Skip to content

821. Shortest Distance to a Character

Description

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.

Solutions

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        int n = s.size();
        const int inf = 1 << 30;
        vector<int> ans(n, inf);
        for (int i = 0, pre = -inf; i < n; ++i) {
            if (s[i] == c) {
                pre = i;
            }
            ans[i] = min(ans[i], i - pre);
        }
        for (int i = n - 1, suf = inf; ~i; --i) {
            if (s[i] == c) {
                suf = i;
            }
            ans[i] = min(ans[i], suf - i);
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
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
    }
}

Comments