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 <= 104s[i] and c are lowercase English letters.- It is guaranteed that
c occurs at least once in s.
Solutions#
Solution 1#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| 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
}
}
|
