Shortest Distance to a Character

Shortest Distance to a Character - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: s = "loveleetcode", c = "e"

› Output: [3,2,1,0,1,0,0,1,2,2,1,0]

💡 Note: Character 'e' appears at indices 3, 5, 6, and 11. For each position, find distance to nearest 'e'.

Example 2 — Single Character

$ Input: s = "aaab", c = "b"

› Output: [3,2,1,0]

💡 Note: Character 'b' only appears at index 3, so distances are [3,2,1,0].

Example 3 — All Same Character

$ Input: s = "abaa", c = "a"

› Output: [0,1,0,0]

💡 Note: Character 'a' appears at indices 0, 2, and 3. Each position finds nearest 'a'.

Constraints

1 ≤ s.length ≤ 10⁴
s[i] and c are lowercase English letters
It is guaranteed that c occurs in s

Visualization

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Asked in

f Facebook 25 G Google 20 a Amazon 15

The key insight is that for each position, the closest character can only be to the left or right. The two-pass approach scans left-to-right tracking distances from the left, then right-to-left tracking distances from the right, taking the minimum. Time: O(n), Space: O(1) excluding output.

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each index i in the string, scan through all positions to find the minimum distance to any occurrence of character c. This requires checking every possible position for every index.

Two Pass Scan

⏱️ Time: O(n) Space: O(n)

Make two passes through the string: first pass tracks distance from nearest character c on the left, second pass tracks distance from nearest character c on the right, then take minimum of both distances.

Brute Force — Algorithm Steps

For each position i in string
Scan all positions j to find character c
Calculate |i - j| and keep minimum distance

Visualization

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Step-by-Step Walkthrough

Position i=0

Check distances to all 'e' positions: |0-1|=1, |0-3|=3, min=1

Position i=1

Check distances to all 'e' positions: |1-1|=0, |1-3|=2, min=0

Continue

Repeat for all positions to build result array

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int* solution(char* s, char* c, int* returnSize) {
    int n = strlen(s);
    int* result = (int*)malloc(n * sizeof(int));
    *returnSize = n;
    char target = c[0];
    
    for (int i = 0; i < n; i++) {
        int minDist = INT_MAX;
        for (int j = 0; j < n; j++) {
            if (s[j] == target) {
                int dist = (i > j) ? (i - j) : (j - i);
                if (dist < minDist) {
                    minDist = dist;
                }
            }
        }
        result[i] = minDist;
    }
    
    return result;
}

int main() {
    char s[10001], c[2];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(c, sizeof(c), stdin);
    c[strcspn(c, "\n")] = 0;
    
    int returnSize;
    int* result = solution(s, c, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops: outer loop n iterations, inner loop n iterations each

✓ Linear Growth

Space Complexity

O(1)

Only using variables, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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