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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Longest Common SubstringGiven two strings 's1' and 's2', find the length of the longest common substring. Example: Input: s1 = "GeeksforGeeks", s2 = "GeeksQuiz" Output : 5 Explanation:The longest common substring is "Geeks" and is of length 5.Input: s1 = "abcdxyz", s2 = "xyzabcd" Output : 4Explanation:The longest common su
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Longest Common Subsequence of two arrays out of which one array consists of distinct elements onlyGiven two arrays firstArr[], consisting of distinct elements only, and secondArr[], the task is to find the length of LCS between these 2 arrays. Examples: Input: firstArr[] = {3, 5, 1, 8}, secondArr[] = {3, 3, 5, 3, 8}Output: 3.Explanation: LCS between these two arrays is {3, 5, 8}. Input : firstAr
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Longest Repeating SubsequenceGiven a string s, the task is to find the length of the longest repeating subsequence, such that the two subsequences don't have the same string character at the same position, i.e. any ith character in the two subsequences shouldn't have the same index in the original string. Examples:Input: s= "ab
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Longest Common Anagram SubsequenceGiven two strings str1 and str2 of length n1 and n2 respectively. The problem is to find the length of the longest subsequence which is present in both the strings in the form of anagrams. Note: The strings contain only lowercase letters. Examples: Input : str1 = "abdacp", str2 = "ckamb" Output : 3
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Length of Longest Common Subsequence with given sum KGiven two arrays a[] and b[] and an integer K, the task is to find the length of the longest common subsequence such that sum of elements is equal to K. Examples: Input: a[] = { 9, 11, 2, 1, 6, 2, 7}, b[] = {1, 2, 6, 9, 2, 3, 11, 7}, K = 18Output: 3Explanation: Subsequence { 11, 7 } and { 9, 2, 7 }
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Longest Common Subsequence with no repeating characterGiven two strings s1 and s2, the task is to find the length of the longest common subsequence with no repeating character. Examples: Input: s1= "aabbcc", s2= "aabc"Output: 3Explanation: "aabc" is longest common subsequence but it has two repeating character 'a'.So the required longest common subsequ
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Find the Longest Common Subsequence (LCS) in given K permutationsGiven K permutations of numbers from 1 to N in a 2D array arr[][]. The task is to find the longest common subsequence of these K permutations. Examples: Input: N = 4, K = 3arr[][] = {{1, 4, 2, 3}, {4, 1, 2, 3}, {1, 2, 4, 3}}Output: 3Explanation: Longest common subsequence is {1, 2, 3} which has leng
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Find length of longest subsequence of one string which is substring of another stringGiven two strings X and Y. The task is to find the length of the longest subsequence of string X which is a substring in sequence Y.Examples: Input : X = "ABCD", Y = "BACDBDCD"Output : 3Explanation: "ACD" is longest subsequence of X which is substring of Y.Input : X = "A", Y = "A"Output : 1Perquisit
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Length of longest common prime subsequence from two given arraysGiven two arrays arr1[] and arr2[] of length N and M respectively, the task is to find the length of the longest common prime subsequence that can be obtained from the two given arrays. Examples: Input: arr1[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, arr2[] = {2, 5, 6, 3, 7, 9, 8} Output: 4 Explanation: The l
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A Space Optimized Solution of LCSGiven two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0.Examples:Input: s1 = “ABCDGHâ€, s2 = “AEDFHRâ€Output: 3Explanation: The longest subsequence present in both strings is "ADH".Input: s1 = “AGGTABâ€, s2 = “GXTXAYBâ€O
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Longest common subarray in the given two arraysGiven two arrays A[] and B[] of N and M integers respectively, the task is to find the maximum length of an equal subarray or the longest common subarray between the two given array. Examples: Input: A[] = {1, 2, 8, 2, 1}, B[] = {8, 2, 1, 4, 7} Output: 3 Explanation: The subarray that is common to b
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Number of ways to insert a character to increase the LCS by oneGiven two strings A and B. The task is to count the number of ways to insert a character in string A to increase the length of the Longest Common Subsequence between string A and string B by 1. Examples: Input : A = "aa", B = "baaa" Output : 4 The longest common subsequence shared by string A and st
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Longest common subsequence with permutations allowedGiven two strings in lowercase, find the longest string whose permutations are subsequences of given two strings. The output longest string must be sorted. Examples: Input : str1 = "pink", str2 = "kite" Output : "ik" The string "ik" is the longest sorted string whose one permutation "ik" is subseque
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Longest subsequence such that adjacent elements have at least one common digitGiven an array arr[], the task is to find the length of the longest sub-sequence such that adjacent elements of the subsequence have at least one digit in common.Examples: Input: arr[] = [1, 12, 44, 29, 33, 96, 89] Output: 5 Explanation: The longest sub-sequence is [1 12 29 96 89]Input: arr[] = [12,
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Longest subsequence with different adjacent charactersGiven string str. The task is to find the longest subsequence of str such that all the characters adjacent to each other in the subsequence are different. Examples:Â Â Input: str = "ababa"Â Output: 5Â Explanation:Â "ababa" is the subsequence satisfying the condition Input: str = "xxxxy"Â Output: 2Â Explan
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Longest subsequence such that difference between adjacents is oneGiven 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: 3Explanation: The three possible subsequences of length 3 are [10, 9, 8], [4, 5, 4], and [4, 5, 6], wher
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Longest Uncommon SubsequenceGiven two strings, find the length of longest uncommon subsequence of the two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings which is not a subsequence of other strings. Examples: Input : "abcd", "abc"Output : 4The longest subsequence is 4 bec
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LCS formed by consecutive segments of at least length KGiven two strings s1, s2 and K, find the length of the longest subsequence formed by consecutive segments of at least length K. Examples: Input : s1 = aggayxysdfa s2 = aggajxaaasdfa k = 4 Output : 8 Explanation: aggasdfa is the longest subsequence that can be formed by taking consecutive segments, m
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Longest Increasing Subsequence using Longest Common Subsequence AlgorithmGiven an array arr[] of N integers, the task is to find and print the Longest Increasing Subsequence.Examples: Input: arr[] = {12, 34, 1, 5, 40, 80} Output: 4 {12, 34, 40, 80} and {1, 5, 40, 80} are the longest increasing subsequences.Input: arr[] = {10, 22, 9, 33, 21, 50, 41, 60, 80} Output: 6 Prer
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