LeetCode - 2370. Longest Ideal Subsequence

Medium

題目

You are given a string s consisting of lowercase letters and an integer k. We call a string t ideal if the following conditions are satisfied:

t is a subsequence of the string s.
The absolute difference in the alphabet order of every two adjacent letters in t is less than or equal to k.
Return the length of the longest ideal string.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Note that the alphabet order is not cyclic. For example, the absolute difference in the alphabet order of 'a' and 'z' is 25, not 1.

Example 1:

Input: s = “acfgbd”, k = 2
Output: 4
Explanation: The longest ideal string is “acbd”. The length of this string is 4, so 4 is returned.
Note that “acfgbd” is not ideal because ‘c’ and ‘f’ have a difference of 3 in alphabet order.

Example 2:

Input: s = “abcd”, k = 3
Output: 4
Explanation: The longest ideal string is “abcd”. The length of this string is 4, so 4 is returned.

Constraints:

1 <= s.length <= 105
0 <= k <= 25
s consists of lowercase English letters.

解題思路

假設 acfb, k = 2
從左至右，依序計算從取出之字母的 +k 或 -k 個字母的 dpLen[] 最大者，然後 +1(加上自己)
dpLen 代表的是，從字串由左至右，已出現的字母，+- k 範圍內的最長 LIS

1 -> a
取 dpLen a, b, c 之中最大者 +1
a c f b
1 0 0 0

2 -> c
取 dpLen a, b, c, d, e 之中最大者 +1，此例取 dpLen[a] = 1 + 1
a c f b
1 2 0 0

3 -> f
取 dpLen d, e, f, g, h 之中最大者 +1
a c f b
1 2 1 0

4 -> b
取 dpLen a, b, c, d 之中最大者 +1，此例取 dpLen[c] = 2 + 1
a c f b
1 2 1 3

maxi = 3

Solution Code

class Solution {
public:
    int longestIdealString(string s, int k) {
        int dpLen[150];
        memset(dpLen, 0, sizeof(dpLen));

        int maxi = -1;
        for(int i = 0; i < s.size(); ++i){
            int idx = s[i] - 'a';
            for(int j = idx - k; j <= idx + k; ++j){
                if(j < 0 || j > 26) continue;
                dpLen[idx] = max(dpLen[idx], dpLen[j]);
            }
            // 有可能遇到 "aa" 的狀況，因此要在全部其他可能做完後才 +1
            // 否則會影響與其他 dpLen[j] 的大小比較
            dpLen[idx]++; 
            maxi = max(maxi, dpLen[idx]);
        }
        return maxi;
    }
};