Longest Path in Matrix

Author: spaceman-dev

Dynamic_Programming/LongestPathInMatrix.cpp

@@ -0,0 +1,72 @@
+// 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;
+}

Dynamic_Programming/README.md

@@ -114,12 +114,22 @@
  Output: 4
  Explanation: Total length is 4, and the cut lengths are 2, 1 and 1. We can make maximum 4 segments each of length 1.
 
+## 15. Longest Path in Matrix:
+ 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 1: 
+ Input: mat[][] = {{1, 2, 9}
+  {5, 3, 8}
+  {4, 6, 7}}
+ Output: 4
+ The longest path is 6-7-8-9. 
 ## 16. Minimum Sum Partition:
  Given an integer array arr of size N, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum and find the minimum difference
  Example 1:
  Input: N = 4, arr[] = {1, 6, 11, 5} 
  Output: 1
  Explanation: Subset1 = {1, 5, 6}, sum of Subset1 = 12 Subset2 = {11}, sum of Subset2 = 11 
+
 ## 17. Count Number of Hops:
  A frog jumps either 1, 2, or 3 steps to go to the top. In how many ways can it reach the top. As the answer will be large find the answer modulo 1000000007.
  Example 1: