LeetCode #2399 — EASY

Check Distances Between Same Letters

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.

Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25).

In a well-spaced string, the number of letters between the two occurrences of the i^th letter is distance[i]. If the i^th letter does not appear in s, then distance[i] can be ignored.

Return true if s is a well-spaced string, otherwise return false.

Example 1:

Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.

Example 2:

Input: s = "aa", distance = [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]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.

Constraints:

2 <= s.length <= 52
s consists only of lowercase English letters.
Each letter appears in s exactly twice.
distance.length == 26
0 <= distance[i] <= 50

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26. Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25). In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored. Return true if s is a well-spaced string, otherwise return false.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

"abaccb"
[1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Example 2

"aa"
[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]

Core Insight

What unlocks the optimal approach

Create an integer array of size 26 to keep track of the first occurrence of each letter.
The number of letters between indices i and j is j - i - 1.

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 #2399: Check Distances Between Same Letters
class Solution {
    public boolean checkDistances(String s, int[] distance) {
        int[] d = new int[26];
        for (int i = 1; i <= s.length(); ++i) {
            int j = s.charAt(i - 1) - 'a';
            if (d[j] > 0 && i - d[j] - 1 != distance[j]) {
                return false;
            }
            d[j] = i;
        }
        return true;
    }
}

// Accepted solution for LeetCode #2399: Check Distances Between Same Letters
func checkDistances(s string, distance []int) bool {
	d := [26]int{}
	for i, c := range s {
		c -= 'a'
		if d[c] > 0 && i-d[c] != distance[c] {
			return false
		}
		d[c] = i + 1
	}
	return true
}

# Accepted solution for LeetCode #2399: Check Distances Between Same Letters
class Solution:
    def checkDistances(self, s: str, distance: List[int]) -> bool:
        d = defaultdict(int)
        for i, c in enumerate(map(ord, s), 1):
            j = c - ord("a")
            if d[j] and i - d[j] - 1 != distance[j]:
                return False
            d[j] = i
        return True

// Accepted solution for LeetCode #2399: Check Distances Between Same Letters
impl Solution {
    pub fn check_distances(s: String, distance: Vec<i32>) -> bool {
        let n = s.len();
        let s = s.as_bytes();
        let mut d = [0; 26];
        for i in 0..n {
            let j = (s[i] - b'a') as usize;
            let i = i as i32;
            if d[j] > 0 && i - d[j] != distance[j] {
                return false;
            }
            d[j] = i + 1;
        }
        true
    }
}

// Accepted solution for LeetCode #2399: Check Distances Between Same Letters
function checkDistances(s: string, distance: number[]): boolean {
    const d: number[] = Array(26).fill(0);
    const n = s.length;
    for (let i = 1; i <= n; ++i) {
        const j = s.charCodeAt(i - 1) - 97;
        if (d[j] && i - d[j] - 1 !== distance[j]) {
            return false;
        }
        d[j] = i;
    }
    return true;
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.