Algorithm

1. Read the values: Accept a, b and c.
2. Start with the first value: Set maximum = a.
3. Compare the second value: If b > maximum, set maximum = b.
4. Compare the third value: If c > maximum, set maximum = c.
5. Display the result: The value in maximum is the largest of the three.

Explanation

After the first assignment, maximum contains a. After comparing b, it contains the larger of a and b. After comparing c, it contains the largest of all three. A replacement is needed only when the new value is larger. Equal values do not cause a problem because the question asks for the maximum value.

Example

For a = −4, b = 9 and c = 9, maximum first contains −4. The comparison with b changes it to 9. Since c is not greater than 9, the last comparison leaves it unchanged. The output is 9.

Special cases

The method also works when all values are negative or all are equal. Do not start maximum at zero unless the input rules justify that choice. For inputs −9, −4 and −12, zero would be an incorrect answer. Starting from the first actual value avoids that mistake.

Time and space complexity

The method uses exactly two comparisons, regardless of the input values. Therefore its time complexity is O(1). It uses a fixed number of variables, so its extra space is also O(1). Finding the maximum of n values is a different problem that requires a repeated scan.

Reference tables

Diagram or flowchart

C program and output

/* CodeBhavya | PPS AR25 | Learning example */
#include <stdio.h>
int main(void)
{
    int a = -4, b = 9, c = 9;
    int maximum = a;
    if (b > maximum) { maximum = b; }
    if (c > maximum) { maximum = c; }
    printf("Maximum = %d\n", maximum);
    return 0;
}

Maximum = 9

Key points

• Keep the maximum seen so far.
• A strict comparison still works when the maximum occurs twice.