Longest Bitonic Subsequence
Last Updated :
23 Jul, 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));
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem