Longest subsequence having difference atmost K
Given a string S of length N and an integer K, the task is to find the length of longest sub-sequence such that the difference between the ASCII values of adjacent characters in the subsequence is not more than K.
Examples:
Input: N = 7, K = 2, S = "afcbedg" Output: 4 Explanation: Longest special sequence present in "afcbedg" is a, c, b, d. It is special because |a - c| <= 2, |c - b| <= 2 and | b-d| <= 2 Input: N = 13, K = 3, S = "zambiatek" Output: 7
Naive approach: A brute force solution is to generate all the possible subsequences of various lengths and compute the maximum length of the valid subsequence. The time complexity will be exponential.
Efficient Approach: An efficient approach is to use the concept Dynamic Programming
- Create an array dp of 0’s with size equal to length of string.
- Create a supporting array max_length with 0’s of size 26.
- Iterate the string character by character and for each character determine the upper and lower bounds.
- Iterate nested loop in the range of lower and upper bounds.
- Fill the dp array with the maximum value between current dp indices and current max_length indices+1.
- Fill the max_length array with the maximum value between current dp indices and current max_length indices.
- Longest sub sequence length is the maximum value in dp array.
- Let us consider an example:
input string s is “afcbedg” and k is 2
- for 1st iteration value of i is ‘a’ and range of j is (0, 2)
and current dp = [1, 0, 0, 0, 0, 0, 0]- for 2nd iteration value of i is ‘f’ and range of j is (3, 7)
and current dp = [1, 1, 0, 0, 0, 0, 0]- for 3rd iteration value of i is ‘c’ and range of j is (0, 4)
and current dp = [1, 1, 2, 0, 0, 0, 0]- for 4th iteration value of i is ‘b’ and range of j is (0, 3)
and current dp = [1, 1, 2, 3, 0, 0, 0]- for 5th iteration value of i is ‘e’ and range of j is (2, 6)
and current dp = [1, 1, 2, 3, 3, 0, 0]- for 6th iteration value of i is ‘d’ and range of j is (1, 5)
and current dp = [1, 1, 2, 3, 3, 4, 0]- for 7th iteration value of i is ‘g’ and range of j is (4, 8)
and current dp = [1, 1, 2, 3, 3, 4, 4]longest length is the maximum value in dp so maximum length is 4
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find // the longest Special Sequence int longest_subseq( int n, int k, string s) { // Creating a list with // all 0's of size // equal to the length of string vector< int > dp(n, 0); // Supporting list with // all 0's of size 26 since // the given string consists // of only lower case alphabets int max_length[26] = {0}; for ( int i = 0; i < n; i++) { // Converting the ascii value to // list indices int curr = s[i] - 'a' ; // Determining the lower bound int lower = max(0, curr - k); // Determining the upper bound int upper = min(25, curr + k); // Filling the dp array with values for ( int j = lower; j < upper + 1; j++) { dp[i] = max(dp[i], max_length[j] + 1); } //Filling the max_length array with max //length of subsequence till now max_length[curr] = max(dp[i], max_length[curr]); } int ans = 0; for ( int i:dp) ans = max(i, ans); // return the max length of subsequence return ans; } // Driver Code int main() { string s = "zambiatek" ; int n = s.size(); int k = 3; cout << (longest_subseq(n, k, s)); return 0; } // This code is contributed by Mohit Kumar |
Java
// Java program for the above approach class GFG { // Function to find // the longest Special Sequence static int longest_subseq( int n, int k, String s) { // Creating a list with // all 0's of size // equal to the length of String int []dp = new int [n]; // Supporting list with // all 0's of size 26 since // the given String consists // of only lower case alphabets int []max_length = new int [ 26 ]; for ( int i = 0 ; i < n; i++) { // Converting the ascii value to // list indices int curr = s.charAt(i) - 'a' ; // Determining the lower bound int lower = Math.max( 0 , curr - k); // Determining the upper bound int upper = Math.min( 25 , curr + k); // Filling the dp array with values for ( int j = lower; j < upper + 1 ; j++) { dp[i] = Math.max(dp[i], max_length[j] + 1 ); } // Filling the max_length array with max // length of subsequence till now max_length[curr] = Math.max(dp[i], max_length[curr]); } int ans = 0 ; for ( int i:dp) ans = Math.max(i, ans); // return the max length of subsequence return ans; } // Driver Code public static void main(String[] args) { String s = "zambiatek" ; int n = s.length(); int k = 3 ; System.out.print(longest_subseq(n, k, s)); } } // This code is contributed by 29AjayKumar |
Python3
# Function to find # the longest Special Sequence def longest_subseq(n, k, s): # Creating a list with # all 0's of size # equal to the length of string dp = [ 0 ] * n # Supporting list with # all 0's of size 26 since # the given string consists # of only lower case alphabets max_length = [ 0 ] * 26 for i in range (n): # Converting the ascii value to # list indices curr = ord (s[i]) - ord ( 'a' ) # Determining the lower bound lower = max ( 0 , curr - k) # Determining the upper bound upper = min ( 25 , curr + k) # Filling the dp array with values for j in range (lower, upper + 1 ): dp[i] = max (dp[i], max_length[j] + 1 ) # Filling the max_length array with max # length of subsequence till now max_length[curr] = max (dp[i], max_length[curr]) # return the max length of subsequence return max (dp) # driver code def main(): s = "zambiatek" n = len (s) k = 3 print (longest_subseq(n, k, s)) main() |
C#
// C# program for the above approach using System; class GFG { // Function to find // the longest Special Sequence static int longest_subseq( int n, int k, String s) { // Creating a list with // all 0's of size // equal to the length of String int []dp = new int [n]; // Supporting list with // all 0's of size 26 since // the given String consists // of only lower case alphabets int []max_length = new int [26]; for ( int i = 0; i < n; i++) { // Converting the ascii value to // list indices int curr = s[i] - 'a' ; // Determining the lower bound int lower = Math.Max(0, curr - k); // Determining the upper bound int upper = Math.Min(25, curr + k); // Filling the dp array with values for ( int j = lower; j < upper + 1; j++) { dp[i] = Math.Max(dp[i], max_length[j] + 1); } // Filling the max_length array with max // length of subsequence till now max_length[curr] = Math.Max(dp[i], max_length[curr]); } int ans = 0; foreach ( int i in dp) ans = Math.Max(i, ans); // return the max length of subsequence return ans; } // Driver Code public static void Main(String[] args) { String s = "zambiatek" ; int n = s.Length; int k = 3; Console.Write(longest_subseq(n, k, s)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program for the above approach // Function to find // the longest Special Sequence function longest_subseq(n, k, s) { // Creating a list with // all 0's of size // equal to the length of String let dp = new Array(n); // Supporting list with // all 0's of size 26 since // the given String consists // of only lower case alphabets let max_length = new Array(26); for (let i = 0; i < 26; i++) { max_length[i] = 0; dp[i] = 0; } for (let i = 0; i < n; i++) { // Converting the ascii value to // list indices let curr = s[i].charCodeAt(0) - 'a' .charCodeAt(0); // Determining the lower bound let lower = Math.max(0, curr - k); // Determining the upper bound let upper = Math.min(25, curr + k); // Filling the dp array with values for (let j = lower; j < upper + 1; j++) { dp[i] = Math.max(dp[i], max_length[j] + 1); } // Filling the max_length array with max // length of subsequence till now max_length[curr] = Math.max(dp[i], max_length[curr]); } let ans = 0; ans = Math.max(...dp) // return the max length of subsequence return ans; } // Driver Code let s = "zambiatek" ; let n = s.length; let k = 3; document.write(longest_subseq(n, k, s)); // This code is contributed by unknown2108 </script> |
Output:
7
Time Complexity: O(N)
Auxiliary Space: O(N)
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