Find indices of all local maxima and local minima in an Array
Last Updated :
12 Aug, 2024
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:
- Create two arrays max[] and min[] to store all the local maxima and local minima.
- 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[].
- Check for the local maxima and minima conditions for the first and last elements separately.
- 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 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 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# 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
<script>
// JavaScript program for the above approach
// Function to find all the local maxima
// and minima in the given array arr[]
function findLocalMaximaMinima(n, arr)
{
// Empty vector to store points of
// local maxima and minima
let mx = [], mn = [];
// Checking whether the first point is
// local maxima or minima or none
if (arr[0] > arr[1])
mx.push(0);
else if (arr[0] < arr[1])
mn.push(0);
// Iterating over all points to check
// local maxima and local minima
for(let i = 1; i < n - 1; i++)
{
// Condition for local minima
if ((arr[i - 1] > arr[i]) &&
(arr[i] < arr[i + 1]))
mn.push(i);
// Condition for local maxima
else if ((arr[i - 1] < arr[i]) &&
(arr[i] > arr[i + 1]))
mx.push(i);
}
// Checking whether the last point is
// local maxima or minima or none
if (arr[n - 1] > arr[n - 2])
mx.push(n - 1);
else if (arr[n - 1] < arr[n - 2])
mn.push(n - 1);
// Print all the local maxima and
// local minima indexes stored
if (mx.length > 0)
{
document.write("Points of Local maxima are : ");
for(let a of mx){
document.write(a," ");
}
document.write("</br>");
}
else
document.write("There are no points of Local Maxima ","</br>");
if (mn.length > 0)
{
document.write("Points of Local minima are : ");
for(let a of mn){
document.write(a," ");
}
document.write("</br>");
}
else
document.write("There are no points of Local Minima","</br>");
}
// Driver Code
let N = 9;
// Given array arr[]
let arr = [ 10, 20, 15, 14, 13, 25, 5, 4, 3 ];
// Function call
findLocalMaximaMinima(N, arr);
// This code is contributed by shinjanpatra
</script>
OutputPoints of Local maxima are : 1 5
Points of Local minima are : 0 4 8
Time Complexity: O(N)
Auxiliary Space: O(N)