# Count of array elements which is smaller than both its adjacent elements

Given an array **arr[]** of size **N**, the task is to find the number of valley points in the array.

Valley Point:Any elements of the array is known as a valley point if it is smaller than both its adjacent elements, i.e. .

**Examples:**

Input:arr[] = {3, 2, 5}

Output:1

Explanation:

There is only one valley point. That is arr[1].

Input:arr[] = {5, 4, 8, 3, 6}

Output:2

Explanation:

There are two valley points. That is arr[1] and arr[3].

**Approach:** The idea is to iterate over the array from 1 to and for each element check for that element is a valley point or not if yes, then increment the count of the valley point by 1.

if (arr[i-1] > arr[i] < arr[i]) count += 1;

Below is the implementation of the above approach:

## Java

`// Java program to count the number ` `// of valley points in the array ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to count the valley points ` ` ` `// in the given character array ` ` ` `static` `int` `countValleys(` `int` `n, ` `int` `arr[]) ` ` ` `{ ` ` ` `int` `count = ` `0` `, temp = ` `0` `; ` ` ` ` ` `// Loop to iterate over the elements ` ` ` `// of the given array ` ` ` `for` `(` `int` `i = ` `1` `; i < n - ` `1` `; i++) { ` ` ` ` ` `// Condition to check if the given ` ` ` `// element is a valley point ` ` ` `if` `(arr[i - ` `1` `] > arr[i] ` ` ` `&& arr[i] < arr[i + ` `1` `]) { ` ` ` `count++; ` ` ` `} ` ` ` `} ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `arr[] = { ` `3` `, ` `2` `, ` `5` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println( ` ` ` `countValleys(n, arr)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1

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