给定长度为N且整数为K的字符串S ，任务是找到最长子序列的长度，以使子序列中相邻字符的ASCII值之差不超过K。
例子：
Input: N = 7, K = 2, S = "afcbedg"
Output: 4
Explantion:
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 = "geeksforgeeks"
Output: 7
天真的方法：蛮力解决方案是生成各种长度的所有可能子序列，并计算有效子序列的最大长度。时间复杂度将是指数级的。
高效的方法：一种有效的方法是使用动态编程的概念
- 创建一个0的数组dp ，其大小等于字符串的长度。
- 创建大小为26的0的支持数组max_length 。
- 迭代由字符和用于每个字符的字符串的字符确定上限和下限。
- 在上下限范围内迭代嵌套循环。
- 用当前dp索引和当前max_length索引+1之间的最大值填充dp数组。
- 用当前dp索引和当前max_length索引之间的最大值填充max_length数组。
- 最长子序列长度是dp数组中的最大值。
- 让我们考虑一个例子：
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
- for 1st iteration value of i is ‘a’ and range of j is (0, 2)
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
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 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 = "geeksforgeeks";
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 = "geeksforgeeks";
int n = s.length();
int k = 3;
System.out.print(longest_subseq(n, k, s));
}
}
// This code is contributed by 29AjayKumarPython3
# 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 = "geeksforgeeks"
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 = "geeksforgeeks";
int n = s.Length;
int k = 3;
Console.Write(longest_subseq(n, k, s));
}
}
// This code is contributed by Rajput-Ji输出：
7