Open In App

Find the longest path in a matrix with given constraints

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Share
Report issue
Report

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. 
Recommended Practice

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. 

Algorithm:

Step 1: Initialize a matrix and set its size to n x n.
Step 2: Define a function “findLongestFromACell” that takes in a cell’s row and column index, the matrix, and a lookup table. If the cell is out of                     bounds or the subproblem has already been solved, return 0 or the previously calculated value in the lookup table, respectively.
Step 3: Define four integer variables to store the length of the path in each of the four possible directions. Check if the adjacent cell in each                          direction satisfies the constraints and if so, recursively call the function for that cell and update the corresponding direction’s length                           variable.
Step 4: Return the maximum length of the four directions plus one, and store it in the lookup table.
Step 5: Define a function “finLongestOverAll” that takes in the matrix. Initialize a result variable to 1 and a lookup table as a two-dimensional array              of size n x n, filled with -1.
Step 6: For each cell in the matrix, call “findLongestFromACell” and update the result as needed.
Step 7: Return the result.

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.



Last Updated : 05 May, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads