Formatted question description: https://leetcode.ca/all/821.html

821. Shortest Distance to a Character (Easy)

Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

Example 1:

Input: S = "loveleetcode", C = 'e'
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

Note:

S string length is in [1, 10000].
C is a single character, and guaranteed to be in string S.
All letters in S and C are lowercase.

Companies:
Bloomberg

Solution 1.

// OJ: https://leetcode.com/problems/shortest-distance-to-a-character/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> shortestToChar(string S, char C) {
        int N = S.size(), prevJ;
        vector<int> ans(N);
        for (int i = 0, j = -1; i < N; ++i) {
            if (i > j) {
                prevJ = j;
                do { ++j; } while (j < N && S[j] != C);
            }
            ans[i] = min(j == N ? INT_MAX : j - i, prevJ == -1 ? INT_MAX : i - prevJ);
        }
        return ans;
    }
};

class Solution {
    public int[] shortestToChar(String S, char C) {
        int length = S.length();
        int[] distances = new int[length];
        List<Integer> indices = new ArrayList<Integer>();
        for (int i = 0; i < length; i++) {
            if (S.charAt(i) == C)
                indices.add(i);
        }
        int size = indices.size();
        int pointer = 0;
        for (int i = 0; i < length; i++) {
            if (indices.contains(i)) {
                distances[i] = 0;
                pointer++;
            } else {
                if (pointer == 0) {
                    int curIndex = indices.get(pointer);
                    distances[i] = Math.abs(curIndex - i);
                } else if (pointer == size) {
                    int prevIndex = indices.get(pointer - 1);
                    distances[i] = Math.abs(i - prevIndex);
                } else {
                    int prevIndex = indices.get(pointer - 1), curIndex = indices.get(pointer);
                    distances[i] = Math.min(Math.abs(i - prevIndex), Math.abs(curIndex - i));
                }
            }
        }
        return distances;
    }
}

############

class Solution {
    public int[] shortestToChar(String s, char c) {
        int n = s.length();
        int[] ans = new int[n];
        for (int i = 0, j = Integer.MAX_VALUE; i < n; ++i) {
            if (s.charAt(i) == c) {
                j = i;
            }
            ans[i] = Math.abs(i - j);
        }
        for (int i = n - 1, j = Integer.MAX_VALUE; i >= 0; --i) {
            if (s.charAt(i) == c) {
                j = i;
            }
            ans[i] = Math.min(ans[i], Math.abs(i - j));
        }
        return ans;
    }
}

// OJ: https://leetcode.com/problems/shortest-distance-to-a-character/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        int N = s.size(), prev = -N, next = 0;
        vector<int> ans(N);
        for (int i = 0; i < N; ++i) {
            while (next < N && s[next] != c) ++next; 
            if (next == N) next = N + N;
            ans[i] = min(i - prev, next - i);
            if (s[i] == c) prev = i, next++;
        }
        return ans;
    }
};

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        n = len(s)
        ans = [0] * n
        j = inf
        for i, ch in enumerate(s):
            if ch == c:
                j = i
            ans[i] = abs(i - j)
        j = inf
        for i in range(n - 1, -1, -1):
            if s[i] == c:
                j = i
            ans[i] = min(ans[i], abs(i - j))
        return ans

############

class Solution:
    def shortestToChar(self, S, C):
        """
        :type S: str
        :type C: str
        :rtype: List[int]
        """
        _len = len(S)
        index = -1000000
        ans = [0] * _len
        for i, s in enumerate(S):
            if s == C:
                index = i
            ans[i] = abs(i - index)
        index = -100000
        for i in range(_len - 1, -1 , -1):
            if S[i] == C:
                index = i
            ans[i] = min(abs(i - index), ans[i])
        return ans

func shortestToChar(s string, c byte) []int {
	n := len(s)
	ans := make([]int, n)
	for i, j := 0, -10000; i < n; i++ {
		if s[i] == c {
			j = i
		}
		ans[i] = i - j
	}
	for i, j := n-1, 10000; i >= 0; i-- {
		if s[i] == c {
			j = i
		}
		if j-i < ans[i] {
			ans[i] = j - i
		}
	}
	return ans
}

function shortestToChar(s: string, c: string): number[] {
    const n = s.length;
    let ans = [];
    let pre = Infinity;
    for (let i = 0; i < n; i++) {
        if (s.charAt(i) == c) pre = i;
        ans[i] = Math.abs(pre - i);
    }
    pre = Infinity;
    for (let i = n - 1; i > -1; i--) {
        if (s.charAt(i) == c) pre = i;
        ans[i] = Math.min(Math.abs(pre - i), ans[i]);
    }
    return ans;
}

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
    }
}

821 - Shortest Distance to a Character

821. Shortest Distance to a Character (Easy)

Solution 1.

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821. Shortest Distance to a Character (Easy)

Solution 1.

All Problems

All Solutions