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 conditionInput: 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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
5
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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
5
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; } |
Java
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)); } } |
Python3
# 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)) |
C#
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 |
Javascript
// 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)); |
Output
5
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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