LeetCode #821 — EASY

Shortest Distance to a Character

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

The Problem

Problem Statement

Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s.

The distance between two indices i and j is abs(i - j), where abs is the absolute value function.

Example 1:

Input: s = "loveleetcode", c = "e"
Output: [3,2,1,0,1,0,0,1,2,2,1,0]
Explanation: The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.

Example 2:

Input: s = "aaab", c = "b"
Output: [3,2,1,0]

Constraints:

1 <= s.length <= 10⁴
s[i] and c are lowercase English letters.
It is guaranteed that c occurs at least once in s.

Step 01

Brute Force Baseline

Problem summary: Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s. The distance between two indices i and j is abs(i - j), where abs is the absolute value function.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

"loveleetcode"
"e"

Example 2

"aaab"
"b"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #821: Shortest Distance to a Character
class Solution {
    public int[] shortestToChar(String s, char c) {
        int n = s.length();
        int[] ans = new int[n];
        final int inf = 1 << 30;
        Arrays.fill(ans, inf);
        for (int i = 0, pre = -inf; i < n; ++i) {
            if (s.charAt(i) == c) {
                pre = i;
            }
            ans[i] = Math.min(ans[i], i - pre);
        }
        for (int i = n - 1, suf = inf; i >= 0; --i) {
            if (s.charAt(i) == c) {
                suf = i;
            }
            ans[i] = Math.min(ans[i], suf - i);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #821: Shortest Distance to a Character
func shortestToChar(s string, c byte) []int {
	n := len(s)
	ans := make([]int, n)
	const inf int = 1 << 30
	pre := -inf
	for i := range s {
		if s[i] == c {
			pre = i
		}
		ans[i] = i - pre
	}
	suf := inf
	for i := n - 1; i >= 0; i-- {
		if s[i] == c {
			suf = i
		}
		ans[i] = min(ans[i], suf-i)
	}
	return ans
}

# Accepted solution for LeetCode #821: Shortest Distance to a Character
class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        n = len(s)
        ans = [n] * n
        pre = -inf
        for i, ch in enumerate(s):
            if ch == c:
                pre = i
            ans[i] = min(ans[i], i - pre)
        suf = inf
        for i in range(n - 1, -1, -1):
            if s[i] == c:
                suf = i
            ans[i] = min(ans[i], suf - i)
        return ans

// Accepted solution for LeetCode #821: Shortest Distance to a Character
impl Solution {
    pub fn shortest_to_char(s: String, c: char) -> Vec<i32> {
        let c = c as u8;
        let s = s.as_bytes();
        let n = s.len();
        let mut res = vec![i32::MAX; n];
        let mut pre = i32::MAX;
        for i in 0..n {
            if s[i] == c {
                pre = i as i32;
            }
            res[i] = i32::abs((i as i32) - pre);
        }
        pre = i32::MAX;
        for i in (0..n).rev() {
            if s[i] == c {
                pre = i as i32;
            }
            res[i] = res[i].min(i32::abs((i as i32) - pre));
        }
        res
    }
}

// Accepted solution for LeetCode #821: Shortest Distance to a Character
function shortestToChar(s: string, c: string): number[] {
    const n = s.length;
    const inf = 1 << 30;
    const ans: number[] = new Array(n).fill(inf);
    for (let i = 0, pre = -inf; i < n; ++i) {
        if (s[i] === c) {
            pre = i;
        }
        ans[i] = i - pre;
    }
    for (let i = n - 1, suf = inf; i >= 0; --i) {
        if (s[i] === c) {
            suf = i;
        }
        ans[i] = Math.min(ans[i], suf - i);
    }
    return ans;
}

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 of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.