## Question

You are given a five-by-five array of cells, some of which contain letters and the others are empty. You need to fill the empty squares with letters so that there is a path moving only horizontally or vertically (not diagonally) at each step that visits all cells with consecutive letters. The letters are lower case ‘a’ through ‘y’, and ‘a’ is present in one of the starting cells.For example, a solution to is Another way to describe the correct output is that if you place the cursor over the‘a’, then using only the up,down,left,and right arrows you can move the cursor over the letters ‘a’ through ‘y’ in order. Complete the recursive backtracking search function given in the Java file. The input to the program consists of five lines, each containing a string of five lower case letters from ‘a’ to ‘z’ with no spaces. An ‘a’ appears exactly once in the five lines. A ‘z’ indicates an empty cell. The output is five lines, each containing five characters (without spaces), and the five lines together contain each of the letters ‘a’ through ‘y’ according to the rules given above.For example, if the input is:

zzzzm

zzzzz

zfzzz

zzzaz

zzzzu

then a correct output is

ijklm

hgpon

efqrs

dcbat

yxwvu

If no solution exists, your program should produce no output. If more than one solution exists, print out just one.

## Solution Preview

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//package edu.njit.;import java.util.Arrays;

import java.util.Scanner;

public class AtoY {

public static void printTable(char[][] t) {

for (int i = 0; i < 5; ++i) {

for (int j = 0; j < 5; ++j)

System.out.print(t[i][j]);

System.out.println();

}

}

// occupied[c] = true if character c is already on the table...

static boolean[] occupied = new boolean[255];

private static boolean solve(char[][] t, int row, int col, char c) {

// table is completely filled, return true

if (c == 'z')

return true;...

By purchasing this solution you'll be able to access the following files:

AtoY.java.