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Find indices of all local maxima and local minima in an Array

  • Difficulty Level : Easy
  • Last Updated : 02 Sep, 2020

Given an array arr[] of integers. The task is to find the indices of all local minima and local maxima in the given array.
Examples:

Input: arr = [100, 180, 260, 310, 40, 535, 695]
Output:
Points of local minima: 0 4 

Points of local maxima: 3 6
Explanation:
Given array can be break as below sub-arrays:
1. first sub array 
[100, 180, 260, 310] 
index of local minima = 0 
index of local maxima = 3
2. second sub array 
[40, 535, 695] 
index of local minima = 4 
index of local maxima = 6

Input: arr = [23, 13, 25, 29, 33, 19, 34, 45, 65, 67]
Output:
Points of local minima: 1 5
Points of local maxima: 0 4 9 
 

 

Approach: The idea is to iterate over the given array arr[] and check if each element of the array is smallest or greatest among their adjacent element. If it is smallest then it is local minima and if it is greatest then it is local maxima. Below are the steps:



  1. Create two arrays max[] and min[] to store all the local maxima and local minima.
  2. Traverse the given array and append the index of the array into the array max[] and min[] according to the below conditions:
    • If arr[i – 1] > arr[i] < arr[i + 1] then append that index to min[].
    • If arr[i – 1] < arr[i] > arr[i + 1] then append that index to max[].
  3. Check for the local maxima and minima conditions for the first and last elements separately.
  4. Print the indexes stored in min[] and max[].

Below is the implementation of the above approach:
 

C++




// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find all the local maxima 
// and minima in the given array arr[] 
void findLocalMaximaMinima(int n, int arr[]) 
      
    // Empty vector to store points of 
    // local maxima and minima 
    vector<int> mx, mn; 
  
    // Checking whether the first point is 
    // local maxima or minima or none 
    if (arr[0] > arr[1]) 
        mx.push_back(0); 
  
    else if (arr[0] < arr[1]) 
        mn.push_back(0); 
  
    // Iterating over all points to check 
    // local maxima and local minima 
    for(int i = 1; i < n - 1; i++) 
    
          
    // Condition for local minima 
    if ((arr[i - 1] > arr[i]) and 
        (arr[i] < arr[i + 1])) 
        mn.push_back(i); 
          
    // Condition for local maxima 
    else if ((arr[i - 1] < arr[i]) and 
                (arr[i] > arr[i + 1])) 
        mx.push_back(i); 
    
  
    // Checking whether the last point is 
    // local maxima or minima or none 
    if (arr[n - 1] > arr[n - 2]) 
        mx.push_back(n - 1); 
  
    else if (arr[n - 1] < arr[n - 2]) 
        mn.push_back(n - 1); 
  
    // Print all the local maxima and 
    // local minima indexes stored 
    if (mx.size() > 0) 
    
        cout << "Points of Local maxima are : "
        for(int a : mx) 
        cout << a << " "
        cout << endl; 
    
    else
        cout << "There are no points of "
            << "Local Maxima \n"
  
    if (mn.size() > 0) 
    
        cout << "Points of Local minima are : "
        for(int a : mn) 
        cout << a << " "
        cout << endl; 
    
    else
        cout << "There are no points of "
            << "Local Minima \n"
  
// Driver Code 
int main() 
    int N = 9; 
  
    // Given array arr[] 
    int arr[] = { 10, 20, 15, 14, 13, 
                25, 5, 4, 3 }; 
  
    // Function call 
    findLocalMaximaMinima(N, arr); 
    return 0;     
  
// This code is contributed by himanshu77 

Java




// Java program for the above approach 
import java.util.*; 
class GFG{ 
      
// Function to find all the local maxima 
// and minima in the given array arr[] 
public static void findLocalMaximaMinima(int n, 
                                        int[] arr) 
      
    // Empty vector to store points of 
    // local maxima and minima 
    Vector<Integer> mx = new Vector<Integer>(); 
    Vector<Integer> mn = new Vector<Integer>(); 
  
    // Checking whether the first point is 
    // local maxima or minima or none 
    if (arr[0] > arr[1]) 
        mx.add(0); 
  
    else if (arr[0] < arr[1]) 
        mn.add(0); 
  
    // Iterating over all points to check 
    // local maxima and local minima 
    for(int i = 1; i < n - 1; i++) 
    
        // Condition for local minima 
        if ((arr[i - 1] > arr[i]) && 
            (arr[i] < arr[i + 1])) 
            mn.add(i); 
              
        // Condition for local maxima 
        else if ((arr[i - 1] < arr[i]) && 
                (arr[i] > arr[i + 1])) 
            mx.add(i); 
    
  
    // Checking whether the last point is 
    // local maxima or minima or none 
    if (arr[n - 1] > arr[n - 2]) 
        mx.add(n - 1); 
  
    else if (arr[n - 1] < arr[n - 2]) 
        mn.add(n - 1); 
  
    // Print all the local maxima and 
    // local minima indexes stored 
    if (mx.size() > 0
    
        System.out.print("Points of Local "
                        "maxima are : "); 
        for(Integer a : mx) 
            System.out.print(a + " "); 
        System.out.println(); 
    
    else
        System.out.println("There are no points "
                        "of Local Maxima "); 
  
    if (mn.size() > 0
    
        System.out.print("Points of Local "
                        "minima are : "); 
        for(Integer a : mn) 
            System.out.print(a + " "); 
        System.out.println(); 
    
    else
        System.out.println("There are no points of "
                        "Local Maxima "); 
  
// Driver code 
public static void main(String[] args) 
    int N = 9
  
    // Given array arr[] 
    int arr[] = { 10, 20, 15, 14, 13
                25, 5, 4, 3 }; 
  
    // Function call 
    findLocalMaximaMinima(N, arr); 
  
// This code is contributed by divyeshrabadiya07 

Python3




# Python3 program for the above approach 
  
# Function to find all the local maxima 
# and minima in the given array arr[] 
  
def findLocalMaximaMinima(n, arr): 
  
    # Empty lists to store points of 
    # local maxima and minima 
    mx = [] 
    mn = [] 
  
    # Checking whether the first point is 
    # local maxima or minima or neither 
    if(arr[0] > arr[1]): 
        mx.append(0
    elif(arr[0] < arr[1]): 
        mn.append(0
  
    # Iterating over all points to check 
    # local maxima and local minima 
    for i in range(1, n-1): 
  
        # Condition for local minima 
        if(arr[i-1] > arr[i] < arr[i + 1]): 
            mn.append(i) 
  
        # Condition for local maxima 
        elif(arr[i-1] < arr[i] > arr[i + 1]): 
            mx.append(i) 
  
    # Checking whether the last point is 
    # local maxima or minima or neither 
    if(arr[-1] > arr[-2]): 
        mx.append(n-1
    elif(arr[-1] < arr[-2]): 
        mn.append(n-1
  
        # Print all the local maxima and 
        # local minima indexes stored 
    if(len(mx) > 0): 
        print("Points of Local maxima"
            " are : ", end ='') 
        print(*mx) 
    else
        print("There are no points of"
            " Local maxima."
  
    if(len(mn) > 0): 
        print("Points of Local minima"
            " are : ", end ='') 
        print(*mn) 
    else
        print("There are no points"
            " of Local minima."
  
# Driver Code 
if __name__ == '__main__'
  
    N = 9
    # Given array arr[] 
    arr = [10, 20, 15, 14, 13, 25, 5, 4, 3
  
    # Function Call 
    findLocalMaximaMinima(N, arr) 

C#




// C# program for the above approach 
using System; 
using System.Collections; 
using System.Collections.Generic; 
  
class GFG{
      
// Function to find all the local maxima 
// and minima in the given array arr[] 
public static void findLocalMaximaMinima(int n,
                                         int[] arr) 
      
    // Empty vector to store points of 
    // local maxima and minima 
    ArrayList mx = new ArrayList();
    ArrayList mn = new ArrayList();
  
    // Checking whether the first point is 
    // local maxima or minima or none 
    if (arr[0] > arr[1]) 
        mx.Add(0); 
  
    else if (arr[0] < arr[1]) 
        mn.Add(0); 
  
    // Iterating over all points to check 
    // local maxima and local minima 
    for(int i = 1; i < n - 1; i++)
    {
          
        // Condition for local minima  
        if ((arr[i - 1] > arr[i]) &&  
            (arr[i] < arr[i + 1])) 
            mn.Add(i); 
              
        // Condition for local maxima 
        else if ((arr[i - 1] < arr[i]) &&  
                 (arr[i] > arr[i + 1]))  
            mx.Add(i); 
    
  
    // Checking whether the last point is 
    // local maxima or minima or none 
    if (arr[n - 1] > arr[n - 2]) 
        mx.Add(n - 1);
      
    else if (arr[n - 1] < arr[n - 2])  
        mn.Add(n - 1);  
  
    // Print all the local maxima and  
    // local minima indexes stored  
    if (mx.Count > 0)  
    
        Console.Write("Points of Local "
                      "maxima are : ");
        foreach(int a in mx) 
            Console.Write(a + " ");
              
        Console.Write("\n");
    
    else
        Console.Write("There are no points "
                      "of Local Maxima ");
  
    if (mn.Count > 0) 
    
        Console.Write("Points of Local " +
                        "minima are : ");
        foreach(int a in mn) 
            Console.Write(a + " ");
              
        Console.Write("\n");
    
    else
        Console.Write("There are no points of " +
                      "Local Maxima ");
  
// Driver code 
public static void Main(string[] args)
{
    int N = 9; 
  
    // Given array arr[] 
    int []arr = { 10, 20, 15, 14, 13, 
                  25, 5, 4, 3 }; 
  
    // Function call 
    findLocalMaximaMinima(N, arr); 
}
}
  
// This code is contributed by rutvik_56
Output: 
Points of Local maxima are : 1 5
Points of Local minima are : 0 4 8

Time Complexity: O(N)
Auxiliary Space: O(N)

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