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Solve the Crossword Puzzle
  • Difficulty Level : Hard
  • Last Updated : 26 May, 2018

A 10 x 10 Crossword grid is provided, along with a set of words (or names of places) which need to be filled into the grid. The cells in the grid are initially, either + signs or signs. Cells marked with a ‘+’ have to be left as they are. Cells marked with a ‘-‘ need to be filled up with an appropriate character.
You are also given an array of words that need to be filled in Crossword grid.

Example :

Input :
+++++++++-
-++++++++-
-------++-
-++++++++-
-++++++++-
-++++-----
------+++-
-++++++++-
+---------
++++++++++

Output :
+++++++++C
P++++++++H
HISTORY++E
Y++++++++M
S++++++++I
I++++MATHS
CIVICS+++T
S++++++++R
+GEOGRAPHY
++++++++++

The approach behind this is to recursively check for each word in the vertical position and in the horizontal position. Then fill the word in the matrix that can be the best fit in the corresponding position of the grid, then update the crossword grid by filling the gap with that word.




// CPP code to fill the crossword puzzle
#include <bits/stdc++.h>
using namespace std;
  
// ways are to calculate the number of
// possible ways to fill the grid
int ways = 0;
  
// this function is used to print
// the resultant matrix
void printMatrix(vector<string>& matrix, int n)
{
    for (int i = 0; i < n; i++)
        cout << matrix[i] << endl;
}
  
// this function checks for the current word
// if it can be placed horizontally or not
// x -> it represent index of row
// y -> it represent index of column
// currentWord -> it represent the
// current word in word array
vector<string> checkHorizontal(int x, int y,
                               vector<string> matrix,
                               string currentWord)
{
    int n = currentWord.length();
  
    for (int i = 0; i < n; i++) {
        if (matrix[x][y + i] == '#' || 
            matrix[x][y + i] == currentWord[i]) {
            matrix[x][y + i] = currentWord[i];
        }
        else {
  
            // this shows that word cannot 
            // be placed horizontally
            matrix[0][0] = '@';
            return matrix;
        }
    }
  
    return matrix;
}
  
// this function checks for the current word
// if it can be placed vertically or not
// x -> it represent index of row
// y -> it represent index of column
// currentWord -> it represent the
// current word in word array
vector<string> checkVertical(int x, int y,
                             vector<string> matrix,
                             string currentWord)
{
    int n = currentWord.length();
  
    for (int i = 0; i < n; i++) {
        if (matrix[x + i][y] == '#' || 
            matrix[x + i][y] == currentWord[i]) {
            matrix[x + i][y] = currentWord[i];
        }
        else {
  
            // this shows that word
            // cannot be placed vertically
            matrix[0][0] = '@';
            return matrix;
        }
    }
    return matrix;
}
  
// this function recursively checks for every
// word that can align vertically in one loop
// and in another loop it checks for those words
// that can align horizontally words -> it
// contains all the words to fill in a crossword
// puzzle matrix -> it contain the current
// state of crossword index -> it represent
// the index of current word n -> it represent
// the length of row or column of the square matrix
void solvePuzzle(vector<string>& words,
                 vector<string> matrix,
                 int index, int n)
{
    if (index < words.size()) {
        string currentWord = words[index];
        int maxLen = n - currentWord.length();
  
        // loop to check the words that can align vertically.
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= maxLen; j++) {
                vector<string> temp = checkVertical(j, i,
                                        matrix, currentWord);
  
                if (temp[0][0] != '@') {
                    solvePuzzle(words, temp, index + 1, n);
                }
            }
        }
  
        // loop to check the words that can align horizontally.
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= maxLen; j++) {
                vector<string> temp = checkHorizontal(i, j,
                                      matrix, currentWord);
  
                if (temp[0][0] != '@') {
                    solvePuzzle(words, temp, index + 1, n);
                }
            }
        }
    }
    else {
        // calling of print function to
        // print the crossword puzzle
        cout << (ways + 1) << " way to solve the puzzle "
             << endl;
        printMatrix(matrix, n);
        cout << endl;
  
        // increase the ways
        ways++;
        return;
    }
}
  
// Driver Code
int main()
{
    // length of grid
    int n1 = 10;
  
    // matrix to hold the grid of puzzle
    vector<string> matrix;
  
    // take input of puzzle in matrix
    // input of grid of size n1 x n1
    matrix.push_back("*#********");
    matrix.push_back("*#********");
    matrix.push_back("*#****#***");
    matrix.push_back("*##***##**");
    matrix.push_back("*#****#***");
    matrix.push_back("*#****#***");
    matrix.push_back("*#****#***");
    matrix.push_back("*#*######*");
    matrix.push_back("*#********");
    matrix.push_back("***#######");
  
    vector<string> words;
  
    // the words matrix will hold all
    // the words need to be filled in the grid
    words.push_back("PUNJAB");
    words.push_back("JHARKHAND");
    words.push_back("MIZORAM");
    words.push_back("MUMBAI");
  
    // initialize the number of ways
    // to solve the puzzle to zero
    ways = 0;
  
    // recursive function to solve the puzzle
    // Here 0 is the initial index of words array
    // n1 is length of grid
    solvePuzzle(words, matrix, 0, n1);
    cout << "Number of ways to fill the grid is "
         << ways << endl;
  
    return 0;
}
Output:
1 way to solve the puzzle 
*J********
*H********
*A****P***
*R#***U#**
*K****N***
*H****J***
*A****A***
*N*MUMBAI*
*D********
***MIZORAM

Number of ways to fill the grid is 1



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