Find Maximum Number without using Conditional Statement or Ternary Operator

Given two numbers, calculate maximum number without using conditional statement or ternary operator.

Approach 1:

We can use (a > b)*b + (b > a)*a expression to find maximum number.

How this works?

Case 1: When a is greater

(a > b)*a + (b > a)*b = 1*a + 0*b = a

Case 2: When b is greater

(a > b)*a + (b > a)*b = 0*a + 1*b = b

C

#include <stdio.h>

int maximum(int a, int b)
{
	int max = (a > b) * a + (b > a) * b;

	return max;
}

// main function
int main()
{
	printf("Maximum number is %d", maximum(9, 4));

	return 0;
}

#include <stdio.h>

int maximum(int a, int b)

{

int max = (a > b) * a + (b > a) * b;

return max;

}

// main function

int main()

{

printf("Maximum number is %d", maximum(9, 4));

return 0;

}

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Approach 2: (Using short-circuiting in Boolean expressions)

We can take advantage of short-circuiting. In Boolean operation such as AND, y is evaluated only if x is true for x && y. y is not evaluated if x is false because the whole expression would be false which can be derived without evaluating y.

The idea is to apply this principle to the below code. Initially max is a. Now if (max < b) is true, then that means, b is greater than a so the second sub-expression (max = b) is evaluated and max is set to b. If however (max < b) is false, then second sub-expression will not be evaluated and max will remain equal to a.

C

#include <stdio.h>

int maximum(int a, int b)
{
	// initialize max with a
	int max = a;

	// set max to b if and only if max is less than b
	(max < b) && (max = b);

	return max;
}

// main function
int main()
{
	printf("Maximum number is %d", maximum(9, 4));

	return 0;
}

#include <stdio.h>

int maximum(int a, int b)

{

// initialize max with a

int max = a;

// set max to b if and only if max is less than b

(max < b) && (max = b);

return max;

}

// main function

int main()

{

printf("Maximum number is %d", maximum(9, 4));

return 0;

}

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Approach 3: (Using array index)

We can also use array indices to find maximum number as shown below -

C

#include <stdio.h>

int maximum(int a, int b)
{
	// arr will contain first two elements
	int arr[] = { a, b };

	// return the maximum element
	return arr[a < b];
}

// main function
int main()
{
	printf("Maximum number is %d", maximum(9, 4));

	return 0;
}

#include <stdio.h>

int maximum(int a, int b)

{

// arr will contain first two elements

int arr[] = { a, b };

// return the maximum element

return arr[a < b];

}

// main function

int main()

{

printf("Maximum number is %d", maximum(9, 4));

return 0;

}

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Approach 4: (Using repeated subtraction)

We can find maximum of two integers by doing repeated subtraction till both numbers becomes negative. The number of times we do subtraction will be the maximum number. This solution won't work on negative numbers.

C

#include <stdio.h>

int maximum(int a, int b)
{
	int max = 0;
	while (a > 0 || b > 0)
	{
		--a;
		--b;
		++max;
	}
	return max;
}

// main function
int main()
{
	printf("Maximum number is %d", maximum(9, 4));

	return 0;
}

#include <stdio.h>

int maximum(int a, int b)

{

int max = 0;

while (a > 0 || b > 0)

{

--a;

--b;

++max;

}

return max;

}

// main function

int main()

{

printf("Maximum number is %d", maximum(9, 4));

return 0;

}

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Below is the recursive version of above solution but it uses conditional statement. Like iterative version, it won't work on negative numbers.

C

int maximum (int a, int b)
{
	if (a < 0 && b < 0)
		return -1;

	return 1 + maximum(--a, --b);
}

int maximum (int a, int b)

{

if (a < 0 && b < 0)

return -1;

return 1 + maximum(--a, --b);

}

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Also See: Find minimum number without using conditional statement or ternary operator

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Approach 1:

C

Approach 2: (Using short-circuiting in Boolean expressions)

C

Approach 3: (Using array index)

C

Approach 4: (Using repeated subtraction)

C

C

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