# 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, max(y, max(z, max(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(n^{2}). 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.

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