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.
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 3using 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// cellint 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 programint 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 constraintsclass 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 constraintsn = 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 celldef 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 programmat = [[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 constraintslet 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 cellfunction 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 programlet 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 constraintsusing 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.
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