Longest subsequence with different adjacent characters
Last Updated :
12 Jul, 2025
Given 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:
Input: str = "ababa"
Output: 5
Explanation:
"ababa" is the subsequence satisfying the condition
Input: str = "xxxxy"
Output: 2
Explanation:
"xy" is the subsequence satisfying the condition
Method 1: Greedy Approach
It can be observed that choosing the first character which is not similar to the previously chosen character given the longest subsequence of the given string with different adjacent characters.
The idea is to keep track of previously picked characters while iterating through the string, and if the current character is different from the previous character, then count the current character to find the longest subsequence.
Implementation:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the longest Subsequence
// with different adjacent character
int longestSubsequence(string s)
{
// Length of the string s
int n = s.length();
int answer = 0;
// Previously picked character
char prev = '-';
for (int i = 0; i < n; i++) {
// If the current character is
// different from the previous
// then include this character
// and update previous character
if (prev != s[i]) {
prev = s[i];
answer++;
}
}
return answer;
}
// Driver Code
int main()
{
string str = "ababa";
// Function call
cout << longestSubsequence(str);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the longest subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
// Length of the String s
int n = s.length();
int answer = 0;
// Previously picked character
char prev = '-';
for (int i = 0; i < n; i++) {
// If the current character is
// different from the previous
// then include this character
// and update previous character
if (prev != s.charAt(i)) {
prev = s.charAt(i);
answer++;
}
}
return answer;
}
// Driver Code
public static void main(String[] args)
{
String str = "ababa";
// Function call
System.out.print(longestSubsequence(str));
}
}
// This code is contributed by sapnasingh4991
# Python3 program for the above approach
# Function to find the longest Subsequence
# with different adjacent character
def longestSubsequence(s):
# Length of the string s
n = len(s)
answer = 0
# Previously picked character
prev = '-'
for i in range(0, n):
# If the current character is
# different from the previous
# then include this character
# and update previous character
if (prev != s[i]):
prev = s[i]
answer += 1
return answer
# Driver Code
str = "ababa"
# Function call
print(longestSubsequence(str))
# This code is contributed by Code_Mech
// C# program for the above approach
using System;
class GFG {
// Function to find the longest subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
// Length of the String s
int n = s.Length;
int answer = 0;
// Previously picked character
char prev = '-';
for (int i = 0; i < n; i++) {
// If the current character is
// different from the previous
// then include this character
// and update previous character
if (prev != s[i]) {
prev = s[i];
answer++;
}
}
return answer;
}
// Driver Code
public static void Main(String[] args)
{
String str = "ababa";
// Function call
Console.Write(longestSubsequence(str));
}
}
// This code is contributed by amal kumar choubey
<script>
// Javascript program for the above approach
// Function to find the longest Subsequence
// with different adjacent character
function longestSubsequence(s)
{
// Length of the string s
var n = s.length;
var answer = 0;
// Previously picked character
var prev = '-';
for (var i = 0; i < n; i++) {
// If the current character is
// different from the previous
// then include this character
// and update previous character
if (prev != s[i]) {
prev = s[i];
answer++;
}
}
return answer;
}
// Driver Code
var str = "ababa";
// Function call
document.write( longestSubsequence(str));
</script>
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Method 2: Dynamic Programming
- For each character in the given string str, do the following:
- Choose the current characters in the string for the resultant subsequence and recur for the remaining string to find the next possible characters for the resultant subsequence.
- Omit the current characters and recur for the remaining string to find the next possible characters for the resultant subsequence.
- The maximum value in the above recursive call will be the longest subsequence with different adjacent elements.
- The recurrence relation is given by:
Let dp[pos][prev] be the length of longest subsequence
till index pos such that alphabet prev was picked previously.a dp[pos][prev] = max(1 + function(pos+1, s[pos] - 'a' + 1, s),
function(pos+1, prev, s));
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// dp table
int dp[100005][27];
// A recursive function to find the
// update the dp[][] table
int calculate(int pos, int prev, string& s)
{
// If we reach end of the string
if (pos == s.length()) {
return 0;
}
// If subproblem has been computed
if (dp[pos][prev] != -1)
return dp[pos][prev];
// Initialise variable to find the
// maximum length
int val = 0;
// Choose the current character
if (s[pos] - 'a' + 1 != prev) {
val = max(val,
1 + calculate(pos + 1,
s[pos] - 'a' + 1,
s));
}
// Omit the current character
val = max(val, calculate(pos + 1, prev, s));
// Return the store answer to the
// current subproblem
return dp[pos][prev] = val;
}
// Function to find the longest Subsequence
// with different adjacent character
int longestSubsequence(string s)
{
// Length of the string s
int n = s.length();
// Initialise the memoisation table
memset(dp, -1, sizeof(dp));
// Return the final ans after every
// recursive call
return calculate(0, 0, s);
}
// Driver Code
int main()
{
string str = "ababa";
// Function call
cout << longestSubsequence(str);
return 0;
}
// Java program for the above approach
class GFG{
// dp table
static int dp[][] = new int[100005][27];
// A recursive function to find the
// update the dp[][] table
static int calculate(int pos, int prev, String s)
{
// If we reach end of the String
if (pos == s.length())
{
return 0;
}
// If subproblem has been computed
if (dp[pos][prev] != -1)
return dp[pos][prev];
// Initialise variable to find the
// maximum length
int val = 0;
// Choose the current character
if (s.charAt(pos) - 'a' + 1 != prev)
{
val = Math.max(val, 1 + calculate(pos + 1,
s.charAt(pos) - 'a' + 1,
s));
}
// Omit the current character
val = Math.max(val, calculate(pos + 1, prev, s));
// Return the store answer to the
// current subproblem
return dp[pos][prev] = val;
}
// Function to find the longest Subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
// Length of the String s
int n = s.length();
// Initialise the memoisation table
for(int i = 0; i < 100005; i++)
{
for (int j = 0; j < 27; j++)
{
dp[i][j] = -1;
}
}
// Return the final ans after every
// recursive call
return calculate(0, 0, s);
}
// Driver Code
public static void main(String[] args)
{
String str = "ababa";
// Function call
System.out.print(longestSubsequence(str));
}
}
// This code is contributed by Rohit_ranjan
# Python3 program for the above approach
# dp table
dp = [[-1 for i in range(27)] for j in range(100005)];
# A recursive function to find the
# update the dp table
def calculate(pos, prev, s):
# If we reach end of the String
if (pos == len(s)):
return 0;
# If subproblem has been computed
if (dp[pos][prev] != -1):
return dp[pos][prev];
# Initialise variable to find the
# maximum length
val = 0;
# Choose the current character
if (ord(s[pos]) - ord('a') + 1 != prev):
val = max(val, 1 + calculate(pos + 1,
ord(s[pos]) -
ord('a') + 1, s));
# Omit the current character
val = max(val, calculate(pos + 1, prev, s));
# Return the store answer to
# the current subproblem
dp[pos][prev] = val;
return dp[pos][prev];
# Function to find the longest Subsequence
# with different adjacent character
def longestSubsequence(s):
# Length of the String s
n = len(s);
# Return the final ans after every
# recursive call
return calculate(0, 0, s);
# Driver Code
if __name__ == '__main__':
str = "ababa";
# Function call
print(longestSubsequence(str));
# This code is contributed by shikhasingrajput
// C# program for the above approach
using System;
public class GFG{
// dp table
static int [,]dp = new int[100005,27];
// A recursive function to find the
// update the [,]dp table
static int calculate(int pos, int prev, String s)
{
// If we reach end of the String
if (pos == s.Length)
{
return 0;
}
// If subproblem has been computed
if (dp[pos,prev] != -1)
return dp[pos,prev];
// Initialise variable to
// find the maximum length
int val = 0;
// Choose the current character
if (s[pos] - 'a' + 1 != prev)
{
val = Math.Max(val, 1 +
calculate(pos + 1,
s[pos] - 'a' + 1,
s));
}
// Omit the current character
val = Math.Max(val, calculate(pos + 1, prev, s));
// Return the store answer to the
// current subproblem
return dp[pos,prev] = val;
}
// Function to find the longest Subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
// Length of the String s
int n = s.Length;
// Initialise the memoisation table
for(int i = 0; i < 100005; i++)
{
for (int j = 0; j < 27; j++)
{
dp[i,j] = -1;
}
}
// Return the readonly ans after every
// recursive call
return calculate(0, 0, s);
}
// Driver Code
public static void Main(String[] args)
{
String str = "ababa";
// Function call
Console.Write(longestSubsequence(str));
}
}
// This code is contributed by shikhasingrajput
<script>
// JavaScript program for the above approach
// dp table
let dp = new Array(100005);
// A recursive function to find the
// update the dp[][] table
function calculate(pos, prev, s)
{
// If we reach end of the String
if (pos == s.length)
{
return 0;
}
// If subproblem has been computed
if (dp[pos][prev] != -1)
return dp[pos][prev];
// Initialise variable to find the
// maximum length
let val = 0;
// Choose the current character
if (s[pos].charCodeAt() - 'a'.charCodeAt() + 1 != prev)
{
val = Math.max(val, 1 + calculate(pos + 1,
s[pos].charCodeAt() - 'a'.charCodeAt() + 1,
s));
}
// Omit the current character
val = Math.max(val, calculate(pos + 1, prev, s));
// Return the store answer to the
// current subproblem
dp[pos][prev] = val;
return dp[pos][prev];
}
// Function to find the longest Subsequence
// with different adjacent character
function longestSubsequence(s)
{
// Length of the String s
let n = s.length;
// Initialise the memoisation table
for(let i = 0; i < 100005; i++)
{
dp[i] = new Array(27);
for (let j = 0; j < 27; j++)
{
dp[i][j] = -1;
}
}
// Return the final ans after every
// recursive call
return calculate(0, 0, s);
}
let str = "ababa";
// Function call
document.write(longestSubsequence(str));
</script>
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(26*N) where N is the length of the given string.
Approach 2: Using DP Tabulation method ( Iterative approach )
The approach to solving this problem is the same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem:
- Create a vector to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the longest Subsequence
// with different adjacent character
int longestSubsequence(string s)
{
int n = s.length();
// dp table
int dp[n + 1][27];
// Initialise the dp table
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= 26; j++) {
dp[i][j] = 0;
}
}
// Fill the dp table bottom-up
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j <= 26; j++) {
if (s[i] - 'a' + 1 != j) {
dp[i][j] = 1 + dp[i + 1][s[i] - 'a' + 1];
}
dp[i][j] = max(dp[i][j], dp[i + 1][j]);
}
}
// Return the final answer
return dp[0][0];
}
// Driver Code
int main()
{
string str = "ababa";
// Function call
cout << longestSubsequence(str);
return 0;
}
import java.util.Arrays;
public class Main {
// Function to find the longest Subsequence
// with different adjacent character
static int longestSubsequence(String s)
{
int n = s.length();
// dp table
int[][] dp = new int[n + 1][27];
// Initialise the dp table
for (int i = 0; i <= n; i++) {
Arrays.fill(dp[i], 0);
}
// Fill the dp table bottom-up
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j <= 26; j++) {
if (s.charAt(i) - 'a' + 1 != j) {
dp[i][j] = 1
+ dp[i + 1]
[s.charAt(i) - 'a' + 1];
}
dp[i][j] = Math.max(dp[i][j], dp[i + 1][j]);
}
}
// Return the final answer
return dp[0][0];
}
// Driver Code
public static void main(String[] args)
{
String str = "ababa";
// Function call
System.out.println(longestSubsequence(str));
}
}
# Function to find the longest Subsequence
# with different adjacent character
def longestSubsequence(s):
n = len(s)
# dp table
dp = [[0 for j in range(27)] for i in range(n+1)]
# Initialise the dp table
for i in range(n+1):
for j in range(27):
dp[i][j] = 0
# Fill the dp table bottom-up
for i in range(n-1, -1, -1):
for j in range(27):
if ord(s[i]) - ord('a') + 1 != j:
dp[i][j] = 1 + dp[i+1][ord(s[i])-ord('a')+1]
dp[i][j] = max(dp[i][j], dp[i+1][j])
# Return the final answer
return dp[0][0]
# Driver Code
if __name__ == '__main__':
str = "ababa"
# Function call
print(longestSubsequence(str))
using System;
public class Solution
{
// Function to find the longest Subsequence
// with different adjacent character
public static int LongestSubsequence(string s)
{
int n = s.Length;
// dp table
int[, ] dp = new int[n + 1, 27];
// Initialise the dp table
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= 26; j++) {
dp[i, j] = 0;
}
}
// Fill the dp table bottom-up
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j <= 26; j++) {
if (s[i] - 'a' + 1 != j) {
dp[i, j]
= 1 + dp[i + 1, s[i] - 'a' + 1];
}
dp[i, j] = Math.Max(dp[i, j], dp[i + 1, j]);
}
}
// Return the final answer
return dp[0, 0];
}
// Driver Code
public static void Main(string[] args)
{
string str = "ababa";
// Function call
Console.WriteLine(LongestSubsequence(str));
}
}
// This code is contributed by user_dtewbxkn77n
// Function to find the longest Subsequence
// with different adjacent character
function longestSubsequence(s) {
let n = s.length;
// dp table
let dp = Array.from(Array(n+1), () => new Array(27).fill(0));
// Fill the dp table bottom-up
for (let i = n-1; i >= 0; i--) {
for (let j = 0; j <= 26; j++) {
if (s.charCodeAt(i) - 'a'.charCodeAt(0) + 1 != j) {
dp[i][j] = 1 + dp[i+1][s.charCodeAt(i)-'a'.charCodeAt(0)+1];
}
dp[i][j] = Math.max(dp[i][j], dp[i+1][j]);
}
}
// Return the final answer
return dp[0][0];
}
// Driver Code
let str = "ababa";
// Function call
console.log(longestSubsequence(str));
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N) where N is the length of the given string.
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