In this post we look at a few more coding questions,. In particular we look at a technique called backtracking.
Backtracking helps over brute force method.
Take the alphabet matrix as follows:
a b c e
s f c s
a d e e
There is a path for the string 'bcced' but not for path abcb. Re-entering is not possible
This technique is called backtracking because it returns back to the previous state and uses pop operations from  a stack. That is when it finds a character it pushes it on to the stack and for all the rejection where the character is not the same as the corresponding character, it backtracks and pops from the stack.
Since all entries can be entered only once, we keep a boolean flag pertaining to each entry.
bool hasPath(char* matrix, int rows, int cols, char* str)
{
// parameter validation
// bool* visited for the matrix of used flags
int pathLength = 0;
for (int row = 0; row < rows; row++){
       for (int col = 0; col < cols; col++){
              if (hasPathStack(matrix, rows, cols, row, col, str, pathLength, visited))
                   return true;
         }
}
//cleanup
return false;
}


bool hasPathStack(char*matrix, int rows, int cols, char* str, int& pathLength, bool* visited)
{
// if the boundary conditions are satisfied &&
// if the path has not been found yet &&
// if the element has not been visited already
    ++ pathLength;
    // visited set to true
    // visit one of the four neighboring cells
    bool hasPath  = hasPathStack(matrix, rows, cols, row, col - 1, str, pathLength, visited) ||
 hasPathStack(matrix, rows, cols, row - 1, col, str, pathLength, visited) ||
 hasPathStack(matrix, rows, cols row, col + 1, str, pathLength, visited) ||
 hasPathStack(matrix, rows, cols row + 1, col, str, pathLength, visited) ;
     if (!hasPathStack) {
        --pathLength; // backtracking
         // visited set to false
        }
}
[from the coding interviews book]