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Longest consecutive path from a given starting character
A matrix of different characters is given. Starting from one character we have to find the longest path by traversing all characters which are greater than the current character. The characters are consecutive to each other.
To find the longest path, we will use the Depth First Search algorithm. During DFS, some subproblems may arise multiple times. To avoid the computation of that, again and again, we will use a dynamic programming approach.
Input and Output
Input: The matrix as shown above. And the starting point. Here the starting point is e. Output: Enter Starting Point (a-i): e Maximum consecutive path: 5
Algorithm
findLongestLen(i, j, prev)
Input: Position i and j and the previous character.
Output: Longest length.
Begin if (i, j) place is valid or prev and matrix[i,j] are adjacent, then return 0 if longestPath[i, j] is already filled, then return longestPath[i, j] len := 0 for all its nearest 8 rooms k, do len := maximum of len and (1 + findLongestLen(i, x[k], j +y[k], matrix[i, j])) done longestPath[i, j] := len return len End
getLen(start)
Input − Start point.
Output − Maximum length.
Begin for all row r of matrix, do for all column c, of matrix, do if matrix[i, j] = start, then for all adjacent room k, do len := maximum of len and (1 + findLongestLen(i, x[k], j +y[k], matrix[i, j]))) done done done return len End
Example
#include<iostream>
#define ROW 3
#define COL 3
using namespace std;
// tool matrices to recur for adjacent cells.
int x[] = {0, 1, 1, -1, 1, 0, -1, -1};
int y[] = {1, 0, 1, 1, -1, -1, 0, -1};
int longestPath[ROW][COL];
char mat[ROW][COL] = {
{'a','c','d'},
{'h','b','a'},
{'i','g','f'}
};
int max(int a, int b) {
return (a>b)?a:b;
}
bool isvalid(int i, int j) {
if (i < 0 || j < 0 || i >= ROW || j >= COL) //when i and j are in range
return false;
return true;
}
bool isadjacent(char previous, char current) {
return ((current - previous) == 1); //check current and previous are adjacent or not
}
int findLongestLen(int i, int j, char prev) {
if (!isvalid(i, j) || !isadjacent(prev, mat[i][j])) //when already included or not adjacent
return 0;
if (longestPath[i][j] != -1)
return longestPath[i][j]; //subproblems are solved already
int len = 0; // Initialize result to 0
for (int k=0; k<8; k++) //find length of the largest path recursively
len = max(len, 1 + findLongestLen(i + x[k], j + y[k], mat[i][j]));
return longestPath[i][j] = len; // save the length and return
}
int getLen(char start) {
for(int i = 0; i<ROW; i++)
for(int j = 0; j<COL; j++)
longestPath[i][j] = -1; //set all elements to -1
int len = 0;
for (int i=0; i<ROW; i++) {
for (int j=0; j<COL; j++) { // check for all possible starting point
if (mat[i][j] == start) {
for (int k=0; k<8; k++) //for all eight adjacent cells
len = max(len, 1 + findLongestLen(i + x[k], j + y[k], start));
}
}
}
return len;
}
int main() {
char start;
cout << "Enter Starting Point (a-i): "; cin >> start;
cout << "Maximum consecutive path: " << getLen(start);
return 0;
}Output
Enter Starting Point (a-i): e Maximum consecutive path: 5
Updated on: 16-Jun-2020
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