Larger than neighbors (Editorial)

Algorithm Analysis

Let the array $m$ contain the input sequence of numbers. The numbering of the array elements starts from 0. The array contains $n$ numbers. The element $m[i]$ has two neighbors if only 1 \leq i \leq n – 1.

To count the number of elements that are greater than their two neighbors, it is necessary to iterate through all such $m[i]$ that 1 \leq i \leq n – 1, and check the fulfillment of two conditions: m[i - 1] < m[i] and m[i] > m[i + 1].

Algorithm Implementation

Declare the working array.

int m[101];

Read the number of input numbers $n$.

scanf("%d", &n);

Read the input array.

for (i = 0; i < n; i++)
    scanf("%d", &m[i]);

In the variable res, count the number of elements that are greater than their two neighbors.

res = 0;

Iterate through all such $m[i]$ that 1 \leq i \leq n – 1 and check the fulfillment of two conditions: m[i - 1] < m[i] and m[i] > m[i + 1].

for (i = 1; i < n - 1; i++)
    if (m[i] > m[i - 1] && m[i] > m[i + 1]) res++;

Output the answer.

printf("%d\n", res);