Maximum number of 2×2 squares that can be fit inside a right isosceles triangle in C

We are given a right isosceles triangle where we need to find the maximum number of 2×2 squares that can fit inside it. An isosceles triangle has two sides of equal length, and in a right isosceles triangle, the base and height are perpendicular and equal to each other.

The key insight is that from the corner triangular areas, we need to subtract 2 units from the base (and height) before calculating the number of squares. The remaining area can be divided into layers of squares.

Syntax

int numofSquares(int base);
// Returns maximum number of 2x2 squares that fit in right isosceles triangle

Algorithm

The approach follows these steps −

• Subtract 2 from the base to account for corner triangular areas
• Divide the remaining base by 2 (since each square has side length 2)
• Use the formula n×(n+1)/2 to count total squares in triangular arrangement

Example

#include <stdio.h>
#include <math.h>

int numofSquares(int b) {
    // Remove the extra triangular parts from corners
    b = b - 2;
    
    // Since each square has side length 2
    b = floor(b / 2);
    
    // Use triangular number formula: n*(n+1)/2
    return b * (b + 1) / 2;
}

int main() {
    int base = 8;
    printf("Base of triangle: %d<br>", base);
    printf("Maximum squares that can fit: %d<br>", numofSquares(base));
    
    // Test with another example
    base = 12;
    printf("\nBase of triangle: %d<br>", base);
    printf("Maximum squares that can fit: %d<br>", numofSquares(base));
    
    return 0;
}

Base of triangle: 8
Maximum squares that can fit: 6

Base of triangle: 12
Maximum squares that can fit: 15

How It Works

For base = 12 −

• After removing corner areas: 12 - 2 = 10
• Number of square units along base: 10 / 2 = 5
• Using triangular arrangement formula: 5 × 6 / 2 = 15 squares

Key Points

• Always subtract 2 from the base before calculation
• Each 2×2 square requires 2 units of space along the base
• Squares are arranged in triangular layers within the triangle
• Time complexity: O(1), Space complexity: O(1)

Conclusion

The maximum number of 2×2 squares in a right isosceles triangle is calculated using the formula (b-2)/2 × ((b-2)/2 + 1) / 2, where b is the base length. This efficiently counts squares arranged in triangular layers.