QUESTION 11 · Array Basics

Why initialise minimum and maximum from the first element?

2 marks

Answer

The first element is an actual value in a non-empty list and is a valid starting candidate.

Zero would give a wrong minimum for some all-positive lists.
Zero would give a wrong maximum for some all-negative lists.

Write a C program to find minimum, maximum and average in an integer array.

5 marks

Answer in brief

The minimum and maximum can be updated while traversing an array. The average is the total divided by the number of elements.

Algorithm

Read a non-empty array.
Set minimum and maximum to a[0] and sum to a[0].
Scan indices 1 through n - 1. Update each extreme when needed and add the current value to sum.
Calculate average as (double)sum / n and display the results.

Worked trace

For 4, -2 and 7, both extremes start at 4 and sum starts at 4. Reading -2 changes minimum to -2 and sum to 2. Reading 7 changes maximum to 7 and sum to 9. Average is 9.0 / 3 = 3.0.

Important cases

A single-element list has the same minimum, maximum and average value. Repeated equal values need no special replacement. The list can contain negative values and need not be sorted. An empty list is rejected because this method needs a first element and the average must not divide by zero.

Preserving fractions

Casting sum to double before division allows a fractional average. Casting after integer division would not restore a discarded fraction. long long stores the bounded sample’s total safely, and the result is displayed with two decimal places.

Complexity

One scan examines each value once. A fixed number of comparisons and an addition are performed per element, giving O(n) time. minimum, maximum, sum and average occupy fixed extra storage, so auxiliary space is O(1), excluding the input array.

C program and output

Download .c Open compiler

/* CodeBhavya | PPS AR25 | Learning example */
#include <stdio.h>

int main(void)
{
    int n, a[20];
    if (scanf("%d", &n) != 1 || n < 1 || n > 20) {
        puts("Unsupported input"); return 1;
    }
    for (int i = 0; i < n; ++i) {
        if (scanf("%d", &a[i]) != 1 || a[i] < -10000 || a[i] > 10000) {
            puts("Unsupported input"); return 1;
        }
    }
    int minimum = a[0], maximum = a[0];
    long long sum = a[0];
    for (int i = 1; i < n; ++i) {
        if (a[i] < minimum) { minimum = a[i]; }
        if (a[i] > maximum) { maximum = a[i]; }
        sum += a[i];
    }
    printf("Minimum = %d, maximum = %d, average = %.2f\n",
           minimum, maximum, (double)sum / n);
    return 0;
}

Sample input and output

Input

3
4 -2 7

Output

Minimum = -2, maximum = 7, average = 3.00