# Number of palindromic paths in a matrix

Given a matrix containing lower alphabetical characters only, we need to count number of palindromic paths in given matrix. A path is defined as a sequence of cells starting from top-left cell and ending at bottom-right cell. We are allowed to move to right and down only from current cell.
Examples:

Input : mat[][] = {"aaab”,
"baaa”
“abba”}
Output : 3

Number of palindromic paths are 3 from top-left to
bottom-right.
aaaaaa (0, 0) -> (0, 1) -> (1, 1) -> (1, 2) ->
(1, 3) -> (2, 3)
aaaaaa (0, 0) -> (0, 1) -> (0, 2) -> (1, 2) ->
(1, 3) -> (2, 3)
abaaba (0, 0) -> (1, 0) -> (1, 1) -> (1, 2) ->
(2, 2) -> (2, 3)

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

We can solve this problem recursively, we start from two corners of a palindromic path(top-left and bottom right). In each recursive call, we maintain a state which will constitute two cells one from starting and one from end which should be equal for palindrome property. If at a state, both cell characters are equal then we call recursively with all possible movements in both directions.
As this can lead to solving same subproblem multiple times, we have taken a map memo in below code which stores the calculated result with key as indices of starting and ending cell so if subproblem with same starting and ending cell is called again, result will be returned by memo directly instead of recalculating again.
Please see below code for better understanding,

 // C++ program to get number of palindrome // paths in matrix #include    using namespace std;    #define R 3 #define C 4    // struct to represent state of recursion // and key of map struct cells {     //  indices of front cell     int rs, cs;        //  indices of end cell     int re, ce;     cells(int rs, int cs, int re, int ce) :         rs(rs), cs(cs), re(re), ce(ce) { }        // operator overloading to compare two     // cells which rs needed for map     bool operator <(const cells& other) const     {         return ((rs != other.rs) || (cs != other.cs) ||                (re != other.re) || (ce != other.ce));     } };    // recursive method to return number of palindromic // paths in matrix // (rs, cs) ==> Indices of current cell from a starting //              point (First Row) // (re, ce) ==> Indices of current cell from a ending //              point (Last Row) // memo     ==> To store results of already computed //              problems int getPalindromicPathsRecur(char mat[R][C], int rs,           int cs, int re, int ce, map& memo) {     // Base Case 1 : if any index rs out of boundary,     // return 0     if (rs < 0 || rs >= R || cs < 0 || cs >= C)         return 0;     if (re < 0 || re < rs || ce < 0 || ce < cs)         return 0;        // Base case 2 : if values are not equal     // then palindrome property rs not satisfied,     // so return 0     if (mat[rs][cs] != mat[re][ce])         return 0;        // If we reach here, then matrix cells are same.        // Base Case 3 : if indices are adjacent then     // return 1     if (abs((rs - re) + (cs - ce)) <= 1)         return 1;        //  if result rs precalculated, return from map     if (memo.find(cells(rs, cs, re, ce)) != memo.end())         return memo[cells(rs, cs, re, ce)];        int ret = 0; // Initialize result        // calling recursively for all possible movements     ret += getPalindromicPathsRecur(mat, rs + 1, cs,                                     re - 1, ce, memo);     ret += getPalindromicPathsRecur(mat, rs + 1, cs, re,                                          ce - 1, memo);     ret += getPalindromicPathsRecur(mat, rs, cs + 1,                                      re - 1, ce, memo);     ret += getPalindromicPathsRecur(mat, rs, cs + 1, re,                                           ce - 1, memo);        // storing the calculated result in map     memo[cells(rs, cs, re, ce)] = ret;        return ret; }    //  method returns number of palindromic paths in matrix int getPalindromicPaths(char mat[R][C]) {     map memo;     return getPalindromicPathsRecur(mat, 0, 0, R - 1,                                           C - 1, memo); }    //  Driver code to test above methods int main() {     char mat[R][C] =     {         'a', 'a', 'a', 'b',         'b', 'a', 'a', 'a',         'a', 'b', 'b', 'a'     };     printf("%d", getPalindromicPaths(mat));        return 0; }

Output:

3

Time Complexity : O(R x C)

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