Longest subsequence such that difference between adjacents is one
Last Updated :
29 Nov, 2024
Given an array arr[] of size n, the task is to find the longest subsequence such that the absolute difference between adjacent elements is 1.
Examples:
Input: arr[] = [10, 9, 4, 5, 4, 8, 6]
Output: 3
Explanation: The three possible subsequences of length 3 are [10, 9, 8], [4, 5, 4], and [4, 5, 6], where adjacent elements have a absolute difference of 1. No valid subsequence of greater length could be formed.
Input: arr[] = [1, 2, 3, 4, 5]
Output: 5
Explanation: All the elements can be included in the valid subsequence.
Using Recursion - O(2^n) Time and O(n) Space
For the recursive approach, we will consider two cases at each step:
- If the element satisfies the condition (the absolute difference between adjacent elements is 1), we include it in the subsequence and move on to the next element.
- else, we skip the current element and move on to the next one.
Mathematically, the recurrence relation will look like the following:
- longestSubseq(arr, idx, prev) = max(longestSubseq(arr, idx + 1, prev), 1 + longestSubseq(arr, idx + 1, idx))
Base Case:
- When idx == arr.size(), we have reached the end of the array, so return 0 (since no more elements can be included).
C++
// C++ program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion.
#include <bits/stdc++.h>
using namespace std;
int subseqHelper(int idx, int prev, vector<int>& arr) {
// Base case: if index reaches the end of the array
if (idx == arr.size()) {
return 0;
}
// Skip the current element and move to the next index
int noTake = subseqHelper(idx + 1, prev, arr);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || abs(arr[idx] - arr[prev]) == 1) {
take = 1 + subseqHelper(idx + 1, idx, arr);
}
// Return the maximum of the two options
return max(take, noTake);
}
// Function to find the longest subsequence
int longestSubseq(vector<int>& arr) {
// Start recursion from index 0
// with no previous element
return subseqHelper(0, -1, arr);
}
int main() {
vector<int> arr = {10, 9, 4, 5, 4, 8, 6};
cout << longestSubseq(arr);
return 0;
}
// Java program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion.
import java.util.ArrayList;
class GfG {
// Helper function to recursively find the subsequence
static int subseqHelper(int idx, int prev,
ArrayList<Integer> arr) {
// Base case: if index reaches the end of the array
if (idx == arr.size()) {
return 0;
}
// Skip the current element and move to the next index
int noTake = subseqHelper(idx + 1, prev, arr);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || Math.abs(arr.get(idx)
- arr.get(prev)) == 1) {
take = 1 + subseqHelper(idx + 1, idx, arr);
}
// Return the maximum of the two options
return Math.max(take, noTake);
}
// Function to find the longest subsequence
static int longestSubseq(ArrayList<Integer> arr) {
// Start recursion from index 0
// with no previous element
return subseqHelper(0, -1, arr);
}
public static void main(String[] args) {
ArrayList<Integer> arr = new ArrayList<>();
arr.add(10);
arr.add(9);
arr.add(4);
arr.add(5);
arr.add(4);
arr.add(8);
arr.add(6);
System.out.println(longestSubseq(arr));
}
}
# Python program to find the longest subsequence such that
# the difference between adjacent elements is one using
# recursion.
def subseq_helper(idx, prev, arr):
# Base case: if index reaches the end of the array
if idx == len(arr):
return 0
# Skip the current element and move to the next index
no_take = subseq_helper(idx + 1, prev, arr)
# Take the current element if the condition is met
take = 0
if prev == -1 or abs(arr[idx] - arr[prev]) == 1:
take = 1 + subseq_helper(idx + 1, idx, arr)
# Return the maximum of the two options
return max(take, no_take)
def longest_subseq(arr):
# Start recursion from index 0
# with no previous element
return subseq_helper(0, -1, arr)
if __name__ == "__main__":
arr = [10, 9, 4, 5, 4, 8, 6]
print(longest_subseq(arr))
// C# program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion.
using System;
using System.Collections.Generic;
class GfG {
// Helper function to recursively find the subsequence
static int SubseqHelper(int idx, int prev,
List<int> arr) {
// Base case: if index reaches the end of the array
if (idx == arr.Count) {
return 0;
}
// Skip the current element and move to the next index
int noTake = SubseqHelper(idx + 1, prev, arr);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || Math.Abs(arr[idx] - arr[prev]) == 1) {
take = 1 + SubseqHelper(idx + 1, idx, arr);
}
// Return the maximum of the two options
return Math.Max(take, noTake);
}
// Function to find the longest subsequence
static int LongestSubseq(List<int> arr) {
// Start recursion from index 0
// with no previous element
return SubseqHelper(0, -1, arr);
}
static void Main(string[] args) {
List<int> arr
= new List<int> { 10, 9, 4, 5, 4, 8, 6 };
Console.WriteLine(LongestSubseq(arr));
}
}
// JavaScript program to find the longest subsequence
// such that the difference between adjacent elements
// is one using recursion.
function subseqHelper(idx, prev, arr) {
// Base case: if index reaches the end of the array
if (idx === arr.length) {
return 0;
}
// Skip the current element and move to the next index
let noTake = subseqHelper(idx + 1, prev, arr);
// Take the current element if the condition is met
let take = 0;
if (prev === -1 || Math.abs(arr[idx] - arr[prev]) === 1) {
take = 1 + subseqHelper(idx + 1, idx, arr);
}
// Return the maximum of the two options
return Math.max(take, noTake);
}
function longestSubseq(arr) {
// Start recursion from index 0
// with no previous element
return subseqHelper(0, -1, arr);
}
const arr = [10, 9, 4, 5, 4, 8, 6];
console.log(longestSubseq(arr));
Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: The solution to finding the longest subsequence such that the difference between adjacent elements is one can be derived from the optimal solutions of smaller subproblems. Specifically, for any given idx (current index) and prev (previous index in the subsequence), we can express the recursive relation as follows:
- subseqHelper(idx, prev) = max(subseqHelper(idx + 1, prev), 1 + subseqHelper(idx + 1, idx))
2. Overlapping Subproblems: When implementing a recursive approach to solve the problem, we observe that many subproblems are computed multiple times. For example, when computing subseqHelper(0, -1) for an array arr = [10, 9, 4, 5], the subproblem subseqHelper(2, -1) may be computed multiple times. To avoid this repetition, we use memoization to store the results of previously computed subproblems.
The recursive solution involves two parameters:
- idx (the current index in the array).
- prev (the index of the last included element in the subsequence).
We need to track both parameters, so we create a 2D array memo of size (n) x (n+1). We initialize the 2D array memo with -1 to indicate that no subproblems have been computed yet. Before computing a result, we check if the value at memo[idx][prev+1] is -1. If it is, we compute and store the result. Otherwise, we return the stored result.
C++
// C++ program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion with memoization.
#include <bits/stdc++.h>
using namespace std;
// Helper function to recursively find the subsequence
int subseqHelper(int idx, int prev, vector<int>& arr,
vector<vector<int>>& memo) {
// Base case: if index reaches the end of the array
if (idx == arr.size()) {
return 0;
}
// Check if the result is already computed
if (memo[idx][prev + 1] != -1) {
return memo[idx][prev + 1];
}
// Skip the current element and move to the next index
int noTake = subseqHelper(idx + 1, prev, arr, memo);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || abs(arr[idx] - arr[prev]) == 1) {
take = 1 + subseqHelper(idx + 1, idx, arr, memo);
}
// Store the result in the memo table
return memo[idx][prev + 1] = max(take, noTake);
}
// Function to find the longest subsequence
int longestSubseq(vector<int>& arr) {
int n = arr.size();
// Create a memoization table initialized to -1
vector<vector<int>> memo(n, vector<int>(n + 1, -1));
// Start recursion from index 0 with no previous element
return subseqHelper(0, -1, arr, memo);
}
int main() {
// Input array of integers
vector<int> arr = {10, 9, 4, 5, 4, 8, 6};
cout << longestSubseq(arr);
return 0;
}
// Java program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion with memoization.
import java.util.ArrayList;
import java.util.Arrays;
class GfG {
// Helper function to recursively find the subsequence
static int subseqHelper(int idx, int prev,
ArrayList<Integer> arr,
int[][] memo) {
// Base case: if index reaches the end of the array
if (idx == arr.size()) {
return 0;
}
// Check if the result is already computed
if (memo[idx][prev + 1] != -1) {
return memo[idx][prev + 1];
}
// Skip the current element and move to the next index
int noTake = subseqHelper(idx + 1, prev, arr, memo);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || Math.abs(arr.get(idx)
- arr.get(prev)) == 1) {
take = 1 + subseqHelper(idx + 1, idx, arr, memo);
}
// Store the result in the memo table
memo[idx][prev + 1] = Math.max(take, noTake);
// Return the stored result
return memo[idx][prev + 1];
}
// Function to find the longest subsequence
static int longestSubseq(ArrayList<Integer> arr) {
int n = arr.size();
// Create a memoization table initialized to -1
int[][] memo = new int[n][n + 1];
for (int[] row : memo) {
Arrays.fill(row, -1);
}
// Start recursion from index 0
// with no previous element
return subseqHelper(0, -1, arr, memo);
}
public static void main(String[] args) {
ArrayList<Integer> arr = new ArrayList<>();
arr.add(10);
arr.add(9);
arr.add(4);
arr.add(5);
arr.add(4);
arr.add(8);
arr.add(6);
System.out.println(longestSubseq(arr));
}
}
# Python program to find the longest subsequence such that
# the difference between adjacent elements is one using
# recursion with memoization.
def subseq_helper(idx, prev, arr, memo):
# Base case: if index reaches the end of the array
if idx == len(arr):
return 0
# Check if the result is already computed
if memo[idx][prev + 1] != -1:
return memo[idx][prev + 1]
# Skip the current element and move to the next index
no_take = subseq_helper(idx + 1, prev, arr, memo)
# Take the current element if the condition is met
take = 0
if prev == -1 or abs(arr[idx] - arr[prev]) == 1:
take = 1 + subseq_helper(idx + 1, idx, arr, memo)
# Store the result in the memo table
memo[idx][prev + 1] = max(take, no_take)
# Return the stored result
return memo[idx][prev + 1]
def longest_subseq(arr):
n = len(arr)
# Create a memoization table initialized to -1
memo = [[-1 for _ in range(n + 1)] for _ in range(n)]
# Start recursion from index 0 with
# no previous element
return subseq_helper(0, -1, arr, memo)
if __name__ == "__main__":
arr = [10, 9, 4, 5, 4, 8, 6]
print(longest_subseq(arr))
// C# program to find the longest subsequence such that
// the difference between adjacent elements is one using
// recursion with memoization.
using System;
using System.Collections.Generic;
class GfG {
// Helper function to recursively find the subsequence
static int SubseqHelper(int idx, int prev,
List<int> arr, int[,] memo) {
// Base case: if index reaches the end of the array
if (idx == arr.Count) {
return 0;
}
// Check if the result is already computed
if (memo[idx, prev + 1] != -1) {
return memo[idx, prev + 1];
}
// Skip the current element and move to the next index
int noTake = SubseqHelper(idx + 1, prev, arr, memo);
// Take the current element if the condition is met
int take = 0;
if (prev == -1 || Math.Abs(arr[idx] - arr[prev]) == 1) {
take = 1 + SubseqHelper(idx + 1, idx, arr, memo);
}
// Store the result in the memoization table
memo[idx, prev + 1] = Math.Max(take, noTake);
// Return the stored result
return memo[idx, prev + 1];
}
// Function to find the longest subsequence
static int LongestSubseq(List<int> arr) {
int n = arr.Count;
// Create a memoization table initialized to -1
int[,] memo = new int[n, n + 1];
for (int i = 0; i < n; i++) {
for (int j = 0; j <= n; j++) {
memo[i, j] = -1;
}
}
// Start recursion from index 0 with no previous element
return SubseqHelper(0, -1, arr, memo);
}
static void Main(string[] args) {
List<int> arr
= new List<int> { 10, 9, 4, 5, 4, 8, 6 };
Console.WriteLine(LongestSubseq(arr));
}
}
// JavaScript program to find the longest subsequence
// such that the difference between adjacent elements
// is one using recursion with memoization.
function subseqHelper(idx, prev, arr, memo) {
// Base case: if index reaches the end of the array
if (idx === arr.length) {
return 0;
}
// Check if the result is already computed
if (memo[idx][prev + 1] !== -1) {
return memo[idx][prev + 1];
}
// Skip the current element and move to the next index
let noTake = subseqHelper(idx + 1, prev, arr, memo);
// Take the current element if the condition is met
let take = 0;
if (prev === -1 || Math.abs(arr[idx] - arr[prev]) === 1) {
take = 1 + subseqHelper(idx + 1, idx, arr, memo);
}
// Store the result in the memoization table
memo[idx][prev + 1] = Math.max(take, noTake);
// Return the stored result
return memo[idx][prev + 1];
}
function longestSubseq(arr) {
let n = arr.length;
// Create a memoization table initialized to -1
let memo =
Array.from({ length: n }, () => Array(n + 1).fill(-1));
// Start recursion from index 0 with no previous element
return subseqHelper(0, -1, arr, memo);
}
const arr = [10, 9, 4, 5, 4, 8, 6];
console.log(longestSubseq(arr));
Using Bottom-Up DP (Tabulation) - O(n) Time and O(n) Space
The approach is similar to the recursive method, but instead of breaking down the problem recursively, we iteratively build the solution in a bottom-up manner.
Instead of using recursion, we utilize a hashmap based dynamic programming table (dp) to store the lengths of the longest subsequences. This helps us efficiently calculate and update the subsequence lengths for all possible values of array elements.
Dynamic Programming Relation:
dp[x] represents the length of the longest subsequence ending with the element x.
For every element arr[i] in the array: If arr[i] + 1 or arr[i] - 1 exists in dp:
- dp[arr[i]] = 1 + max(dp[arr[i] + 1], dp[arr[i] - 1]);
This means we can extend the subsequences ending with arr[i] + 1 or arr[i] - 1 by including arr[i].
Otherwise, start a new subsequence:
C++
// C++ program to find the longest subsequence such that
// the difference between adjacent elements is one using
// Tabulation.
#include <bits/stdc++.h>
using namespace std;
int longestSubseq(vector<int>& arr) {
int n = arr.size();
// Base case: if the array has only
// one element
if (n == 1) {
return 1;
}
// Map to store the length of the longest subsequence
unordered_map<int, int> dp;
int ans = 1;
// Loop through the array to fill the map
// with subsequence lengths
for (int i = 0; i < n; ++i) {
// Check if the current element is adjacent
// to another subsequence
if (dp.count(arr[i] + 1) > 0
|| dp.count(arr[i] - 1) > 0) {
dp[arr[i]] = 1 +
max(dp[arr[i] + 1], dp[arr[i] - 1]);
}
else {
dp[arr[i]] = 1;
}
// Update the result with the maximum
// subsequence length
ans = max(ans, dp[arr[i]]);
}
return ans;
}
int main() {
vector<int> arr = {10, 9, 4, 5, 4, 8, 6};
cout << longestSubseq(arr);
return 0;
}
// Java code to find the longest subsequence such that
// the difference between adjacent elements
// is one using Tabulation.
import java.util.HashMap;
import java.util.ArrayList;
class GfG {
static int longestSubseq(ArrayList<Integer> arr) {
int n = arr.size();
// Base case: if the array has only one element
if (n == 1) {
return 1;
}
// Map to store the length of the longest subsequence
HashMap<Integer, Integer> dp = new HashMap<>();
int ans = 1;
// Loop through the array to fill the map
// with subsequence lengths
for (int i = 0; i < n; ++i) {
// Check if the current element is adjacent
// to another subsequence
if (dp.containsKey(arr.get(i) + 1)
|| dp.containsKey(arr.get(i) - 1)) {
dp.put(arr.get(i), 1 +
Math.max(dp.getOrDefault(arr.get(i) + 1, 0),
dp.getOrDefault(arr.get(i) - 1, 0)));
}
else {
dp.put(arr.get(i), 1);
}
// Update the result with the maximum
// subsequence length
ans = Math.max(ans, dp.get(arr.get(i)));
}
return ans;
}
public static void main(String[] args) {
ArrayList<Integer> arr = new ArrayList<>();
arr.add(10);
arr.add(9);
arr.add(4);
arr.add(5);
arr.add(4);
arr.add(8);
arr.add(6);
System.out.println(longestSubseq(arr));
}
}
# Python code to find the longest subsequence such that
# the difference between adjacent elements is
# one using Tabulation.
def longestSubseq(arr):
n = len(arr)
# Base case: if the array has only one element
if n == 1:
return 1
# Dictionary to store the length of the
# longest subsequence
dp = {}
ans = 1
for i in range(n):
# Check if the current element is adjacent to
# another subsequence
if arr[i] + 1 in dp or arr[i] - 1 in dp:
dp[arr[i]] = 1 + max(dp.get(arr[i] + 1, 0), \
dp.get(arr[i] - 1, 0))
else:
dp[arr[i]] = 1
# Update the result with the maximum
# subsequence length
ans = max(ans, dp[arr[i]])
return ans
if __name__ == "__main__":
arr = [10, 9, 4, 5, 4, 8, 6]
print(longestSubseq(arr))
// C# code to find the longest subsequence such that
// the difference between adjacent elements
// is one using Tabulation.
using System;
using System.Collections.Generic;
class GfG {
static int longestSubseq(List<int> arr) {
int n = arr.Count;
// Base case: if the array has only one element
if (n == 1) {
return 1;
}
// Map to store the length of the longest subsequence
Dictionary<int, int> dp = new Dictionary<int, int>();
int ans = 1;
// Loop through the array to fill the map with
// subsequence lengths
for (int i = 0; i < n; ++i) {
// Check if the current element is adjacent to
// another subsequence
if (dp.ContainsKey(arr[i] + 1) || dp.ContainsKey(arr[i] - 1)) {
dp[arr[i]] = 1 + Math.Max(dp.GetValueOrDefault(arr[i] + 1, 0),
dp.GetValueOrDefault(arr[i] - 1, 0));
}
else {
dp[arr[i]] = 1;
}
// Update the result with the maximum
// subsequence length
ans = Math.Max(ans, dp[arr[i]]);
}
return ans;
}
static void Main(string[] args) {
List<int> arr
= new List<int> { 10, 9, 4, 5, 4, 8, 6 };
Console.WriteLine(longestSubseq(arr));
}
}
// Function to find the longest subsequence such that
// the difference between adjacent elements
// is one using Tabulation.
function longestSubseq(arr) {
const n = arr.length;
// Base case: if the array has only one element
if (n === 1) {
return 1;
}
// Object to store the length of the
// longest subsequence
let dp = {};
let ans = 1;
// Loop through the array to fill the object
// with subsequence lengths
for (let i = 0; i < n; i++) {
// Check if the current element is adjacent to
// another subsequence
if ((arr[i] + 1) in dp || (arr[i] - 1) in dp) {
dp[arr[i]] = 1 + Math.max(dp[arr[i] + 1]
|| 0, dp[arr[i] - 1] || 0);
} else {
dp[arr[i]] = 1;
}
// Update the result with the maximum
// subsequence length
ans = Math.max(ans, dp[arr[i]]);
}
return ans;
}
const arr = [10, 9, 4, 5, 4, 8, 6];
console.log(longestSubseq(arr));
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