Longest Bitonic Subsequence
Last Updated :
07 Mar, 2025
Given an array arr[] containing n positive integers, a subsequence of numbers is called bitonic if it is first strictly increasing, then strictly decreasing. The task is to find the length of the longest bitonic subsequence.
Note: Only strictly increasing (no decreasing part) or a strictly decreasing sequence should not be considered as a bitonic sequence.
Examples:
Input: arr[]= [12, 11, 40, 5, 3, 1]
Output: 5
Explanation: The Longest Bitonic Subsequence is {12, 40, 5, 3, 1} which is of length 5.
Input: arr[] = [80, 60, 30]
Output: 0
Explanation: There is no possible Bitonic Subsequence.
Using Recursion - O(n*(2^n)) Time and O(n) Space
The recursive approach for finding the longest bitonic subsequence relies on breaking the problem into two parts for each potential peak:
- Calculate the Longest Increasing Subsequence (LIS) ending at the current element.
- Calculate the Longest Decreasing Subsequence (LDS) starting from the current element.
At each step, the algorithm explores two choices:
- Include the current element in the subsequence if it satisfies the increasing or decreasing condition.
- Skip the current element and move to the next index.
Recurrence Relation:
For any index i, the solution involves:
- Finding the LIS to the left: if nums[i] < nums[prev]
LIS(i, prev) = max(LIS(i-1, prev),1 + LIS(i-1, i))
- Finding the LDS to the right: if nums[i] < nums[prev]
LDS(i, prev) = max(LDS(i+1, prev), 1 + LDS(i+1, i))
Base Cases:
- For LIS: LIS(i, prev) = 0, if i < 0
- For LDS: LDS(i, prev) = 0, if i >= n
C++
// C++ implementation to find longest Bitonic
// subsequence using Recursion
#include <bits/stdc++.h>
using namespace std;
// Function to find the longest decreasing subsequence
// to the left
int left(int prev, int idx, vector<int>& arr) {
if (idx < 0) {
return 0;
}
// Check if nums[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + left(idx, idx - 1, arr);
}
// Return the maximum of including
// or excluding nums[idx]
return max(include, left(prev, idx - 1, arr));
}
// Function to find the longest decreasing
// subsequence to the right
int right(int prev, int idx, vector<int>& arr) {
if (idx >= arr.size()) {
return 0;
}
// Check if nums[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + right(idx, idx + 1, arr);
}
// Return the maximum of including or
// excluding nums[idx]
return max(include, right(prev, idx + 1, arr));
}
// Function to find the longest bitonic sequence
int LongestBitonicSequence(vector<int>& arr) {
int maxLength = 0;
// Iterate over potential peaks in the array
for (int i = 1; i < arr.size() - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
int leftLen = left(i, i - 1, arr);
int rightLen = right(i, i + 1, arr);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = max(maxLength, leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return (maxLength < 3) ? 0 : maxLength;
}
int main() {
vector<int> arr = {12, 11, 40, 5, 3, 1};
cout << LongestBitonicSequence(arr) << endl;
return 0;
}
// Java implementation to find longest Bitonic
// subsequence using Recursion
import java.util.ArrayList;
import java.util.*;
class GfG {
// Function to find the longest decreasing subsequence
// to the left
static int left(int prev, int idx, ArrayList<Integer> arr) {
if (idx < 0) {
return 0;
}
// Check if nums[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr.get(idx) < arr.get(prev)) {
include = 1 + left(idx, idx - 1, arr);
}
// Return the maximum of including
// or excluding nums[idx]
return Math.max(include, left(prev, idx - 1, arr));
}
// Function to find the longest decreasing
// subsequence to the right
static int right(int prev, int idx, ArrayList<Integer> arr) {
if (idx >= arr.size()) {
return 0;
}
// Check if nums[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr.get(idx) < arr.get(prev)) {
include = 1 + right(idx, idx + 1, arr);
}
// Return the maximum of including or
// excluding nums[idx]
return Math.max(include, right(prev, idx + 1, arr));
}
// Function to find the longest bitonic sequence
static int LongestBitonicSequence(ArrayList<Integer> arr) {
int maxLength = 0;
// Iterate over potential peaks in the array
for (int i = 1; i < arr.size() - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of nums[i]
int leftLen = left(i, i - 1, arr);
int rightLen = right(i, i + 1, arr);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.max(maxLength, leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return (maxLength < 3) ? 0 : maxLength;
}
public static void main(String[] args) {
ArrayList<Integer> arr
= new ArrayList<>(Arrays.asList(12, 11, 40, 5, 3, 1));
System.out.println(LongestBitonicSequence(arr));
}
}
# Python implementation to find longest Bitonic
# subsequence using Recursion
# Function to find the longest decreasing
# subsequence to the left
def left(prev, idx, arr):
if idx < 0:
return 0
# Check if arr[idx] can be included in the
# decreasing subsequence
include = 0
if arr[idx] < arr[prev]:
include = 1 + left(idx, idx - 1, arr)
# Return the maximum of including or excluding arr[idx]
return max(include, left(prev, idx - 1, arr))
# Function to find the longest decreasing subsequence
# to the right
def right(prev, idx, arr):
if idx >= len(arr):
return 0
# Check if nums[idx] can be included in the
# decreasing subsequence
include = 0
if arr[idx] < arr[prev]:
include = 1 + right(idx, idx + 1, arr)
# Return the maximum of including or excluding arr[idx]
return max(include, right(prev, idx + 1, arr))
# Function to find the longest bitonic sequence
def LongestBitonicSequence(arr):
max_length = 0
# Iterate over potential peaks in the array
for i in range(1, len(arr) - 1):
# Find the longest decreasing subsequences
# on both sides of arr[i]
left_len = left(i, i - 1, arr)
right_len = right(i, i + 1, arr)
# Ensure both left and right subsequences are valid
if left_len == 0 or right_len == 0:
left_len = 0
right_len = 0
# Update maximum bitonic sequence length
max_length = max(max_length, left_len + right_len + 1)
# If no valid bitonic sequence, return 0
return 0 if max_length < 3 else max_length
if __name__ == "__main__":
arr = [12, 11, 40, 5, 3, 1]
print(LongestBitonicSequence(arr))
// C# implementation to find longest Bitonic
// subsequence using Recursion
using System;
using System.Collections.Generic;
class GfG {
// Function to find the longest decreasing
// subsequence to the left
static int Left(int prev, int idx, List<int> arr) {
if (idx < 0) {
return 0;
}
// Check if nums[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + Left(idx, idx - 1, arr);
}
// Return the maximum of including or excluding nums[idx]
return Math.Max(include, Left(prev, idx - 1, arr));
}
// Function to find the longest decreasing
// subsequence to the right
static int Right(int prev, int idx, List<int> arr) {
if (idx >= arr.Count) {
return 0;
}
// Check if nums[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + Right(idx, idx + 1, arr);
}
// Return the maximum of including or excluding nums[idx]
return Math.Max(include, Right(prev, idx + 1, arr));
}
// Function to find the longest bitonic sequence
static int LongestBitonicSequence(List<int> arr) {
int maxLength = 0;
// Iterate over potential peaks in the array
for (int i = 1; i < arr.Count - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of nums[i]
int leftLen = Left(i, i - 1, arr);
int rightLen = Right(i, i + 1, arr);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.Max(maxLength,
leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return (maxLength < 3) ? 0 : maxLength;
}
static void Main() {
List<int> arr
= new List<int> { 12, 11, 40, 5, 3, 1 };
Console.WriteLine(LongestBitonicSequence(arr));
}
}
// Javascript implementation to find longest Bitonic
// subsequence using Recursion<script>
// Function to find the longest decreasing subsequence to the left
function left(prev, idx, arr) {
if (idx < 0) {
return 0;
}
// Check if arr[idx] can be included in decreasing subsequence
let include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + left(idx, idx - 1, arr);
}
// Return the maximum of including or excluding arr[idx]
return Math.max(include, left(prev, idx - 1, arr));
}
// Function to find the longest decreasing
// subsequence to the right
function right(prev, idx, arr) {
if (idx >= arr.length) {
return 0;
}
// Check if arr[idx] can be included in
// decreasing subsequence
let include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + right(idx, idx + 1, arr);
}
// Return the maximum of including or
// excluding arr[idx]
return Math.max(include, right(prev, idx + 1, arr));
}
// Function to find the longest bitonic sequence
function longestBitonicSequence(nums) {
let maxLength = 0;
// Iterate over potential peaks in the array
for (let i = 1; i < arr.length - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
let leftLen = left(i, i - 1, arr);
let rightLen = right(i, i + 1, arr);
// Ensure both left and right subsequences are valid
if (leftLen === 0 || rightLen === 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.max(maxLength,
leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
const arr = [12, 11, 40, 5, 3, 1];
console.log(longestBitonicSequence(arr));
Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space
1. Optimal Substructure: The solution for finding the longest bitonic subsequence around a peak element can be derived from solutions to smaller subproblems. Specifically:
- The left subsequence is determined by comparing each element on the left to the current peak, recursively finding the longest decreasing sequence.
- left(prev, idx) = max(1 + left(idx, idx - 1), left(prev, idx - 1)) if arr[idx] < arr[prev]. Otherwise, it is left(prev, idx - 1).
- Base case: left(prev, idx) = 0 if idx < 0.
- The right subsequence is similarly found by exploring decreasing sequences to the right of the peak.
- right(prev, idx) = max(1 + right(idx, idx + 1), right(prev, idx + 1)) if arr[idx] < arr[prev]. Otherwise, it is right(prev, idx + 1).
- Base case: right(prev, idx) = 0 if idx >= arr.size().
Total length at a peak nums[i] is given by: Total Length = leftLen + rightLen + 1.
2. Overlapping Subproblems: Subproblems like left(prev, idx) or right(prev, idx) for specific values of prev and idx are repeatedly computed, memoization avoids redundant calculations by storing results in 2D arrays leftMemo and rightMemo.
- If leftMemo[prev][idx] != -1 or rightMemo[prev][idx] != -1 , return the stored result.
C++
// C++ implementation to find longest Bitonic
// subsequence using Recursion and Memoization
#include <bits/stdc++.h>
using namespace std;
// Function to find the longest decreasing subsequence
// to the left with memoization
int left(int prev, int idx, vector<int>& arr,
vector<vector<int>>& leftMemo) {
if (idx < 0) {
return 0;
}
if (leftMemo[prev][idx] != -1) {
return leftMemo[prev][idx];
}
// Check if arr[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + left(idx, idx - 1, arr, leftMemo);
}
// Store and return the result
return leftMemo[prev][idx]
= max(include, left(prev, idx - 1, arr, leftMemo));
}
// Function to find the longest decreasing subsequence
// to the right with memoization
int right(int prev, int idx, vector<int>& arr,
vector<vector<int>>& rightMemo) {
if (idx >= arr.size()) {
return 0;
}
if (rightMemo[prev][idx] != -1) {
return rightMemo[prev][idx];
}
// Check if arr[idx] can be included
// in decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + right(idx, idx + 1, arr, rightMemo);
}
// Store and return the result
return rightMemo[prev][idx]
= max(include, right(prev, idx + 1, arr, rightMemo));
}
// Function to find the longest bitonic sequence
int LongestBitonicSequence(vector<int>& arr) {
int n = arr.size();
int maxLength = 0;
// Initialize memoization tables
vector<vector<int>> leftMemo(n, vector<int>(n, -1));
vector<vector<int>> rightMemo(n, vector<int>(n, -1));
// Iterate over potential peaks in the array
for (int i = 1; i < n - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
int leftLen = left(i, i - 1, arr, leftMemo);
int rightLen = right(i, i + 1, arr, rightMemo);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = max(maxLength, leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return (maxLength < 3) ? 0 : maxLength;
}
int main() {
vector<int> arr = {12, 11, 40, 5, 3, 1};
cout << LongestBitonicSequence(arr) << endl;
return 0;
}
// Java implementation to find longest Bitonic
// subsequence using Recursion and Memoization
import java.util.*;
class GfG {
// Function to find the longest decreasing subsequence
// to the left with memoization
static int left(int prev, int idx, ArrayList<Integer> arr,
int[][] leftMemo) {
if (idx < 0) {
return 0;
}
if (leftMemo[prev][idx] != -1) {
return leftMemo[prev][idx];
}
// Check if arr[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr.get(idx) < arr.get(prev)) {
include = 1 + left(idx, idx - 1, arr, leftMemo);
}
// Store and return the result
return leftMemo[prev][idx]
= Math.max(include, left(prev, idx - 1, arr, leftMemo));
}
// Function to find the longest decreasing subsequence
// to the right with memoization
static int right(int prev, int idx, ArrayList<Integer> arr,
int[][] rightMemo) {
if (idx >= arr.size()) {
return 0;
}
if (rightMemo[prev][idx] != -1) {
return rightMemo[prev][idx];
}
// Check if arr[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr.get(idx) < arr.get(prev)) {
include = 1 + right(idx, idx + 1, arr, rightMemo);
}
// Store and return the result
return rightMemo[prev][idx]
= Math.max(include, right(prev, idx + 1, arr, rightMemo));
}
// Function to find the longest bitonic sequence
static int LongestBitonicSequence(ArrayList<Integer> arr) {
int n = arr.size();
int maxLength = 0;
// Initialize memoization tables
int[][] leftMemo = new int[n][n];
int[][] rightMemo = new int[n][n];
// Fill memoization tables with -1
for (int i = 0; i < n; i++) {
Arrays.fill(leftMemo[i], -1);
Arrays.fill(rightMemo[i], -1);
}
// Iterate over potential peaks in the array
for (int i = 1; i < n - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
int leftLen = left(i, i - 1, arr, leftMemo);
int rightLen = right(i, i + 1, arr, rightMemo);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.max(maxLength, leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return (maxLength < 3) ? 0 : maxLength;
}
public static void main(String[] args) {
ArrayList<Integer> arr
= new ArrayList<>(Arrays.asList(12, 11, 40, 5, 3, 1));
System.out.println(LongestBitonicSequence(arr));
}
}
# Python implementation to find longest Bitonic
# subsequence using Recursion and Memoization
# Function to find the longest decreasing subsequence
# to the left
def left(prev, idx, arr, leftMemo):
if idx < 0:
return 0
if leftMemo[prev][idx] != -1:
return leftMemo[prev][idx]
# Check if arr[idx] can be included in
# decreasing subsequence
include = 0
if arr[idx] < arr[prev]:
include = 1 + left(idx, idx - 1, arr, leftMemo)
# Store and return the result
leftMemo[prev][idx] = max(include,
left(prev, idx - 1, arr, leftMemo))
return leftMemo[prev][idx]
# Function to find the longest decreasing subsequence
# to the right
def right(prev, idx, arr, rightMemo):
if idx >= len(arr):
return 0
if rightMemo[prev][idx] != -1:
return rightMemo[prev][idx]
# Check if arr[idx] can be included in decreasing
# subsequence
include = 0
if arr[idx] < arr[prev]:
include = 1 + right(idx, idx + 1, arr, rightMemo)
# Store and return the result
rightMemo[prev][idx] = max(include,
right(prev, idx + 1, arr, rightMemo))
return rightMemo[prev][idx]
def LongestBitonicSequence(arr):
n = len(arr)
maxLength = 0
# Initialize memoization tables
leftMemo = [[-1 for i in range(n)] for i in range(n)]
rightMemo = [[-1 for i in range(n)] for i in range(n)]
# Iterate over potential peaks in the array
for i in range(1, n - 1):
# Find the longest decreasing subsequences
# on both sides of arr[i]
leftLen = left(i, i - 1, arr, leftMemo)
rightLen = right(i, i + 1, arr, rightMemo)
# Ensure both left and right subsequences are valid
if leftLen == 0 or rightLen == 0:
leftLen = 0
rightLen = 0
# Update maximum bitonic sequence length
maxLength = max(maxLength, leftLen + rightLen + 1)
# If no valid bitonic sequence, return 0
return 0 if maxLength < 3 else maxLength
if __name__ == "__main__":
arr = [12, 11, 40, 5, 3, 1]
print(LongestBitonicSequence(arr))
// C# implementation to find longest Bitonic
// subsequence using Recursion and Memoization
using System;
using System.Collections.Generic;
class GfG {
// Function to find the longest decreasing subsequence
// to the left with memoization
static int Left(int prev, int idx,
List<int> arr, int[,] leftMemo) {
if (idx < 0) {
return 0;
}
if (leftMemo[prev, idx] != -1) {
return leftMemo[prev, idx];
}
// Check if arr[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + Left(idx, idx - 1, arr, leftMemo);
}
// Store and return the result
leftMemo[prev, idx]
= Math.Max(include, Left(prev,
idx - 1, arr, leftMemo));
return leftMemo[prev, idx];
}
// Function to find the longest decreasing subsequence
// to the right with memoization
static int Right(int prev, int idx,
List<int> arr, int[,] rightMemo) {
if (idx >= arr.Count) {
return 0;
}
if (rightMemo[prev, idx] != -1) {
return rightMemo[prev, idx];
}
// Check if arr[idx] can be included in
// decreasing subsequence
int include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + Right(idx, idx + 1,
arr, rightMemo);
}
// Store and return the result
rightMemo[prev, idx]
= Math.Max(include, Right(prev,
idx + 1, arr, rightMemo));
return rightMemo[prev, idx];
}
// Function to find the longest bitonic sequence
static int LongestBitonicSequence(List<int> arr) {
int n = arr.Count;
int maxLength = 0;
// Initialize memoization tables
int[,] leftMemo = new int[n, n];
int[,] rightMemo = new int[n, n];
// Fill memoization tables with -1
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
leftMemo[i, j] = -1;
rightMemo[i, j] = -1;
}
}
// Iterate over potential peaks in the array
for (int i = 1; i < n - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
int leftLen = Left(i, i - 1, arr, leftMemo);
int rightLen = Right(i, i + 1, arr, rightMemo);
// Ensure both left and right subsequences are valid
if (leftLen == 0 || rightLen == 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.Max(maxLength,
leftLen + rightLen + 1);
}
return (maxLength < 3) ? 0 : maxLength;
}
static void Main(string[] args) {
List<int> arr
= new List<int>{12, 11, 40, 5, 3, 1};
Console.WriteLine(LongestBitonicSequence(arr));
}
}
// Javascript implementation to find longest Bitonic
// subsequence using Recursion and Memoization
// Function to find the longest decreasing subsequence
// to the left with memoization
function left(prev, idx, arr, leftMemo) {
if (idx < 0) {
return 0;
}
if (leftMemo[prev][idx] !== -1) {
return leftMemo[prev][idx];
}
// Check if arr[idx] can be included in decreasing subsequence
let include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + left(idx, idx - 1, arr, leftMemo);
}
// Store and return the result
return leftMemo[prev][idx]
= Math.max(include, left(prev,
idx - 1, arr, leftMemo));
}
// Function to find the longest decreasing
// subsequence to the right with memoization
function right(prev, idx, arr, rightMemo) {
if (idx >= arr.length) {
return 0;
}
if (rightMemo[prev][idx] !== -1) {
return rightMemo[prev][idx];
}
// Check if arr[idx] can be included in decreasing subsequence
let include = 0;
if (arr[idx] < arr[prev]) {
include = 1 + right(idx, idx + 1, arr, rightMemo);
}
// Store and return the result
return rightMemo[prev][idx]
= Math.max(include, right(prev,
idx + 1, arr, rightMemo));
}
// Function to find the longest bitonic sequence
function LongestBitonicSequence(arr) {
const n = arr.length;
let maxLength = 0;
// Initialize memoization tables
const leftMemo = Array.from({ length: n },
() => Array(n).fill(-1));
const rightMemo = Array.from({ length: n },
() => Array(n).fill(-1));
// Iterate over potential peaks in the array
for (let i = 1; i < n - 1; i++) {
// Find the longest decreasing subsequences
// on both sides of arr[i]
let leftLen = left(i, i - 1, arr, leftMemo);
let rightLen = right(i, i + 1, arr, rightMemo);
// Ensure both left and right subsequences are valid
if (leftLen === 0 || rightLen === 0) {
leftLen = 0;
rightLen = 0;
}
// Update maximum bitonic sequence length
maxLength = Math.max(maxLength, leftLen + rightLen + 1);
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
const arr = [12, 11, 40, 5, 3, 1];
console.log(LongestBitonicSequence(arr));
Using Bottom-Up DP (Tabulation) - O(n^2) Time and O(n) Space
This approach iteratively builds the solution from smaller subproblems in a bottom-up manner, avoiding recursion. This problem is a variation of standard Longest Increasing Subsequence (LIS) problem.
We create two 1D arrays, left and right, of size n.
- left[i] stores the length of the Longest Increasing Subsequence (LIS) ending at index i.
- right[i] stores the length of the Longest Decreasing Subsequence (LDS) starting at index i.
The dynamic programming relations are as follows:
For LIS:
If arr[i] > arr[j], then
left[i] = max(left[i], left[j] + 1)
This means the LIS at i can be extended by the LIS ending at j.
For LDS:
If arr[i] > arr[j], then
right[i] = max(right[i], right[j] + 1)
This means the LDS at i can be extended by the LDS starting at j.
The final bitonic length for a peak at index i is calculated as: maxLength = max(maxLength, left[i] + right[i] - 1)
The subtraction of 1 accounts for the peak element being counted in both LIS and LDS.
C++
// C++ implementation to find the longest
// bitonic subsequence using tabulation
#include <bits/stdc++.h>
using namespace std;
int LongestBitonicSequence(vector<int>& arr) {
int n = arr.size();
// If there are less than 3 elements,
// no bitonic subsequence exists
if (n < 3) return 0;
// Create tables for longest increasing subsequence
// (LIS) and longest decreasing subsequence (LDS)
vector<int> left(n, 1), right(n, 1);
// Fill left table for LIS
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
// If arr[i] is greater than arr[j],
// update LIS value at i
if (arr[i] > arr[j]) {
left[i] = max(left[i], left[j] + 1);
}
}
}
// Fill right table for LDS
for (int i = n - 2; i >= 0; i--) {
// Compare each element with subsequent
// ones to build LDS
for (int j = n - 1; j > i; j--) {
// If arr[i] is greater than arr[j],
// update LDS value at i
if (arr[i] > arr[j]) {
right[i] = max(right[i], right[j] + 1);
}
}
}
// Calculate the maximum length of bitonic subsequence
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Check if both LIS and LDS are valid
// for the current index
if (left[i] > 1 && right[i] > 1) {
// Update maxLength considering both LIS
// and LDS, subtracting 1 for the peak element
maxLength = max(maxLength, left[i] + right[i] - 1);
}
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
int main() {
vector<int> arr = {12, 11, 40, 5, 3, 1};
cout << LongestBitonicSequence(arr) << endl;
return 0;
}
// Java implementation to find the longest
// bitonic subsequence using tabulation
import java.util.*;
class GfG {
// Function to find the longest bitonic subsequence
static int LongestBitonicSequence(ArrayList<Integer> arr) {
int n = arr.size();
// If there are less than 3 elements,
// no bitonic subsequence exists
if (n < 3) return 0;
// Create ArrayLists for longest increasing subsequence
// (LIS) and longest decreasing subsequence (LDS)
ArrayList<Integer> left
= new ArrayList<>(Collections.nCopies(n, 1));
ArrayList<Integer> right
= new ArrayList<>(Collections.nCopies(n, 1));
// Fill left ArrayList for LIS
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
// If arr[i] is greater than arr[j],
// update LIS value at i
if (arr.get(i) > arr.get(j)) {
left.set(i, Math.max(left.get(i),
left.get(j) + 1));
}
}
}
// Fill right ArrayList for LDS
for (int i = n - 2; i >= 0; i--) {
// Compare each element with subsequent
// ones to build LDS
for (int j = n - 1; j > i; j--) {
// If arr[i] is greater than arr[j],
// update LDS value at i
if (arr.get(i) > arr.get(j)) {
right.set(i, Math.max(right.get(i),
right.get(j) + 1));
}
}
}
// Calculate the maximum length of bitonic subsequence
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Check if both LIS and LDS are valid
// for the current index
if (left.get(i) > 1 && right.get(i) > 1) {
// Update maxLength considering both LIS
// and LDS, subtracting 1 for the peak element
maxLength = Math.max(maxLength,
left.get(i) + right.get(i) - 1);
}
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
public static void main(String[] args) {
ArrayList<Integer> arr
= new ArrayList<>(Arrays.asList(12, 11, 40, 5, 3, 1));
System.out.println(LongestBitonicSequence(arr));
}
}
# Python implementation to find the longest
# bitonic subsequence using tabulation
def LongestBitonicSequence(arr):
n = len(arr)
# If there are less than 3 elements,
# no bitonic subsequence exists
if n < 3:
return 0
# Create lists for longest increasing subsequence
# (LIS) and longest decreasing subsequence (LDS)
left = [1] * n
right = [1] * n
# Fill left list for LIS
for i in range(1, n):
for j in range(i):
# If arr[i] is greater than arr[j],
# update LIS value at i
if arr[i] > arr[j]:
left[i] = max(left[i], left[j] + 1)
# Fill right list for LDS
for i in range(n - 2, -1, -1):
# Compare each element with subsequent
# ones to build LDS
for j in range(n - 1, i, -1):
# If arr[i] is greater than arr[j],
# update LDS value at i
if arr[i] > arr[j]:
right[i] = max(right[i], right[j] + 1)
# Calculate the maximum length of bitonic subsequence
maxLength = 0
for i in range(n):
# Check if both LIS and LDS are valid
# for the current index
if left[i] > 1 and right[i] > 1:
# Update maxLength considering both LIS
# and LDS, subtracting 1 for the peak element
maxLength = max(maxLength, left[i] + right[i] - 1)
# If no valid bitonic sequence, return 0
return maxLength if maxLength >= 3 else 0
if __name__ == "__main__":
arr = [12, 11, 40, 5, 3, 1]
print(LongestBitonicSequence(arr))
// C# implementation to find the longest
// bitonic subsequence using tabulation
using System;
using System.Collections.Generic;
class GfG {
static int LongestBitonicSequence(List<int> arr) {
int n = arr.Count;
// If there are less than 3 elements,
// no bitonic subsequence exists
if (n < 3) return 0;
// Create lists for longest increasing subsequence
// (LIS) and longest decreasing subsequence (LDS)
List<int> left = new List<int>(new int[n]);
List<int> right = new List<int>(new int[n]);
// Initialize lists with 1
for (int i = 0; i < n; i++) {
left[i] = 1;
right[i] = 1;
}
// Fill left list for LIS
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
// If arr[i] is greater than arr[j],
// update LIS value at i
if (arr[i] > arr[j]) {
left[i] = Math.Max(left[i], left[j] + 1);
}
}
}
// Fill right list for LDS
for (int i = n - 2; i >= 0; i--) {
// Compare each element with subsequent
// ones to build LDS
for (int j = n - 1; j > i; j--) {
// If arr[i] is greater than arr[j],
// update LDS value at i
if (arr[i] > arr[j]) {
right[i] = Math.Max(right[i], right[j] + 1);
}
}
}
// Calculate the maximum length of bitonic subsequence
int maxLength = 0;
for (int i = 0; i < n; i++) {
// Check if both LIS and LDS are valid
// for the current index
if (left[i] > 1 && right[i] > 1) {
// Update maxLength considering both LIS
// and LDS, subtracting 1 for the peak element
maxLength = Math.Max(maxLength, left[i] + right[i] - 1);
}
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
static void Main() {
List<int> arr = new List<int> { 12, 11, 40, 5, 3, 1 };
Console.WriteLine(LongestBitonicSequence(arr));
}
}
// Javascript implementation to find the longest
// bitonic subsequence using tabulation
function LongestBitonicSequence(arr) {
const n = arr.length;
// If there are less than 3 elements,
// no bitonic subsequence exists
if (n < 3) return 0;
// Create arrays for longest increasing subsequence
// (LIS) and longest decreasing subsequence (LDS)
const left = Array(n).fill(1);
const right = Array(n).fill(1);
// Fill left array for LIS
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
// If arr[i] is greater than arr[j],
// update LIS value at i
if (arr[i] > arr[j]) {
left[i] = Math.max(left[i], left[j] + 1);
}
}
}
// Fill right array for LDS
for (let i = n - 2; i >= 0; i--) {
// Compare each element with subsequent
// ones to build LDS
for (let j = n - 1; j > i; j--) {
// If arr[i] is greater than arr[j],
// update LDS value at i
if (arr[i] > arr[j]) {
right[i] = Math.max(right[i], right[j] + 1);
}
}
}
// Calculate the maximum length of bitonic subsequence
let maxLength = 0;
for (let i = 0; i < n; i++) {
// Check if both LIS and LDS are valid
// for the current index
if (left[i] > 1 && right[i] > 1) {
// Update maxLength considering both LIS
// and LDS, subtracting 1 for the peak element
maxLength = Math.max(maxLength, left[i] + right[i] - 1);
}
}
// If no valid bitonic sequence, return 0
return maxLength < 3 ? 0 : maxLength;
}
const arr = [12, 11, 40, 5, 3, 1];
console.log(LongestBitonicSequence(arr));
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