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
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++
// C++ implementation of the approach#include <bits/stdc++.h>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 codeint main(){ int p[] = { 2, 5, 1, 3, 4 }; int n = sizeof(p) / sizeof(p[0]); cout << countElements(p, n); return 0;} |
Java
// Java implementation of the approachclass 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 codepublic 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
# 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#
// C# implementation of the approachusing 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 codepublic 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 |
Javascript
<script>// JavaScript 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] function countElements(p , n) { // To store the required answer var ans = 0; // Traverse from the second element // to the second last element for (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 var p = [ 2, 5, 1, 3, 4 ]; var n = p.length; document.write(countElements(p, n));// This code contributed by Rajput-Ji </script> |
Output:
1
Time Complexity: O(n)
Auxiliary Space: O(1)
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