACell(i, j - 1, mat, dp);
        }

        if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j])) {
            z = dp[i][j]
                = 1
                  + findLongestFromACell(i - 1, j, mat, dp);
        }

        if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j])) {
            w = dp[i][j]
                = 1
                  + findLongestFromACell(i + 1, j, mat, dp);
        }

        // If none of the adjacent fours is one greater we
        // will take 1 otherwise we will pick maximum from
        // all the four directions
        dp[i][j] = Math.Max(
            x, Math.Max(y, Math.Max(z, Math.Max(w, 1))));
        return dp[i][j];
    }

    // Function that returns length of the
    // longest path beginning with any cell
    public static int finLongestOverAll(int[][] mat)
    {
        // Initialize result
        int result = 1;

        // Create a lookup table and fill
        // all entries in it as -1
        int[][] dp
            = RectangularArrays.ReturnRectangularIntArray(
                n, n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = -1;
            }
        }

        // Compute longest path beginning
        // from all cells
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dp[i][j] == -1) {
                    findLongestFromACell(i, j, mat, dp);
                }

                // Update result if needed
                result = Math.Max(result, dp[i][j]);
            }
        }

        return result;
    }

    public static class RectangularArrays {
        public static int[][] ReturnRectangularIntArray(
            int size1, int size2)
        {
            int[][] newArray = new int[size1][];
            for (int array1 = 0; array1 < size1; array1++) {
                newArray[array1] = new int[size2];
            }

            return newArray;
        }
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[][] mat = new int[][] { new int[] { 1, 2, 9 },
                                    new int[] { 5, 3, 8 },
                                    new int[] { 4, 6, 7 } };
        Console.WriteLine("Length of the longest path is "
                          + finLongestOverAll(mat));
    }
}

// This code is contributed by Shrikant13

Output

Length of the longest path is 4

Time complexity of the above solution is O(n²). It may seem more at first look. If we take a closer look, we can notice that all values of dp[i][j] are computed only once.
Auxiliary Space: O(N x N), since N x N extra space has been taken.

This article is contributed by Aarti_Rathi and Ekta Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.