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.
Naive Approach: Below is the idea to solve the problem
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) |
C
// 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) |
Java
// 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) |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // 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; document.write("Index of a peak point is " + findPeak(arr, n)); // This code is contributed by rdtank. </script> |
Output
Index 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
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
// 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; } |
Java
// 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)); } } |
Python3
# 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 |
C#
// 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 |
Javascript
// 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
<?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 ?> |
Output
Index 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
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) |
C
// 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) |
Java
// 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) |
Python3
# 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) |
C#
// 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) |
Javascript
<script> // 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; document.write("Index of a peak point is " +findPeak(arr, n)); // This code is contributed by Rajdeep Mallick (rajdeep999) </script> |
Output
Index 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.
References:
http://courses.csail.mit.edu/6.006/spring11/lectures/lec02.pdf
Asked in: Amazon
Related Problem:
Find local minima in an array
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