Check if given string satisfies the following conditions

Given a string S of length L, where L is a perfect square, the task is to check if the given string satisfies the following conditions:

• Insert the characters of the string into a square matrix A[][] of dimensions √L x √L in row-wise manner.
• Initialize another matrix M[][] with 0s. Fill the left diagonal with 1. Now, for 1 present in the left diagonal, fill its corresponding right diagonal with 1s.

• Now, check if all the indices in matrix M[][] which contains 1, contains the same character in A[][].

If the condition is satisfied, print “Yes”. Otherwise, print “No”.

Examples:

Input: S = ”abacdaeaafaghaia”
Output: Yes
Explanation:

Input: S = ”abacdaeabfaghaia” 
Output: No 

Approach: The idea is to traverse the matrix A[][] where its corresponding character in the matrix M[][] is 1. Follow the steps below to solve the problem: 

• Calculate the dimensions of the matrix as N = √L.
• Iterate over the left diagonal by visiting each cell, A[i][i] where 1<= i<= N.
• For every element of the left diagonal at cell A[i][i], initialize variables x and y with i and traverse its corresponding right diagonal by visiting the character S[x*N + y] and S[y*N + x] and decrement x each time by 1 and increment y each time by 1 to move along the next cells in the right diagonals, while x is not smaller than 0 and y is smaller than N.
• If all characters are found to be same in the above step, print “Yes”. Otherwise, print “No” if any mismatch is found.

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 given string
// satisfies the given conditions
void isValid(string s)
{

    // Dimensions
    int n = sqrt(s.length());

    char check = s[0];

    // Left diagonal
    for (int i = 0; i < n; i++) {
        int x = i, y = i;

        // Right diagonal
        while (x >= 0 && y < n) {
            if (s[(n * x) + y] != check
                || s[(n * y) + x] != check) {

                // Conditions not satisfied
                cout << "No" << endl;
                return;
            }
            x--;
            y++;
        }
    }

    // Print Yes
    cout << "Yes" << endl;
}

// Driver Code
int main()
{

    // Given String
    string str = "abacdaeaafaghaia";

    // Function call
    isValid(str);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if given string
// satisfies the given conditions
static void isValid(String s)
{

    // Dimensions
    int n = (int)Math.sqrt(s.length());

    char check = s.charAt(0);

    // Left diagonal
    for(int i = 0; i < n; i++)
    {
        int x = i, y = i;

        // Right diagonal
        while (x >= 0 && y < n)
        {
            if (s.charAt((n * x) + y) != check ||
                s.charAt((n * y) + x) != check)
            {

                // Conditions not satisfied
                System.out.print("No");
                return;
            }
            x--;
            y++;
        }
    }

    // Print Yes
    System.out.print("Yes");
}

// Driver Code
public static void main(String[] args)
{

    // Given String
    String str = "abacdaeaafaghaia";

    // Function call
    isValid(str);
}
}

// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
import math

# Function to check if given string
# satisfies the given conditions
def isValid(s):

    # Dimensions
    n = int(math.sqrt(len(s)))
    check = s[0]

    # Left diagonal
    for i in range(n):
        x = i
        y = i

        # Right diagonal
        while (x >= 0 and y < n):
            if (s[n * x + y] != check or
                s[n * x + x] != check):

                # Conditions not satisfied
                print("No")
                return

            x -= 1
            y += 1

    # Print Yes
    print("Yes")

# Driver Code

# Given String
str = "abacdaeaafaghaia"

# Function call
isValid(str)

# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if given string
// satisfies the given conditions
static void isValid(string s)
{

    // Dimensions
    int n = (int)Math.Sqrt(s.Length);

    char check = s[0];

    // Left diagonal
    for(int i = 0; i < n; i++)
    {
        int x = i, y = i;

        // Right diagonal
        while (x >= 0 && y < n)
        {
            if (s[(n * x) + y] != check ||
                s[(n * y) + x] != check)
            {

                // Conditions not satisfied
                Console.Write("No");
                return;
            }
            x--;
            y++;
        }
    }

    // Print Yes
    Console.Write("Yes");
}

// Driver code
public static void Main()
{

    // Given String
    string str = "abacdaeaafaghaia";

    // Function call
    isValid(str);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to check if given string
// satisfies the given conditions
function isValid(s)
{

    // Dimensions
    let n = Math.sqrt(s.length);

    let check = s[0];

    // Left diagonal
    for(let i = 0; i < n; i++)
    {
        let x = i, y = i;

        // Right diagonal
        while (x >= 0 && y < n)
        {
            if (s[(n * x) + y]!= check ||
                s[(n * y) + x] != check)
            {

                // Conditions not satisfied
                document.write("No");
                return;
            }
            x--;
            y++;
        }
    }

    // Print Yes
    document.write("Yes");
}

    // Driver Code

    // Given String
    let str = "abacdaeaafaghaia";

    // Function call
    isValid(str);

// This code is contributed by souravghosh0416.
</script>

Output:

Yes

Time Complexity: O(L^1/2) where L is the length of the given string.
Auxiliary Space: O(1) since constant space has been used