CSES Solutions Number Spiral (original) (raw)

Last Updated : 27 Mar, 2024

A number spiral is an infinite grid whose upper-left square has the number 1. The task is to find the number in row **Y and column **X. Here are the first five layers of the spiral:

NumberSpiral

**Examples:

**Input: Y = 2, X = 3
**Output: 8
**Explanation: The 2nd row, 3rd column contains 8.

**Input: Y = 4, X = 2
**Output: 15
**Explanation: The 4th row, 2nd column contains 15.

**Approach: To solve the problem, follow the below idea:

It can be observed that grid consists of many squares and the values at boundaries of the square is either in increasing or decreasing order. The answer lies at the boundary of the square whose side is the maximum of Y or X. So now to get the value at Yth row and Xth column we can compute the area of inner square whose side is just one less than the side of square whose boundary contains the answer. The remaining value can be added to the area of inner square by checking the parity of Minimum of Y or X. It can be observed that for even row in the grid, the numbers are in decreasing order, and for the odd row the numbers are in increasing order in the anticlockwise direction.

NumberSpiral2

**Step-by-step algorithm:

Compare Y and X:
If **Y>X:
- Compute the area of inner square which (Y-1)*(Y-1). This represents the numbers which are not on the boundary of the square which contains the answer.
- Now to add the remaining numbers which are on the boundary, check the parity of Yth row, if the Yth row is odd then simply add X to the area of inner square or else add (2*Y - X) to the area.
Else:
- Similarly, compute the area of inner square which (X-1)*(X-1). This represents the numbers which are not on the boundary of the square which contains the answer.
- Now to add the remaining numbers which are on the boundary, check the parity of Xth column, if the Xth column is even then simply add Y to the area of inner square or else add (2*X - Y) to the area. This will give the final answer.

Below is the implementation of above approach:

C++ `

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

// Function to compute the value at position (Y, X) in the // number spiral void NumberSpiral(ll Y, ll X) { // If Y is greater than X, implying Yth row is the outer // boundary if (Y > X) { // Compute the area of the inner square ll ans = (Y - 1) * (Y - 1); ll add = 0;

    // Check parity of Y to determine if numbers are in
    // increasing or decreasing order
    if (Y % 2 != 0) {
        // Add X to the area if Yth row is odd
        add = X;
    }
    else {
        // Add 2*Y - X to the area if Yth row is even
        add = 2 * Y - X;
    }
    // Print the final result
    cout << ans + add << "\n";
}
// If X is greater than or equal to Y, implying Xth
// column is the outer boundary
else {

    // Compute the area of the inner square
    ll ans = (X - 1) * (X - 1);
    ll add = 0;

    // Check parity of X to determine if numbers are in
    // increasing or decreasing order
    if (X % 2 == 0) {
        // Add Y to the area if Xth column is even
        add = Y;
    }
    else {
        // Add 2*X - Y to the area if Xth column is odd
        add = 2 * X - Y;
    }
    // Print the final result
    cout << ans + add << "\n";
}

}

// Driver Code int main() { ll Y = 2, X = 3; NumberSpiral(Y, X); }

Java

import java.util.*;

public class NumberSpiral { // Function to compute the value at position (Y, X) in the number spiral static void numberSpiral(long Y, long X) { // If Y is greater than X, implying Yth row is the outer boundary if (Y > X) { // Compute the area of the inner square long ans = (Y - 1) * (Y - 1); long add;

        // Check parity of Y to determine if numbers are in increasing or decreasing order
        if (Y % 2 != 0) {
            // Add X to the area if Yth row is odd
            add = X;
        } else {
            // Add 2*Y - X to the area if Yth row is even
            add = 2 * Y - X;
        }

        // Print the final result
        System.out.println(ans + add);
    }
    // If X is greater than or equal to Y, implying Xth column is the outer boundary
    else {
        // Compute the area of the inner square
        long ans = (X - 1) * (X - 1);
        long add;

        // Check parity of X to determine if numbers are in increasing or decreasing order
        if (X % 2 == 0) {
            // Add Y to the area if Xth column is even
            add = Y;
        } else {
            // Add 2*X - Y to the area if Xth column is odd
            add = 2 * X - Y;
        }

        // Print the final result
        System.out.println(ans + add);
    }
}

// Driver Code
public static void main(String[] args) {
    long Y = 2, X = 3;
    numberSpiral(Y, X);
}

}

// This code is contributed by akshitaguprzj3

Python

def number_spiral(Y, X): # If Y is greater than X, implying Yth row is the outer boundary if Y > X: # Compute the area of the inner square ans = (Y - 1) * (Y - 1) # Check parity of Y to determine if numbers are in increasing or decreasing order if Y % 2 != 0: # Add X to the area if Yth row is odd add = X else: # Add 2Y - X to the area if Yth row is even add = 2 * Y - X # Print the final result print(ans + add) # If X is greater than or equal to Y, implying Xth column is the outer boundary else: # Compute the area of the inner square ans = (X - 1) * (X - 1) # Check parity of X to determine if numbers are in increasing or decreasing order if X % 2 == 0: # Add Y to the area if Xth column is even add = Y else: # Add 2X - Y to the area if Xth column is odd add = 2 * X - Y # Print the final result print(ans + add)

Driver Code

Y = 2 X = 3 number_spiral(Y, X)

C#

using System;

class GFG { // Function to compute the value at position (Y, X) in the // number spiral static void NumberSpiral(long Y, long X) { // If Y is greater than X, implying Yth row is the outer // boundary if (Y > X) { // Compute the area of the inner square long ans = (Y - 1) * (Y - 1); long add = 0;

        // Check parity of Y to determine if numbers are in
        // increasing or decreasing order
        if (Y % 2 != 0)
        {
            // Add X to the area if Yth row is odd
            add = X;
        }
        else
        {
            // Add 2*Y - X to the area if Yth row is even
            add = 2 * Y - X;
        }

        // Print the final result
        Console.WriteLine(ans + add);
    }
    // If X is greater than or equal to Y, implying Xth
    // column is the outer boundary
    else
    {
        // Compute the area of the inner square
        long ans = (X - 1) * (X - 1);
        long add = 0;

        // Check parity of X to determine if numbers are in
        // increasing or decreasing order
        if (X % 2 == 0)
        {
            // Add Y to the area if Xth column is even
            add = Y;
        }
        else
        {
            // Add 2*X - Y to the area if Xth column is odd
            add = 2 * X - Y;
        }

        // Print the final result
        Console.WriteLine(ans + add);
    }
}

// Driver Code
static void Main()
{
    long Y = 2, X = 3;
    NumberSpiral(Y, X);
}

}

JavaScript

// Function to compute the value at position (Y, X) in the number spiral function numberSpiral(Y, X) { // If Y is greater than X, implying Yth row is the outer boundary if (Y > X) { // Compute the area of the inner square let ans = (Y - 1) * (Y - 1); let add = 0;

    // Check parity of Y to determine if numbers are in increasing or decreasing order
    if (Y % 2 !== 0) {
        // Add X to the area if Yth row is odd
        add = X;
    } else {
        // Add 2*Y - X to the area if Yth row is even
        add = 2 * Y - X;
    }

    // Print the final result
    console.log(ans + add);
}
// If X is greater than or equal to Y, implying Xth column is the outer boundary
else {
    // Compute the area of the inner square
    let ans = (X - 1) * (X - 1);
    let add = 0;

    // Check parity of X to determine if numbers are in increasing or decreasing order
    if (X % 2 === 0) {
        // Add Y to the area if Xth column is even
        add = Y;
    } else {
        // Add 2*X - Y to the area if Xth column is odd
        add = 2 * X - Y;
    }

    // Print the final result
    console.log(ans + add);
}

}

// Driver Code let Y = 2, X = 3; numberSpiral(Y, X);

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