# Count of elements which are second smallest among three consecutive elements

Given a permutation P of first N natural numbers. The task is to find the number of elements Pi such that Pi is second smallest among Pi – 1, Pi and Pi + 1.

Examples:

Input: P[] = {2, 5, 1, 3, 4}
Output: 1
3 is the only such element.

Input: P[] = {1, 2, 3, 4}
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Traverse the permutation from 1 to N – 2 ( zero-based indexing) and check the below two conditions. If anyone of these conditions satisfy then increment the required answer.

• If P[i – 1] < P[i] < P[i + 1].
• If P[i – 1] > P[i] > P[i + 1].

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of elements ` `// P[i] such that P[i] is the second smallest ` `// among P[i – 1], P[i] and P[i + 1] ` `int` `countElements(``int` `p[], ``int` `n) ` `{ ` `    ``// To store the required answer ` `    ``int` `ans = 0; ` ` `  `    ``// Traverse from the second element ` `    ``// to the second last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` `        ``if` `(p[i - 1] > p[i] and p[i] > p[i + 1]) ` `            ``ans++; ` `        ``else` `if` `(p[i - 1] < p[i] and p[i] < p[i + 1]) ` `            ``ans++; ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `p[] = { 2, 5, 1, 3, 4 }; ` `    ``int` `n = ``sizeof``(p) / ``sizeof``(p); ` ` `  `    ``cout << countElements(p, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the count of elements ` `// P[i] such that P[i] is the second smallest ` `// among P[i-1], P[i] and P[i + 1] ` `static` `int` `countElements(``int` `p[], ``int` `n) ` `{ ` `    ``// To store the required answer ` `    ``int` `ans = ``0``; ` ` `  `    ``// Traverse from the second element ` `    ``// to the second last element ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++)  ` `    ``{ ` `        ``if` `(p[i - ``1``] > p[i] && p[i] > p[i + ``1``]) ` `            ``ans++; ` `        ``else` `if` `(p[i - ``1``] < p[i] && p[i] < p[i + ``1``]) ` `            ``ans++; ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args)  ` `{ ` `    ``int` `p[] = { ``2``, ``5``, ``1``, ``3``, ``4` `}; ` `    ``int` `n = p.length; ` ` `  `    ``System.out.println(countElements(p, n)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of elements  ` `# P[i] such that P[i] is the second smallest  ` `# among P[i – 1], P[i] and P[i + 1]  ` `def` `countElements(p, n) :  ` ` `  `    ``# To store the required answer  ` `    ``ans ``=` `0``;  ` ` `  `    ``# Traverse from the second element  ` `    ``# to the second last element  ` `    ``for` `i ``in` `range``(``1``, n ``-` `1``) : ` `         `  `        ``if` `(p[i ``-` `1``] > p[i] ``and` `p[i] > p[i ``+` `1``]) : ` `            ``ans ``+``=` `1``;  ` `        ``elif` `(p[i ``-` `1``] < p[i] ``and` `p[i] < p[i ``+` `1``]) : ` `            ``ans ``+``=` `1``;  ` `     `  `    ``# Return the required answer  ` `    ``return` `ans;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``p ``=` `[ ``2``, ``5``, ``1``, ``3``, ``4` `];  ` `    ``n ``=` `len``(p);  ` ` `  `    ``print``(countElements(p, n));  ` ` `  `# This code is contributed by AnkitRai01 `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the count of elements ` `// P[i] such that P[i] is the second smallest ` `// among P[i-1], P[i] and P[i + 1] ` `static` `int` `countElements(``int` `[]p, ``int` `n) ` `{ ` `    ``// To store the required answer ` `    ``int` `ans = 0; ` ` `  `    ``// Traverse from the second element ` `    ``// to the second last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++)  ` `    ``{ ` `        ``if` `(p[i - 1] > p[i] && p[i] > p[i + 1]) ` `            ``ans++; ` `        ``else` `if` `(p[i - 1] < p[i] && p[i] < p[i + 1]) ` `            ``ans++; ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `[]p = { 2, 5, 1, 3, 4 }; ` `    ``int` `n = p.Length; ` ` `  `    ``Console.WriteLine(countElements(p, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:
```1
```

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