Count Rectangle Formations

Given an integer n representing the length of a stick, split it into four positive integer parts. Return the number of distinct ways to split the stick such that the four parts can form a rectangle but cannot form a square. Two splits are considered the same if they contain the same four side lengths, regardless of their order.

Examples:

Input: n = 10
Output: 2
Explanation: The valid splits are: {1, 1, 4, 4}, {2, 2, 3, 3}.

Input: n = 20
Output: 4
Explanation: The valid splits are: {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7}, {4, 4, 6, 6}.The split {5, 5, 5, 5} forms a square, so it is not counted.

[Naive Approach] Iterating Over Possible Lengths - O(n) Time and O(1) Space

The idea is to observe that a rectangle must have side lengths of the form {x, x, y, y}. Hence, the total length of the stick satisfies 2x + 2y = n, or equivalently, x + y = n / 2.

If n is odd, no valid split is possible. Otherwise, iterate over all possible values of x and compute y = n / 2 - x. Count only the pairs where x < y, since x = y forms a square and x > y represents a duplicate of an already counted split.

#include <iostream>
using namespace std;

int countRectangles(int n) {
    if (n % 2 != 0) return 0;

    int half = n / 2;
    int res = 0;

    for (int x = 1; x < half - x; x++) {
        
        // x and half - x form unequal sides.
        res++;
    }

    return res;
}

int main() {
    int n = 10;
    cout << countRectangles(n) << endl;

    n = 20;
    cout << countRectangles(n) << endl;
}

class GFG {
    static int countRectangles(int n) {
        if (n % 2 != 0) return 0;

        int half = n / 2;
        int res = 0;

        for (int x = 1; x < half - x; x++) {
            
            // x and half - x form unequal sides.
            res++;
        }

        return res;
    }

    public static void main(String[] args) {
        int n = 10;
        System.out.println(countRectangles(n));

        n = 20;
        System.out.println(countRectangles(n));
    }
}

def countRectangles(n):
    if n % 2 != 0:
        return 0

    half = n // 2
    res = 0

    for x in range(1, half):
        
        # x and half - x form unequal sides.
        if x < half - x:
            res += 1

    return res


if __name__ == "__main__":
    n = 10
    print(countRectangles(n))

    n = 20
    print(countRectangles(n))

using System;

class GFG {
    static int countRectangles(int n) {
        if (n % 2 != 0) return 0;

        int half = n / 2;
        int res = 0;

        for (int x = 1; x < half - x; x++) {
            
            // x and half - x form unequal sides.
            res++;
        }

        return res;
    }

    static void Main() {
        int n = 10;
        Console.WriteLine(countRectangles(n));

        n = 20;
        Console.WriteLine(countRectangles(n));
    }
}

function countRectangles(n) {
    if (n % 2 !== 0) return 0;

    const half = Math.floor(n / 2);
    let res = 0;

    for (let x = 1; x < half - x; x++) {
        
        // x and half - x form unequal sides.
        res++;
    }

    return res;
}

// Driver Code
let n = 10;
console.log(countRectangles(n));

n = 20;
console.log(countRectangles(n));

2
4

[Expected Approach] Using Direct Formula - O(1) Time and O(1) Space

The idea is to use the fact that a rectangle must have side lengths of the form {x, x, y, y}. Thus, the total length satisfies 2x + 2y = n, or equivalently, x + y = n / 2.

If n is odd, no valid split is possible. Otherwise, the number of positive pairs (x, y) with x < y is n / 4. If n / 2 is even, one of these pairs is (n / 4, n / 4), which forms a square. Exclude this case by subtracting 1 from the count.

#include <iostream>
using namespace std;

int countRectangles(int n) {
    
    // Rectangle cannot be formed if n is odd.
    if (n % 2 != 0) {
        return 0;
    }

    int res = n / 4;

    // Exclude the square case.
    if ((n / 2) % 2 == 0) {
        res--;
    }

    return res;
}

int main() {
    int n = 10;
    cout << countRectangles(n) << endl;

    n = 20;
    cout << countRectangles(n) << endl;
}

class GFG {
    static int countRectangles(int n) {
        
        // Rectangle cannot be formed if n is odd.
        if ((n & 1) == 1) {
            return 0;
        }

        int res = n / 4;

        // Exclude the square case.
        if (((n / 2) & 1) == 0) {
            res--;
        }

        return res;
    }

    public static void main(String[] args) {
        int n = 10;
        System.out.println(countRectangles(n));

        n = 20;
        System.out.println(countRectangles(n));
    }
}

def countRectangles(n):
    
    # Rectangle cannot be formed if n is odd.
    if n % 2 != 0:
        return 0

    res = n // 4

    # Exclude the square case.
    if (n // 2) % 2 == 0:
        res -= 1

    return res


if __name__ == "__main__":
    n = 10
    print(countRectangles(n))

    n = 20
    print(countRectangles(n))

using System;

class GFG {
    static int countRectangles(int n) {
        
        // Rectangle cannot be formed if n is odd.
        if ((n & 1) == 1) {
            return 0;
        }

        int res = n / 4;

        // Exclude the square case.
        if (((n / 2) & 1) == 0) {
            res--;
        }

        return res;
    }

    static void Main() {
        int n = 10;
        Console.WriteLine(countRectangles(n));

        n = 20;
        Console.WriteLine(countRectangles(n));
    }
}

function countRectangles(n) {
    
    // Rectangle cannot be formed if n is odd.
    if ((n & 1) === 1) {
        return 0;
    }

    let res = Math.floor(n / 4);

    // Exclude the square case.
    if (((Math.floor(n / 2)) & 1) === 0) {
        res--;
    }

    return res;
}

// Driver Code
let n = 10;
console.log(countRectangles(n));

n = 20;
console.log(countRectangles(n));

2
4