Maximum frequencies in each M-length subarray

Maximum Length Bitonic Subarray

Last Updated : 25 Mar, 2025

Try it on GfG Practice

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.

Table of Content

[Naive Approach] Using Nested Loops – O(n^3) time and O(1) space
[Expected Approach] Using Peak-Based Mountain – O(n) time and O(n) space
[Optimized Approach] Single Traversal – O(n) time and O(1) space

[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;
}

Java

// 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));
    }
}

Python

# 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))

C#

// 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));
    }
}

JavaScript

// 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));

Output

[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; 
}

C

#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;
}

Java

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));
    }
}

Python

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))

C#

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));
    }
}

JavaScript

// 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));

Output

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 frequencies in each M-length subarray

kartik

News

Improve

Article Tags :

Arrays
DSA

Practice Tags :

Arrays

Similar Reads

Maximum sum bitonic subarray

Given an array containing n numbers. The problem is to find the maximum sum bitonic subarray. A bitonic subarray is a subarray in which elements are first increasing and then decreasing. A strictly increasing or strictly decreasing subarray is also considered a bitonic subarray. Time Complexity of O

Longest subarray having maximum sum

Given an array arr[] containing n integers. The problem is to find the length of the subarray having maximum sum. If there exists two or more subarrays with maximum sum then print the length of the longest subarray.Examples: Input : arr[] = {5, -2, -1, 3, -4}Output : 4There are two subarrays with ma

Length of maximum product subarray

Given an integer array arr[] of size N, the task is to find the maximum length subarray whose products of element is non zero. . Examples: Input: arr[] = [1, 1, 0, 2, 1, 0, 1, 6, 1] Output: 3 Explanation Possible subarray whose product are non zero are [1, 1], [2, 1] and [1, 6, 1] So maximum possibl

Maximum Sum Alternating Subarray

Given an array arr[] of size N, the task is to find the maximum alternating sum of a subarray possible for a given array. Alternating Subarray Sum: Considering a subarray {arr[i], arr[j]}, alternating sum of the subarray is arr[i] - arr[i + 1] + arr[i + 2] - ........ (+ / -) arr[j]. Examples: Input:

Maximum Length Bitonic Subarray | Set 2 (O(n) time and O(1) Space)

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 â‰¤ k â‰¤ j) such that arr[i] â‰¤ arr[i+1] â‰¤ ... â‰¤ arr[k] and arr[k] â‰¥ arr[k+1] â‰¥ ... â‰¥ arr[j].Find the length of the longest bitonic subarray. Examples Input: arr[] = [12, 4, 78, 90, 45, 2

Maximum Length Bitonic Subarray | Set 1 (O(n) time and O(n) space)

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 â‰¤ k â‰¤ j) such that arr[i] â‰¤ arr[i+1] â‰¤ ... â‰¤ arr[k] and arr[k] â‰¥ arr[k+1] â‰¥ ... â‰¥ arr[j].Find the length of the longest bitonic subarray. Examples Input: arr[] = [12, 4, 78, 90, 45,

Maximum frequencies in each M-length subarray

Given an array arr[] consisting of N integers and a positive integer M, the task is to find the maximum frequency for each M-length subarray ( 0 < M ? N). Examples: Input: arr[] = {1, 2, 3, 1, 2, 4, 1, 4, 4}, M = 4Output: 2 2 1 2 2 3Explanation:All the M length sub-arrays with the maximum frequen

Maximum MEX from all subarrays of length K

Given an array arr[] consisting of N distinct integers and an integer K, the task is to find the maximum MEX from all subarrays of length K. The MEX is the smallest positive integer that is not present in the array. Examples: Input: arr[] = {3, 2, 1, 4}, K = 2Output: 3Explanation:All subarrays havin

Maximum XOR subarray of K distinct integer

Given an array A consisting of N positive integers, the task is to calculate the maximum XOR of the subarray of size K consisting of all distinct integers. If such subarray is not present then Print "-1". Examples: Input: N = 10, A[] = [2, 3, 4, 2, 4, 5, 7, 4, 3, 9], K = 4Output: 9Explanation : All

Maximum GCD of all subarrays of length at least 2

Given an array arr[] of N numbers. The task is to find the maximum GCD of all subarrays of size greater than 1. Examples: Input: arr[] = { 3, 18, 9, 9, 5, 15, 8, 7, 6, 9 } Output: 9 Explanation: GCD of the subarray {18, 9, 9} is maximum which is 9.Input: arr[] = { 4, 8, 12, 16, 20, 24 } Output: 4 Ex