LeetCode #2370 — MEDIUM

Longest Ideal Subsequence

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Dynamic Programming

Example 1

"acfgbd"
2

Example 2

"abcd"
3

Core Insight

What unlocks the optimal approach

How can you calculate the longest ideal subsequence that ends at a specific index i?
Can you calculate it for all positions i? How can you use previously calculated answers to calculate the answer for the next position?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2370: Longest Ideal Subsequence
class Solution {
    public int longestIdealString(String s, int k) {
        int n = s.length();
        int ans = 1;
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        Map<Character, Integer> d = new HashMap<>(26);
        d.put(s.charAt(0), 0);
        for (int i = 1; i < n; ++i) {
            char a = s.charAt(i);
            for (char b = 'a'; b <= 'z'; ++b) {
                if (Math.abs(a - b) > k) {
                    continue;
                }
                if (d.containsKey(b)) {
                    dp[i] = Math.max(dp[i], dp[d.get(b)] + 1);
                }
            }
            d.put(a, i);
            ans = Math.max(ans, dp[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2370: Longest Ideal Subsequence
func longestIdealString(s string, k int) int {
	n := len(s)
	ans := 1
	dp := make([]int, n)
	for i := range dp {
		dp[i] = 1
	}
	d := map[byte]int{s[0]: 0}
	for i := 1; i < n; i++ {
		a := s[i]
		for b := byte('a'); b <= byte('z'); b++ {
			if int(a)-int(b) > k || int(b)-int(a) > k {
				continue
			}
			if v, ok := d[b]; ok {
				dp[i] = max(dp[i], dp[v]+1)
			}
		}
		d[a] = i
		ans = max(ans, dp[i])
	}
	return ans
}

# Accepted solution for LeetCode #2370: Longest Ideal Subsequence
class Solution:
    def longestIdealString(self, s: str, k: int) -> int:
        n = len(s)
        ans = 1
        dp = [1] * n
        d = {s[0]: 0}
        for i in range(1, n):
            a = ord(s[i])
            for b in ascii_lowercase:
                if abs(a - ord(b)) > k:
                    continue
                if b in d:
                    dp[i] = max(dp[i], dp[d[b]] + 1)
            d[s[i]] = i
        return max(dp)

// Accepted solution for LeetCode #2370: Longest Ideal Subsequence
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2370: Longest Ideal Subsequence
// class Solution {
//     public int longestIdealString(String s, int k) {
//         int n = s.length();
//         int ans = 1;
//         int[] dp = new int[n];
//         Arrays.fill(dp, 1);
//         Map<Character, Integer> d = new HashMap<>(26);
//         d.put(s.charAt(0), 0);
//         for (int i = 1; i < n; ++i) {
//             char a = s.charAt(i);
//             for (char b = 'a'; b <= 'z'; ++b) {
//                 if (Math.abs(a - b) > k) {
//                     continue;
//                 }
//                 if (d.containsKey(b)) {
//                     dp[i] = Math.max(dp[i], dp[d.get(b)] + 1);
//                 }
//             }
//             d.put(a, i);
//             ans = Math.max(ans, dp[i]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2370: Longest Ideal Subsequence
function longestIdealString(s: string, k: number): number {
    const dp = new Array(26).fill(0);
    for (const c of s) {
        const x = c.charCodeAt(0) - 'a'.charCodeAt(0);
        let t = 0;
        for (let i = 0; i < 26; i++) {
            if (Math.abs(x - i) <= k) {
                t = Math.max(t, dp[i] + 1);
            }
        }
        dp[x] = Math.max(dp[x], t);
    }

    return dp.reduce((r, c) => Math.max(r, c), 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.