LC2370. Longest Ideal Subsequence

Approach 1: Bottom-Up DP

class Solution:
    def longestIdealString(self, s: str, k: int) -> int:
        dp = [0] * 26
 
        for c in s:
            curr = ord(c) - ord('a')
            longest = 1
 
            for prev in range(26):
                if abs(curr - prev) <= k:
                    longest = max(longest, 1 + dp[prev])
            
            dp[curr] = max(longest, dp[curr])
        
        return max(dp)

Complexity

Time: $O(n)$
Space: $O(1)$

Notes

Visualizing tabulation helps to understand how this works. If we include the following lines:

print(" " + ", ".join(chr(ord('a') + x) for x in range(26)))
print(dp)
print()

And for the input s = "abcd" and k = 3:

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 2, 3, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

For the input s = "acfgbd" and k = 2:

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 0, 2, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 3, 2, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

 a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
[1, 3, 2, 4, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Approach 3: Recursion, TLE

class Solution:
    def longestIdealString(self, s: str, k: int) -> int:
        cache = {}
 
        def helper(i, prev):
            if i == len(s):
                return 0
            
            if (i, prev) in cache:
                return cache[(i, prev)]
            
            # The case where we skip ith char in subsequence
            res = helper(i + 1, prev)
 
            if prev is None or abs(ord(s[i]) - ord(s[prev])) <= k:
                res = max(res, 1 + helper(i + 1, i))
 
            cache[(i, prev)] = res
 
            return res
        
        return helper(0, None)

Complexity

Time: TODO
Space: TODO

Notes

This will TLE on some of the test cases on LC.

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