Maximum Length Bitonic Subarray
Last Updated :
23 Jul, 2025
Given an array arr[0 ... n-1] of n positive integers, a subarray arr[i ... j] is bitonic if there exists an index k (i ≤ k ≤ j) such that arr[i] ≤ arr[i+1] ≤ ... ≤ arr[k] and arr[k] ≥ arr[k+1] ≥ ... ≥ arr[j].Find the length of the longest bitonic subarray.
Examples
Input: arr[] = [12, 4, 78, 90, 45, 23]
Output: 5
Explanation: The longest bitonic subarray is [4, 78, 90, 45, 23]. It starts increasing at 4, peaks at 90, and decreases to 23, giving a length of 5.
Input: arr[] = [10, 20, 30, 40]
Output: 4
Explanation: The array [10, 20, 30, 40] is strictly increasing with no decreasing part, so the longest bitonic subarray is the entire array itself, giving a length of 4.
[Naive Approach] Using Nested Loops - O(n^3) time and O(1) space
We check all possible subarrays, trying each index as a peak. If a subarray follows the bitonic pattern, we update the longest length found.
C++
#include <bits/stdc++.h>
using namespace std;
// Check if subarray arr[s...e] is bitonic
bool isBit(const vector<int>& arr, int s, int e) {
// Try each index as potential peak
for (int p = s; p <= e; p++) {
bool nd = true, ni = true;
// Check non-decreasing from s to p
for (int i = s + 1; i <= p; i++) {
if (arr[i] < arr[i - 1]) { nd = false; break; }
}
// Check non-increasing from p to e
for (int i = p + 1; i <= e; i++) {
if (arr[i] > arr[i - 1]) { ni = false; break; }
}
if (nd && ni)
return true;
}
return false;
}
// Find the length of the longest bitonic subarray
int findLB(const vector<int>& arr) {
int n = arr.size(), maxL = 0;
// Enumerate all subarrays arr[i...j]
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (isBit(arr, i, j))
maxL = max(maxL, j - i + 1);
}
}
return maxL;
}
int main() {
vector<int> arr = {12, 4, 78, 90, 45, 23};
cout << findLB(arr) << endl;
return 0;
}
// Check if subarray arr[s...e] is bitonic
public class GfG {
public static boolean isBit(int[] arr, int s, int e) {
// Try each index as potential peak
for (int p = s; p <= e; p++) {
boolean nd = true, ni = true;
// Check non-decreasing from s to p
for (int i = s + 1; i <= p; i++) {
if (arr[i] < arr[i - 1]) { nd = false; break; }
}
// Check non-increasing from p to e
for (int i = p + 1; i <= e; i++) {
if (arr[i] > arr[i - 1]) { ni = false; break; }
}
if (nd && ni)
return true;
}
return false;
}
// Find the length of the longest bitonic subarray
public static int findLB(int[] arr) {
int n = arr.length, maxL = 0;
// Enumerate all subarrays arr[i...j]
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (isBit(arr, i, j))
maxL = Math.max(maxL, j - i + 1);
}
}
return maxL;
}
public static void main(String[] args) {
int[] arr = {12, 4, 78, 90, 45, 23};
System.out.println(findLB(arr));
}
}
# Check if subarray arr[s...e] is bitonic
def is_bit(arr, s, e):
# Try each index as potential peak
for p in range(s, e + 1):
nd = True
ni = True
# Check non-decreasing from s to p
for i in range(s + 1, p + 1):
if arr[i] < arr[i - 1]:
nd = False
break
# Check non-increasing from p to e
for i in range(p + 1, e + 1):
if arr[i] > arr[i - 1]:
ni = False
break
if nd and ni:
return True
return False
# Find the length of the longest bitonic subarray
def find_lb(arr):
n = len(arr)
max_l = 0
# Enumerate all subarrays arr[i...j]
for i in range(n):
for j in range(i, n):
if is_bit(arr, i, j):
max_l = max(max_l, j - i + 1)
return max_l
arr = [12, 4, 78, 90, 45, 23]
print(find_lb(arr))
// Check if subarray arr[s...e] is bitonic
using System;
using System.Linq;
class GfG {
public static bool IsBit(int[] arr, int s, int e) {
// Try each index as potential peak
for (int p = s; p <= e; p++) {
bool nd = true, ni = true;
// Check non-decreasing from s to p
for (int i = s + 1; i <= p; i++) {
if (arr[i] < arr[i - 1]) { nd = false; break; }
}
// Check non-increasing from p to e
for (int i = p + 1; i <= e; i++) {
if (arr[i] > arr[i - 1]) { ni = false; break; }
}
if (nd && ni)
return true;
}
return false;
}
// Find the length of the longest bitonic subarray
public static int FindLB(int[] arr) {
int n = arr.Length, maxL = 0;
// Enumerate all subarrays arr[i...j]
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (IsBit(arr, i, j))
maxL = Math.Max(maxL, j - i + 1);
}
}
return maxL;
}
public static void Main() {
int[] arr = {12, 4, 78, 90, 45, 23};
Console.WriteLine(FindLB(arr));
}
}
// Check if subarray arr[s...e] is bitonic
function isBit(arr, s, e) {
// Try each index as potential peak
for (let p = s; p <= e; p++) {
let nd = true, ni = true;
// Check non-decreasing from s to p
for (let i = s + 1; i <= p; i++) {
if (arr[i] < arr[i - 1]) { nd = false; break; }
}
// Check non-increasing from p to e
for (let i = p + 1; i <= e; i++) {
if (arr[i] > arr[i - 1]) { ni = false; break; }
}
if (nd && ni)
return true;
}
return false;
}
// Find the length of the longest bitonic subarray
function findLB(arr) {
const n = arr.length;
let maxL = 0;
// Enumerate all subarrays arr[i...j]
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
if (isBit(arr, i, j))
maxL = Math.max(maxL, j - i + 1);
}
}
return maxL;
}
const arr = [12, 4, 78, 90, 45, 23];
console.log(findLB(arr));
[Expected Approach] Using Peak-Based Mountain - O(n) time and O(n) space
For solving this problem we can imagine each number as the peak of a mountain. For every element, count how many consecutive numbers before it are rising (or flat) and how many after it are falling (or flat). Add these two counts and subtract one (to avoid counting the peak twice). The longest mountain found this way is answer.
C++
#include <bits/stdc++.h>
using namespace std;
int bitonic(int arr[], int n)
{
// Length of increasing subarray
// ending at all indexes
int inc[n];
// Length of decreasing subarray
// starting at all indexes
int dec[n];
int i, max;
// length of increasing sequence
// ending at first index is 1
inc[0] = 1;
// length of increasing sequence
// starting at first index is 1
dec[n-1] = 1;
// Step 1) Construct increasing sequence array
for (i = 1; i < n; i++)
inc[i] = (arr[i] >= arr[i-1])? inc[i-1] + 1: 1;
// Step 2) Construct decreasing sequence array
for (i = n-2; i >= 0; i--)
dec[i] = (arr[i] >= arr[i+1])? dec[i+1] + 1: 1;
// Step 3) Find the length of
// maximum length bitonic sequence
max = inc[0] + dec[0] - 1;
for (i = 1; i < n; i++)
if (inc[i] + dec[i] - 1 > max)
max = inc[i] + dec[i] - 1;
return max;
}
int main()
{
int arr[] = {12, 4, 78, 90, 45, 23};
int n = sizeof(arr)/sizeof(arr[0]);
cout <<bitonic(arr, n);
return 0;
}
#include<stdio.h>
#include<stdlib.h>
int bitonic(int arr[], int n)
{
int inc[n]; // Length of increasing subarray ending at all indexes
int dec[n]; // Length of decreasing subarray starting at all indexes
int i, max;
// length of increasing sequence ending at first index is 1
inc[0] = 1;
// length of increasing sequence starting at first index is 1
dec[n-1] = 1;
// Step 1) Construct increasing sequence array
for (i = 1; i < n; i++)
inc[i] = (arr[i] >= arr[i-1])? inc[i-1] + 1: 1;
// Step 2) Construct decreasing sequence array
for (i = n-2; i >= 0; i--)
dec[i] = (arr[i] >= arr[i+1])? dec[i+1] + 1: 1;
// Step 3) Find the length of maximum length bitonic sequence
max = inc[0] + dec[0] - 1;
for (i = 1; i < n; i++)
if (inc[i] + dec[i] - 1 > max)
max = inc[i] + dec[i] - 1;
return max;
}
int main()
{
int arr[] = {12, 4, 78, 90, 45, 23};
int n = sizeof(arr)/sizeof(arr[0]);
printf("%d",
bitonic(arr, n));
return 0;
}
import java.io.*;
import java.util.*;
class GFG
{
static int bitonic(int arr[], int n)
{
int[] inc = new int[n]; // Length of increasing subarray ending
// at all indexes
int[] dec = new int[n]; // Length of decreasing subarray starting
// at all indexes
int max;
// Length of increasing sequence ending at first index is 1
inc[0] = 1;
// Length of increasing sequence starting at first index is 1
dec[n-1] = 1;
// Step 1) Construct increasing sequence array
for (int i = 1; i < n; i++)
inc[i] = (arr[i] >= arr[i-1])? inc[i-1] + 1: 1;
// Step 2) Construct decreasing sequence array
for (int i = n-2; i >= 0; i--)
dec[i] = (arr[i] >= arr[i+1])? dec[i+1] + 1: 1;
// Step 3) Find the length of maximum length bitonic sequence
max = inc[0] + dec[0] - 1;
for (int i = 1; i < n; i++)
if (inc[i] + dec[i] - 1 > max)
max = inc[i] + dec[i] - 1;
return max;
}
/*Driver function to check for above function*/
public static void main (String[] args)
{
int arr[] = {12, 4, 78, 90, 45, 23};
int n = arr.length;
System.out.println(bitonic(arr, n));
}
}
def bitonic(arr, n):
# Length of increasing subarray ending at all indexes
# Initialize with 1 because each element is a bitonic sequence of length 1 by itself
inc = [1] * n
# Length of decreasing subarray starting at all indexes
dec = [1] * n # Initialize with 1 for the same reason
# Step 1) Construct increasing sequence array
for i in range(1, n):
if arr[i] >= arr[i-1]:
inc[i] = inc[i-1] + 1
else:
inc[i] = 1
# Step 2) Construct decreasing sequence array
for i in range(n-2, -1, -1):
if arr[i] >= arr[i+1]:
dec[i] = dec[i+1] + 1
else:
dec[i] = 1
# Step 3) Find the length of the maximum length bitonic sequence
max_len = inc[0] + dec[0] - 1
for i in range(1, n):
if inc[i] + dec[i] - 1 > max_len:
max_len = inc[i] + dec[i] - 1
return max_len
# Driver program to test the function
arr = [12, 4, 78, 90, 45, 23]
n = len(arr)
print(bitonic(arr, n))
using System;
class Program
{
// Function to find length of the longest bitonic subarray
static int Bitonic(int[] arr, int n)
{
// Length of increasing subarray ending at all indexes
int[] inc = new int[n];
// Length of decreasing subarray starting at all indexes
int[] dec = new int[n];
int i, max;
// Length of increasing sequence ending at first index is 1
inc[0] = 1;
// Length of decreasing sequence starting at last index is 1
dec[n - 1] = 1;
// Step 1) Construct increasing sequence array
for (i = 1; i < n; i++)
{
inc[i] = (arr[i] >= arr[i - 1]) ? inc[i - 1] + 1 : 1;
}
// Step 2) Construct decreasing sequence array
for (i = n - 2; i >= 0; i--)
{
dec[i] = (arr[i] >= arr[i + 1]) ? dec[i + 1] + 1 : 1;
}
// Step 3) Find the length of maximum length bitonic sequence
max = inc[0] + dec[0] - 1;
for (i = 1; i < n; i++)
{
if (inc[i] + dec[i] - 1 > max)
{
max = inc[i] + dec[i] - 1;
}
}
return max;
}
static void Main()
{
int[] arr = {12, 4, 78, 90, 45, 23};
int n = arr.Length;
Console.WriteLine(Bitonic(arr, n));
}
}
// longest bitonic subarray
function bitonic(arr, n)
{
// Length of increasing subarray ending
// at all indexes
let inc = new Array(n);
// Length of decreasing subarray starting
// at all indexes
let dec = new Array(n);
let max;
// Length of increasing sequence
// ending at first index is 1
inc[0] = 1;
// Length of increasing sequence
// starting at first index is 1
dec[n - 1] = 1;
// Step 1) Construct increasing sequence array
for (let i = 1; i < n; i++)
inc[i]
= (arr[i] >= arr[i - 1]) ? inc[i - 1] + 1 : 1;
// Step 2) Construct decreasing sequence array
for (let i = n - 2; i >= 0; i--)
dec[i]
= (arr[i] >= arr[i + 1]) ? dec[i + 1] + 1 : 1;
// Step 3) Find the length of maximum
// length bitonic sequence
max = inc[0] + dec[0] - 1;
for (let i = 1; i < n; i++)
if (inc[i] + dec[i] - 1 > max)
max = inc[i] + dec[i] - 1;
return max;
}
let arr = [12, 4, 78, 90, 45, 23];
let n = arr.length;
console.log(bitonic(arr, n));
Time Complexity : O(n)
Auxiliary Space : O(n)
[Optimized Approach] Single Traversal - O(n) time and O(1) space
The idea is to check longest bitonic subarray starting at arr[i]. From arr[i], first we will check for end of ascent and then end of descent.Overlapping of bitonic subarrays is taken into account by recording a nextStart position when it finds two equal values when going down the slope of the current subarray. If length of this subarray is greater than max_len, we will update max_len. We continue this process till end of array is reached. Please refer Maximum Length Bitonic Subarray | Set 2 (O(n) time and O(1) Space) for implementation.
Maximum Length Bitonic Subarray
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