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Check whether jigsaw puzzle solveable or not

  • Difficulty Level : Easy
  • Last Updated : 19 Apr, 2021
Geek Week

Given a special Jigsaw puzzle consisting of N rows and M columns all identical pieces. Every piece has three tabs and one blank. The task is to check if the puzzle is solvable by placing the pieces in such a way that the tab of one piece fits perfectly into a blank of other piece.

Note: Rotate and Translate the pieces to solve the puzzle. 

Examples:

Input: N = 2, M = 2
Output: Yes



Input: N = 1, M = 3
Output: Yes
 

Approach: The key observation in the problem is that:

  • If the Puzzle has only one row or only one column. Then it is possible to solve the puzzle by placing a blank tab on that shared side itself.
  • If the Puzzle has two rows and two columns. Then The puzzle is solvable by placing the blank Tabs in a circular chain.
  • Otherwise, It is not possible to solve the Puzzle.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the jigsaw
// Puzzle is solveable or not
void checkSolveable(int n, int m)
{
   
    // Base Case
    if (n == 1 or m == 1)
        cout << "YES";
   
    // By placing the blank tabs
    // as a chain
    else if (m == 2 and n == 2)
        cout << "YES";
    else
        cout << "NO";
}
  
// Driver Code
int main()
{
    int n = 1, m = 3;
   
    checkSolveable(n, m);
}
 
// This code is contributed by jana_sayantan

Java




// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check if the jigsaw
// Puzzle is solveable or not
static void checkSolveable(int n, int m)
{
    
    // Base Case
    if (n == 1 || m == 1)
       System.out.print("YES");
    
    // By placing the blank tabs
    // as a chain
    else if (m == 2 && n == 2)
        System.out.print("YES");
    else
        System.out.print("NO");
}
 
// Driver Code
public static void main(String[] args)
{
     int n = 1, m = 3;
    
    checkSolveable(n, m);
}
}
 
// This code is contributed by sanjoy_62

Python




# Python program for the above approach
 
# Function to check if the jigsaw
# Puzzle is solveable or not
def checkSolveable(n, m):
 
    # Base Case
    if n == 1 or m == 1:
        print("YES")
         
    # By placing the blank tabs
    # as a chain
    elif m == 2 and n == 2:
        print("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == "__main__":
    n = 1
    m = 3
    checkSolveable(n, m)

C#




// C# program for the above approach
using System;
   
class GFG{
   
// Function to check if the jigsaw
// Puzzle is solveable or not
static void checkSolveable(int n, int m)
{
     
    // Base Case
    if (n == 1 || m == 1)
       Console.WriteLine("YES");
     
    // By placing the blank tabs
    // as a chain
    else if (m == 2 && n == 2)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
  
// Driver Code
public static void Main()
{
    int n = 1, m = 3;
   
    checkSolveable(n, m);
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function to check if the jigsaw
// Puzzle is solveable or not
function checkSolveable(n, m)
{
     
    // Base Case
    if (n == 1 || m == 1)
       document.write("YES");
     
    // By placing the blank tabs
    // as a chain
    else if (m == 2 && n == 2)
        document.write("YES");
    else
        document.write("NO");
}
 
// Driver code
         
    let n = 1, m = 3;
     
    checkSolveable(n, m);
     
    // This code is contributed by code_hunt.
</script>
Output: 
YES

 

Time Complexity: O(1) 
Auxiliary Space: O(1)
 

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