Find the maximum element in an array which is first increasing and then decreasing
Given an array of integers which is initially increasing and then decreasing, find the maximum value in the array.
Examples :
Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1}
Output: 500
Input: arr[] = {1, 3, 50, 10, 9, 7, 6}
Output: 50
Corner case (No decreasing part)
Input: arr[] = {10, 20, 30, 40, 50}
Output: 50
Corner case (No increasing part)
Input: arr[] = {120, 100, 80, 20, 0}
Output: 120
Method 1 (Linear Search)
We can traverse the array and keep track of maximum and element. And finally return the maximum element.
C++
// C++ program to find maximum
// element
#include <bits/stdc++.h>
using namespace std;
// function to find the maximum element
int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low + 1; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
// break when once an element is smaller than
// the max then it will go on decreasing
// and no need to check after that
else
break;
}
return max;
}
/* Driver code*/
int main()
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "The maximum element is " << findMaximum(arr, 0, n-1);
return 0;
}
// This is code is contributed by rathbhupendra
C
// C program to find maximum
// element
#include <stdio.h>
// function to find the maximum element
int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low+1; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
// break when once an element is smaller than
// the max then it will go on decreasing
// and no need to check after that
else
break;
}
return max;
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The maximum element is %d", findMaximum(arr, 0, n-1));
getchar();
return 0;
}
Java
// java program to find maximum
// element
class Main
{
// function to find the
// maximum element
static int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
// main function
public static void main (String[] args)
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = arr.length;
System.out.println("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
Python3
# Python3 program to find
# maximum element
def findMaximum(arr, low, high):
max = arr[low]
i = low
for i in range(high+1):
if arr[i] > max:
max = arr[i]
return max
# Driver program to check above functions */
arr = [1, 30, 40, 50, 60, 70, 23, 20]
n = len(arr)
print ("The maximum element is %d"%
findMaximum(arr, 0, n-1))
# This code is contributed by Shreyanshi Arun.
C#
// C# program to find maximum
// element
using System;
class GFG
{
// function to find the
// maximum element
static int findMaximum(int []arr, int low, int high)
{
int max = arr[low];
int i;
for (i = low; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
// Driver code
public static void Main ()
{
int []arr = {1, 30, 40, 50, 60, 70, 23, 20};
int n = arr.Length;
Console.Write("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP program to Find the maximum
// element in an array which is first
// increasing and then decreasing
function findMaximum($arr, $low, $high)
{
$max = $arr[$low];
$i;
for ($i = $low; $i <= $high; $i++)
{
if ($arr[$i] > $max)
$max = $arr[$i];
}
return $max;
}
// Driver Code
$arr = array(1, 30, 40, 50,
60, 70, 23, 20);
$n = count($arr);
echo "The maximum element is ",
findMaximum($arr, 0, $n-1);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// Javascript program to find maximum
// element
// function to find the maximum element
function findMaximum(arr, low, high)
{
var max = arr[low];
var i;
for (i = low + 1; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
// break when once an element is smaller than
// the max then it will go on decreasing
// and no need to check after that
else
break;
}
return max;
}
/* Driver code*/
var arr = [1, 30, 40, 50, 60, 70, 23, 20];
var n = arr.length;
document.write("The maximum element is " + findMaximum(arr, 0, n-1));
</script>
Output
The maximum element is 70
Time Complexity : O(n)
Method 2 (Binary Search)
We can modify the standard Binary Search algorithm for the given type of arrays.
i) If the mid element is greater than both of its adjacent elements, then mid is the maximum.
ii) If mid element is greater than its next element and smaller than the previous element then maximum lies on left side of mid. Example array: {3, 50, 10, 9, 7, 6}
iii) If mid element is smaller than its next element and greater than the previous element then maximum lies on right side of mid. Example array: {2, 4, 6, 8, 10, 3, 1}
C++
#include <bits/stdc++.h>
using namespace std;
int findMaximum(int arr[], int low, int high)
{
/* Base Case: Only one element is present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and first is greater then
the first element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and second is greater then
the second element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* If we reach a point where arr[mid] is greater than both of
its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next
element and smaller than the previous
element then maximum lies on left side of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
// when arr[mid] is greater than arr[mid-1]
// and smaller than arr[mid+1]
else
return findMaximum(arr, mid + 1, high);
}
/* Driver code */
int main()
{
int arr[] = {1, 3, 50, 10, 9, 7, 6};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "The maximum element is " << findMaximum(arr, 0, n-1);
return 0;
}
// This is code is contributed by rathbhupendra
C
#include <stdio.h>
int findMaximum(int arr[], int low, int high)
{
/* Base Case: Only one element is present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and first is greater then
the first element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and second is greater then
the second element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* If we reach a point where arr[mid] is greater than both of
its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next element and smaller than the previous
element then maximum lies on left side of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
else // when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1]
return findMaximum(arr, mid + 1, high);
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 3, 50, 10, 9, 7, 6};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The maximum element is %d", findMaximum(arr, 0, n-1));
getchar();
return 0;
}
Java
// java program to find maximum
// element
class Main
{
// function to find the
// maximum element
static int findMaximum(int arr[], int low, int high)
{
/* Base Case: Only one element is
present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and
first is greater then the first
element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and
second is greater then the second
element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
/*low + (high - low)/2;*/
int mid = (low + high)/2;
/* If we reach a point where arr[mid]
is greater than both of its adjacent
elements arr[mid-1] and arr[mid+1],
then arr[mid] is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next
element and smaller than the previous
element then maximum lies on left side
of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
else
return findMaximum(arr, mid + 1, high);
}
// main function
public static void main (String[] args)
{
int arr[] = {1, 3, 50, 10, 9, 7, 6};
int n = arr.length;
System.out.println("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
Python3
def findMaximum(arr, low, high):
# Base Case: Only one element is present in arr[low..high]*/
if low == high:
return arr[low]
# If there are two elements and first is greater then
# the first element is maximum */
if high == low + 1 and arr[low] >= arr[high]:
return arr[low];
# If there are two elements and second is greater then
# the second element is maximum */
if high == low + 1 and arr[low] < arr[high]:
return arr[high]
mid = (low + high)//2 #low + (high - low)/2;*/
# If we reach a point where arr[mid] is greater than both of
# its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
# is the maximum element*/
if arr[mid] > arr[mid + 1] and arr[mid] > arr[mid - 1]:
return arr[mid]
# If arr[mid] is greater than the next element and smaller than the previous
# element then maximum lies on left side of mid */
if arr[mid] > arr[mid + 1] and arr[mid] < arr[mid - 1]:
return findMaximum(arr, low, mid-1)
else: # when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1]
return findMaximum(arr, mid + 1, high)
# Driver program to check above functions */
arr = [1, 3, 50, 10, 9, 7, 6]
n = len(arr)
print ("The maximum element is %d"% findMaximum(arr, 0, n-1))
# This code is contributed by Shreyanshi Arun.
C#
// C# program to find maximum
// element
using System;
class GFG
{
// function to find the
// maximum element
static int findMaximum(int []arr, int low, int high)
{
/* Base Case: Only one element is
present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and
first is greater then the first
element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and
second is greater then the second
element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
/*low + (high - low)/2;*/
int mid = (low + high)/2;
/* If we reach a point where arr[mid]
is greater than both of its adjacent
elements arr[mid-1] and arr[mid+1],
then arr[mid] is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next
element and smaller than the previous
element then maximum lies on left side
of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
else
return findMaximum(arr, mid + 1, high);
}
// main function
public static void Main()
{
int []arr = {1, 3, 50, 10, 9, 7, 6};
int n = arr.Length;
Console.Write("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP program to Find the maximum
// element in an array which is
// first increasing and then decreasing
function findMaximum($arr, $low, $high)
{
/* Base Case: Only one element
is present in arr[low..high]*/
if ($low == $high)
return $arr[$low];
/* If there are two elements
and first is greater then
the first element is maximum */
if (($high == $low + 1) &&
$arr[$low] >= $arr[$high])
return $arr[$low];
/* If there are two elements
and second is greater then
the second element is maximum */
if (($high == $low + 1) &&
$arr[$low] < $arr[$high])
return $arr[$high];
/*low + (high - low)/2;*/
$mid = ($low + $high) / 2;
/* If we reach a point where
arr[mid] is greater than
both of its adjacent elements
arr[mid-1] and arr[mid+1],
then arr[mid] is the maximum
element */
if ( $arr[$mid] > $arr[$mid + 1] &&
$arr[$mid] > $arr[$mid - 1])
return $arr[$mid];
/* If arr[mid] is greater than
the next element and smaller
than the previous element then
maximum lies on left side of mid */
if ($arr[$mid] > $arr[$mid + 1] &&
$arr[$mid] < $arr[$mid - 1])
return findMaximum($arr, $low, $mid - 1);
// when arr[mid] is greater than
// arr[mid-1] and smaller than
// arr[mid+1]
else
return findMaximum($arr,
$mid + 1, $high);
}
// Driver Code
$arr = array(1, 3, 50, 10, 9, 7, 6);
$n = sizeof($arr);
echo("The maximum element is ");
echo(findMaximum($arr, 0, $n-1));
// This code is contributed by nitin mittal.
?>
Javascript
<script>
function findMaximum( arr, low, high)
{
/* Base Case: Only one element is present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and first is greater then
the first element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and second is greater then
the second element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
mid = (low + high)/2;
/*low + (high - low)/2;*/
/* If we reach a point where arr[mid] is greater than both of
its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next
element and smaller than the previous
element then maximum lies on left side of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
// when arr[mid] is greater than arr[mid-1]
// and smaller than arr[mid+1]
return findMaximum(arr, mid + 1, high);
}
/* Driver code */
arr = new Array(1, 3, 50, 10, 9, 7, 6);
n = arr.length;
document.write("The maximum element is" + "\n" + findMaximum(arr, 0, n-1));
// This code is contributed by simranarora5sos
</script>
Output
The maximum element is 50
Time Complexity : O(Logn)
This method works only for distinct numbers. For example, it will not work for an array like {0, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 5, 3, 3, 2, 2, 1, 1}.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Method 3 (Binary Search – Iterative Solution)
The iterative approach of Binary search to find the maximum element in an array which is first increasing and then decreasing.
We can modify the standard Binary Search algorithm for the given type of arrays.
i) If the mid element is greater than both of its adjacent elements, then mid is the maximum.
ii) If mid element is greater than its next element and smaller than the previous element then maximum lies on left side of mid.
Example array: {3, 50, 10, 9, 7, 6}
iii) If mid element is smaller than its next element and greater than the previous element then maximum lies on right side of mid. Example array: {2, 4, 6, 8, 10, 3, 1}
C++
#include <iostream>
using namespace std;
int maxInBitonic(int arr[], int l, int r)
{
while (l <= r) {
int m = l + (r - l) / 2; // m = (l + r) / 2
/****Base Cases Starts*****/
/* If there are two elements and first is greater
then the first element is maximum */
if ((r == l + 1) && arr[l] >= arr[r])
return arr[l];
/* If there are two elements and second is greater
then the second element is maximum */
if ((r == l + 1) && arr[l] < arr[r])
return arr[r];
/* If we reach a point where arr[mid] is greater
than both of its adjacent elements arr[mid-1] and
arr[mid+1], then arr[mid] is the maximum
element*/
if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
return arr[m];
/****Base Case ends *****/
// move to left with l and r=m-1
if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
r = m - 1;
else
l = m + 1; // move to right with l=m+1 and r
}
// if we reach here, then element was
// not present
return -1;
}
// Driver function
int main()
{
int arr[] = { 1, 3, 50, 10, 9, 7, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "The maximum element is "
<< maxInBitonic(arr, 0, n - 1);
return 0;
}
Java
import java.util.*;
class GFG{
static int maxInBitonic(int arr[], int l, int r)
{
while (l <= r) {
int m = l + (r - l) / 2; // m = (l + r) / 2
/****Base Cases Starts*****/
/* If there are two elements and first is greater
then the first element is maximum */
if ((r == l + 1) && arr[l] >= arr[r])
return arr[l];
/* If there are two elements and second is greater
then the second element is maximum */
if ((r == l + 1) && arr[l] < arr[r])
return arr[r];
/* If we reach a point where arr[mid] is greater
than both of its adjacent elements arr[mid-1] and
arr[mid+1], then arr[mid] is the maximum
element*/
if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
return arr[m];
/****Base Case ends *****/
// move to left with l and r=m-1
if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
r = m - 1;
else
l = m + 1; // move to right with l=m+1 and r
}
// if we reach here, then element was
// not present
return -1;
}
// Driver function
public static void main(String[] args)
{
int arr[] = { 1, 3, 50, 10, 9, 7, 6 };
int n = arr.length;
System.out.print("The maximum element is "
+ maxInBitonic(arr, 0, n - 1));
}
}
// This code is contributed by todaysgaurav
Python3
# Python 3 program for the above approach
def maxInBitonic(arr, l, r) :
while (l <= r) :
m = int(l + (r - l) / 2) # m = (l + r) / 2
#Base Cases Starts*****/
# If there are two elements and first is greater
# then the first element is maximum */
if ((r == l + 1) and arr[l] >= arr[r]):
return arr[l]
# If there are two elements and second is greater
# then the second element is maximum */
if ((r == l + 1) and arr[l] < arr[r]):
return arr[r]
# If we reach a powhere arr[mid] is greater
# than both of its adjacent elements arr[mid-1] and
# arr[mid+1], then arr[mid] is the maximum
# element*/
if (arr[m] > arr[m + 1] and arr[m] > arr[m - 1]):
return arr[m]
#***Base Case ends *****/
# move to left with l and r=m-1
if (arr[m] > arr[m + 1] and arr[m] < arr[m - 1]) :
r = m - 1
else :
l = m + 1 # move to right with l=m+1 and r
# if we reach here, then element was
# not present
return -1
# Driver function
arr = [ 1, 3, 50, 10, 9, 7, 6 ]
n = len(arr)
print("The maximum element is ", maxInBitonic(arr, 0, n - 1))
# This code is contributed by splevel62.
C#
using System;
class GFG{
static int maxInBitonic(int []arr, int l, int r)
{
while (l <= r) {
int m = l + (r - l) / 2; // m = (l + r) / 2
/****Base Cases Starts*****/
/* If there are two elements and first is greater
then the first element is maximum */
if ((r == l + 1) && arr[l] >= arr[r])
return arr[l];
/* If there are two elements and second is greater
then the second element is maximum */
if ((r == l + 1) && arr[l] < arr[r])
return arr[r];
/* If we reach a point where arr[mid] is greater
than both of its adjacent elements arr[mid-1] and
arr[mid+1], then arr[mid] is the maximum
element*/
if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
return arr[m];
/****Base Case ends *****/
// move to left with l and r=m-1
if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
r = m - 1;
else
l = m + 1; // move to right with l=m+1 and r
}
// if we reach here, then element was
// not present
return -1;
}
// Driver function
public static void Main(String[] args)
{
int []arr = { 1, 3, 50, 10, 9, 7, 6 };
int n = arr.Length;
Console.Write("The maximum element is "
+ maxInBitonic(arr, 0, n - 1));
}
}
// This code is contributed by shivanisinghss2110
Javascript
<script>
// JavaScript program
function maxInBitonic(arr, l, r)
{
while (l <= r) {
var m = l + (r - l) / 2; // m = (l + r) / 2
/****Base Cases Starts*****/
/* If there are two elements and first is greater
then the first element is maximum */
if ((r == l + 1) && arr[l] >= arr[r])
return arr[l];
/* If there are two elements and second is greater
then the second element is maximum */
if ((r == l + 1) && arr[l] < arr[r])
return arr[r];
/* If we reach a point where arr[mid] is greater
than both of its adjacent elements arr[mid-1] and
arr[mid+1], then arr[mid] is the maximum
element*/
if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
return arr[m];
/****Base Case ends *****/
// move to left with l and r=m-1
if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
r = m - 1;
else
l = m + 1; // move to right with l=m+1 and r
}
// if we reach here, then element was
// not present
return -1;
}
// Driver function
var arr = [ 1, 3, 50, 10, 9, 7, 6 ];
var n = arr.length;
document.write("The maximum element is "
+ maxInBitonic(arr, 0, n - 1));
// This code is contributed by shivanisinghss2110
</script>
Output
The maximum element is 50
Time Complexity: O(log n)
Auxiliary Space: O(1)
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