Find a peak element which is not smaller than its neighbours
Last Updated :
31 May, 2024
Given an array arr of n elements that is first strictly increasing and then maybe strictly decreasing, find the maximum element in the array.
Note: If the array is increasing then just print the last element will be the maximum value.
Example:
Input: array[]= {5, 10, 20, 15}
Output: 20
Explanation: The element 20 has neighbors 10 and 15, both of them are less than 20.
Input: array[] = {10, 20, 15, 2, 23, 90, 67}
Output: 20 or 90
Explanation: The element 20 has neighbors 10 and 15, both of them are less than 20, similarly 90 has neighbors 23 and 67.
The following corner cases give a better idea about the problem.
- If the input array is sorted in a strictly increasing order, the last element is always a peak element. For example, 50 is peak element in {10, 20, 30, 40, 50}.
- If the input array is sorted in a strictly decreasing order, the first element is always a peak element. 100 is the peak element in {100, 80, 60, 50, 20}.
- If all elements of the input array are the same, every element is a peak element.
It is clear from the above examples that there is always a peak element in the input array.
The array can be traversed and the element whose neighbors are less than that element can be returned.
Follow the below steps to Implement the idea:
- If the first element is greater than the second or the last element is greater than the second last, print the respective element and terminate the program.
- Else traverse the array from the second index to the second last index i.e. 1 to N – 1
- If for an element array[i] is greater than both its neighbors, i.e., array[i] > =array[i-1] and array[i] > =array[i+1] , then print that element and terminate.
Below is the implementation of above idea.
C++
// A C++ program to find a peak element
#include <bits/stdc++.h>
using namespace std;
// Find the peak element in the array
int findPeak(int arr[], int n)
{
// first or last element is peak element
if (n == 1)
return 0;
if (arr[0] >= arr[1])
return 0;
if (arr[n - 1] >= arr[n - 2])
return n - 1;
// check for every other element
for (int i = 1; i < n - 1; i++) {
// check if the neighbors are smaller
if (arr[i] >= arr[i - 1] && arr[i] >= arr[i + 1])
return i;
}
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Index of a peak point is " << findPeak(arr, n);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A C program to find a peak element
#include <stdio.h>
// Find the peak element in the array
int findPeak(int arr[], int n)
{
// first or last element is peak element
if (n == 1)
return 0;
if (arr[0] >= arr[1])
return 0;
if (arr[n - 1] >= arr[n - 2])
return n - 1;
// check for every other element
for (int i = 1; i < n - 1; i++) {
// check if the neighbors are smaller
if (arr[i] >= arr[i - 1] && arr[i] >= arr[i + 1])
return i;
}
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Index of a peak point is %d",findPeak(arr, n));
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A Java program to find a peak element
import java.util.*;
class GFG {
// Find the peak element in the array
static int findPeak(int arr[], int n)
{
// First or last element is peak element
if (n == 1)
return 0;
if (arr[0] >= arr[1])
return 0;
if (arr[n - 1] >= arr[n - 2])
return n - 1;
// Check for every other element
for (int i = 1; i < n - 1; i++) {
// Check if the neighbors are smaller
if (arr[i] >= arr[i - 1] && arr[i] >= arr[i + 1])
return i;
}
return 0;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = arr.length;
System.out.print("Index of a peak point is " + findPeak(arr, n));
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# A Python3 program to find a peak element
# Find the peak element in the array
def findPeak(arr, n) :
# first or last element is peak element
if (n == 1) :
return 0
if (arr[0] >= arr[1]) :
return 0
if (arr[n - 1] >= arr[n - 2]) :
return n - 1
# check for every other element
for i in range(1, n - 1) :
# check if the neighbors are smaller
if (arr[i] >= arr[i - 1] and arr[i] >= arr[i + 1]) :
return i
# Driver code.
arr = [ 1, 3, 20, 4, 1, 0 ]
n = len(arr)
print("Index of a peak point is", findPeak(arr, n))
# This code is contributed by divyeshrabadiya07
// A C# program to find a peak element
using System;
public class GFG{
// Find the peak element in the array
static int findPeak(int []arr, int n)
{
// First or last element is peak element
if (n == 1)
return 0;
if (arr[0] >= arr[1])
return 0;
if (arr[n - 1] >= arr[n - 2])
return n - 1;
// Check for every other element
for(int i = 1; i < n - 1; i++)
{
// Check if the neighbors are smaller
if (arr[i] >= arr[i - 1] &&
arr[i] >= arr[i + 1])
return i;
}
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 3, 20, 4, 1, 0 };
int n = arr.Length;
Console.Write("Index of a peak point is " +
findPeak(arr, n));
}
}
// This code is contributed by 29AjayKumar
// A JavaScript program to find a peak element
// Find the peak element in the array
function findPeak(arr, n)
{
// first or last element is peak element
if (n == 1) return 0;
if (arr[0] >= arr[1]) return 0;
if (arr[n - 1] >= arr[n - 2]) return n - 1;
// check for every other element
for (var i = 1; i < n - 1; i++)
{
// check if the neighbors are smaller
if (arr[i] >= arr[i - 1] && arr[i] >= arr[i + 1]) return i;
}
}
// Driver Code
var arr = [1, 3, 20, 4, 1, 0];
var n = arr.length;
console.log("Index of a peak point is " + findPeak(arr, n));
// This code is contributed by rdtank.
OutputIndex of a peak point is 2
Time complexity: O(n), One traversal is needed so the time complexity is O(n)
Auxiliary Space: O(1), No extra space is needed, so space complexity is constant
Find a peak element using recursive Binary Search
Below is the idea to solve the problem.
Using Binary Search, check if the middle element is the peak element or not. If the middle element is not the peak element, then check if the element on the right side is greater than the middle element then there is always a peak element on the right side. If the element on the left side is greater than the middle element then there is always a peak element on the left side.
Follow the steps below to implement the idea:
- Create two variables, l and r, initialize l = 0 and r = n-1
- Recursively perform the below steps till l <= r, i.e. lowerbound is less than the upperbound
- Check if the mid value or index mid = low + (high – low) / 2, is the peak element or not, if yes then print the element and terminate.
- Else if the element on the left side of the middle element is greater then check for peak element on the left side, i.e. update r = mid – 1
- Else if the element on the right side of the middle element is greater then check for peak element on the right side, i.e. update [Tex]l = mid + 1
[/Tex]
Below is the implementation of the above approach
C++
// A C++ program to find a peak element
// using divide and conquer
#include <bits/stdc++.h>
using namespace std;
// A binary search based function
// that returns index of a peak element
int findPeakUtil(int arr[], int low,
int high, int n)
{
// Find index of middle element
// low + (high - low) / 2
int mid = low + (high - low) / 2;
// Compare middle element with its
// neighbours (if neighbours exist)
if ((mid == 0 || arr[mid - 1] <= arr[mid]) &&
(mid == n - 1 || arr[mid + 1] <= arr[mid]))
return mid;
// If middle element is not peak and its
// left neighbour is greater than it,
// then left half must have a peak element
else if (mid > 0 && arr[mid - 1] > arr[mid])
return findPeakUtil(arr, low, (mid - 1), n);
// If middle element is not peak and its
// right neighbour is greater than it,
// then right half must have a peak element
else
return findPeakUtil(
arr, (mid + 1), high, n);
}
// A wrapper over recursive function findPeakUtil()
int findPeak(int arr[], int n)
{
return findPeakUtil(arr, 0, n - 1, n);
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Index of a peak point is "
<< findPeak(arr, n);
return 0;
}
// This code is contributed by rajdeep999
// C program to find a peak
// element using divide and conquer
#include <stdio.h>
// A binary search based function that
// returns index of a peak element
int findPeakUtil(
int arr[], int low, int high, int n)
{
// Find index of middle element
// low + (high - low) / 2
int mid = low + (high - low) / 2;
// Compare middle element with
// its neighbours (if neighbours exist)
if ((mid == 0 || arr[mid - 1] <= arr[mid]) && (mid == n - 1 || arr[mid + 1] <= arr[mid]))
return mid;
// If middle element is not peak and
// its left neighbour is greater
// than it, then left half must have a peak element
else if (mid > 0 && arr[mid - 1] > arr[mid])
return findPeakUtil(arr, low, (mid - 1), n);
// If middle element is not peak and
// its right neighbour is greater
// than it, then right half must have a peak element
else
return findPeakUtil(arr, (mid + 1), high, n);
}
// A wrapper over recursive function findPeakUtil()
int findPeak(int arr[], int n)
{
return findPeakUtil(arr, 0, n - 1, n);
}
/* Driver program to check above functions */
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
printf(
"Index of a peak point is %d", findPeak(arr, n));
return 0;
}
// A Java program to find a peak
// element using divide and conquer
import java.util.*;
import java.lang.*;
import java.io.*;
class PeakElement {
// A binary search based function
// that returns index of a peak element
static int findPeakUtil(
int arr[], int low, int high, int n)
{
// Find index of middle element
// low + (high - low) / 2
int mid = low + (high - low) / 2;
// Compare middle element with its
// neighbours (if neighbours exist)
if ((mid == 0 || arr[mid - 1] <= arr[mid])
&& (mid == n - 1 || arr[mid + 1] <= arr[mid]))
return mid;
// If middle element is not peak
// and its left neighbor is
// greater than it, then left half
// must have a peak element
else if (mid > 0 && arr[mid - 1] > arr[mid])
return findPeakUtil(arr, low, (mid - 1), n);
// If middle element is not peak
// and its right neighbor
// is greater than it, then right
// half must have a peak
// element
else
return findPeakUtil(
arr, (mid + 1), high, n);
}
// A wrapper over recursive function
// findPeakUtil()
static int findPeak(int arr[], int n)
{
return findPeakUtil(arr, 0, n - 1, n);
}
// Driver method
public static void main(String[] args)
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = arr.length;
System.out.println(
"Index of a peak point is " + findPeak(arr, n));
}
}
# A python3 program to find a peak
# element using divide and conquer
# A binary search based function
# that returns index of a peak element
def findPeakUtil(arr, low, high, n):
# Find index of middle element
# low + (high - low) / 2
mid = low + (high - low)/2
mid = int(mid)
# Compare middle element with its
# neighbours (if neighbours exist)
if ((mid == 0 or arr[mid - 1] <= arr[mid]) and
(mid == n - 1 or arr[mid + 1] <= arr[mid])):
return mid
# If middle element is not peak and
# its left neighbour is greater
# than it, then left half must
# have a peak element
elif (mid > 0 and arr[mid - 1] > arr[mid]):
return findPeakUtil(arr, low, (mid - 1), n)
# If middle element is not peak and
# its right neighbour is greater
# than it, then right half must
# have a peak element
else:
return findPeakUtil(arr, (mid + 1), high, n)
# A wrapper over recursive
# function findPeakUtil()
def findPeak(arr, n):
return findPeakUtil(arr, 0, n - 1, n)
# Driver code
arr = [1, 3, 20, 4, 1, 0]
n = len(arr)
print("Index of a peak point is", findPeak(arr, n))
# This code is contributed by
# Smitha Dinesh Semwal
// A C# program to find
// a peak element
// using divide and conquer
using System;
class GFG {
// A binary search based
// function that returns
// index of a peak element
static int findPeakUtil(int[] arr, int low,
int high, int n)
{
// Find index of
// middle element mid = low + (high - low) / 2
int mid = low + (high - low) / 2;
// Compare middle element with
// its neighbours (if neighbours
// exist)
if ((mid == 0 || arr[mid - 1] <= arr[mid]) && (mid == n - 1 || arr[mid + 1] <= arr[mid]))
return mid;
// If middle element is not
// peak and its left neighbor
// is greater than it, then
// left half must have a
// peak element
else if (mid > 0 && arr[mid - 1] > arr[mid])
return findPeakUtil(arr, low,
(mid - 1), n);
// If middle element is not
// peak and its right neighbor
// is greater than it, then
// right half must have a peak
// element
else
return findPeakUtil(arr, (mid + 1),
high, n);
}
// A wrapper over recursive
// function findPeakUtil()
static int findPeak(int[] arr,
int n)
{
return findPeakUtil(arr, 0,
n - 1, n);
}
// Driver Code
static public void Main()
{
int[] arr = { 1, 3, 20,
4, 1, 0 };
int n = arr.Length;
Console.WriteLine("Index of a peak "
+ "point is " + findPeak(arr, n));
}
}
// This code is contributed by ajit
// A Javascript program to find a peak element
// using divide and conquer
// A binary search based function
// that returns index of a peak element
function findPeakUtil(arr, low, high, n)
{
// Find index of middle element
// low + (high - low) / 2
var mid = low + parseInt((high - low) / 2);
// Compare middle element with its
// neighbours (if neighbours exist)
if ((mid == 0 || arr[mid - 1] <= arr[mid]) &&
(mid == n - 1 || arr[mid + 1] <= arr[mid]))
return mid;
// If middle element is not peak and its
// left neighbour is greater than it,
// then left half must have a peak element
else if (mid > 0 && arr[mid - 1] > arr[mid])
return findPeakUtil(arr, low, (mid - 1), n);
// If middle element is not peak and its
// right neighbour is greater than it,
// then right half must have a peak element
else
return findPeakUtil(
arr, (mid + 1), high, n);
}
// A wrapper over recursive function findPeakUtil()
function findPeak(arr, n)
{
return findPeakUtil(arr, 0, n - 1, n);
}
// Driver Code
var arr = [ 1, 3, 20, 4, 1, 0 ];
var n = arr.length;
console.log("Index of a peak point is "
+ findPeak(arr, n));
<?php
// A PHP program to find a
// peak element using
// divide and conquer
// A binary search based function
// that returns index of a peak
// element
function findPeakUtil($arr, $low,
$high, $n)
{
// Find index of middle element
$mid = $low + ($high - $low) / 2; // (low + high)/2
// Compare middle element with
// its neighbours (if neighbours exist)
if (($mid == 0 ||
$arr[$mid - 1] <= $arr[$mid]) &&
($mid == $n - 1 ||
$arr[$mid + 1] <= $arr[$mid]))
return $mid;
// If middle element is not peak
// and its left neighbour is greater
// than it, then left half must
// have a peak element
else if ($mid > 0 &&
$arr[$mid - 1] > $arr[$mid])
return findPeakUtil($arr, $low,
($mid - 1), $n);
// If middle element is not peak
// and its right neighbour is
// greater than it, then right
// half must have a peak element
else return(findPeakUtil($arr, ($mid + 1),
$high, $n));
}
// A wrapper over recursive
// function findPeakUtil()
function findPeak($arr, $n)
{
return floor(findPeakUtil($arr, 0,
$n - 1, $n));
}
// Driver Code
$arr = array(1, 3, 20, 4, 1, 0);
$n = sizeof($arr);
echo "Index of a peak point is ",
findPeak($arr, $n);
// This code is contributed by ajit
?>
OutputIndex of a peak point is 2
Time Complexity: O(log N), Where N is the number of elements in the input array.
Auxiliary Space: O(log N), As recursive call is there, hence implicit stack is used.
Find a peak element using iterative Binary search
Below is the idea to solve the problem.
Using Binary Search, check if the middle element is the peak element or not. If the middle element the peak element terminate the while loop and print middle element, then check if the element on the right side is greater than the middle element then there is always a peak element on the right side. If the element on the left side is greater than the middle element then there is always a peak element on the left side.
Follow the steps below to implement the idea:
- Create two variables, l and r, initialize l = 0 and r = n-1
- Run a while loop till l <= r, lowerbound is less than the upperbound
- Check if the mid value or index mid = low + (high – low) / 2, is the peak element or not, if yes then print the element and terminate.
- Else if the element on the left side of the middle element is greater then check for peak element on the left side, i.e. update r = mid – 1
- Else if the element on the right side of the middle element is greater then check for peak element on the right side, i.e. update [Tex]l = mid + 1
[/Tex]
The below-given code is the iterative version of the above explained and demonstrated recursive based divide and conquer technique.
C++
// A C++ program to find a peak element
// using divide and conquer
#include <bits/stdc++.h>
using namespace std;
// A binary search based function
// that returns index of a peak element
int findPeak(int arr[], int n)
{
int l = 0;
int r = n-1;
int mid;
while (l <= r) {
// finding mid by binary right shifting.
mid = (l + r) >> 1;
// first case if mid is the answer
if ((mid == 0 || arr[mid - 1] <= arr[mid])
and (mid == n - 1 || arr[mid + 1] <= arr[mid]))
break;
// move the right pointer
if (mid > 0 and arr[mid - 1] > arr[mid])
r = mid - 1;
// move the left pointer
else
l = mid + 1;
}
return mid;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << "Index of a peak point is " << findPeak(arr, N);
return 0;
}
// This code is contributed by Rajdeep Mallick (rajdeep999)
// A C program to find a peak element using divide and
// conquer
#include <stdio.h>
// A binary search based function
// that returns index of a peak element
int findPeak(int arr[], int n)
{
int l = 0;
int r = n-1;
int mid;
while (l <= r) {
// finding mid by binary right shifting.
mid = (l + r) >> 1;
// first case if mid is the answer
if ((mid == 0 || arr[mid - 1] <= arr[mid])
&& (mid == n - 1 || arr[mid + 1] <= arr[mid]))
break;
// move the right pointer
if (mid > 0 && arr[mid - 1] > arr[mid])
r = mid - 1;
// move the left pointer
else
l = mid + 1;
}
return mid;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Index of a peak point is %d", findPeak(arr, n));
return 0;
}
// This code is contributed by Rajdeep Mallick (rajdeep999)
// A Java program to find a peak element using divide and
// conquer
import java.io.*;
class GFG {
// A binary search based function that returns index of
// a peak element
static int findPeak(int arr[], int n)
{
int l = 0;
int r = n-1;
int mid = 0;
while (l <= r) {
// finding mid by binary right shifting.
mid = (l + r) >> 1;
// first case if mid is the answer
if ((mid == 0
|| arr[mid - 1] <= arr[mid])
&& (mid == n - 1
|| arr[mid + 1] <= arr[mid]))
break;
// move the right pointer
if (mid > 0 && arr[mid - 1] > arr[mid])
r = mid - 1;
// move the left pointer
else
l = mid + 1;
}
return mid;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 1, 3, 20, 4, 1, 0 };
int n = arr.length;
System.out.println("Index of a peak point is "
+ findPeak(arr, n));
}
}
// This code is contributed by Rajdeep Mallick (rajdeep999)
# A Python program to find a peak element
# using divide and conquer
# A binary search based function
# that returns index of a peak element
def findPeak(arr, n):
l = 0
r = n-1
while(l <= r):
# finding mid by binary right shifting.
mid = (l + r) >> 1
# first case if mid is the answer
if((mid == 0 or arr[mid - 1] <= arr[mid]) and (mid == n - 1 or arr[mid + 1] <= arr[mid])):
break
# move the right pointer
if(mid > 0 and arr[mid - 1] > arr[mid]):
r = mid - 1
# move the left pointer
else:
l = mid + 1
return mid
# Driver Code
arr = [1, 3, 20, 4, 1, 0]
n = len(arr)
print(f"Index of a peak point is {findPeak(arr, n)}")
# This code is contributed by Rajdeep Mallick (rajdeep999)
// A C# program to find a peak element
// using divide and conquer
using System;
public class GFG {
// A binary search based function
// that returns index of a peak element
static int findPeak(int[] arr, int n)
{
int l = 0;
int r = n-1;
int mid = 0;
while (l <= r) {
// finding mid by binary right shifting.
mid = (l + r) >> 1;
// first case if mid is the answer
if ((mid == 0
|| arr[mid - 1] <= arr[mid]
&& (mid == n - 1
|| arr[mid + 1] <= arr[mid])))
break;
// move the right pointer
if (mid > 0 && arr[mid - 1] > arr[mid])
r = mid - 1;
// move the left pointer
else
l = mid + 1;
}
return mid;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 3, 20, 4, 1, 0 };
int n = arr.Length;
Console.WriteLine("Index of a peak point is "
+ findPeak(arr, n));
}
}
// This code is contributed by Rajdeep Mallick (rajdeep999)
// A JavaScript program to find a peak element
// using divide and conquer
// A binary search based function
// that returns index of a peak element
function findPeakUtil(arr, low, high, n)
{
let l = low;
let r = high - 1;
let mid;
while(l <= r)
{
// finding the mid index by right shifting
mid = (l + r) >> 1;
// first case if mid is the answer
if((mid == 0 || arr[mid - 1] <= arr[mid]) &&
(mid == n - 1 || arr[mid + 1] <= arr[mid]))
break;
// change the right pointer to mid-1
if(mid > 0 && arr[mid - 1] > arr[mid])
r = mid - 1;
// change the left pointer to mid+1
else
l = mid + 1;
}
return mid;
}
// A wrapper over recursive function findPeakUtil()
function findPeak(arr,n)
{
return findPeakUtil(arr, 0, n, n);
}
// Driver Code
let arr = [ 1, 3, 20, 4, 1, 0 ];
let n = arr.length;
console.log("Index of a peak point is " +findPeak(arr, n));
// This code is contributed by Rajdeep Mallick (rajdeep999)
OutputIndex of a peak point is 2
Time Complexity: O(log N), Where n is the number of elements in the input array. In each step our search becomes half. So it can be compared to Binary search, So the time complexity is O(log N)
Auxiliary Space: O(1), No extra space is required, so the space complexity is constant.
Exercise:
Consider the following modified definition of peak element. An array element is a peak if it is greater than its neighbors. Note that an array may not contain a peak element with this modified definition.
Related Problem:
Find local minima in an array
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