Question

2399. Check Distances Between Same Letters

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Description

You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.

Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25).

In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.

Return true_ if _s_ is a well-spaced string, otherwise return _false.

 

Example 1:

Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.

Example 2:

Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.

 

Constraints:

• 2 <= s.length <= 52
• s consists only of lowercase English letters.
• Each letter appears in s exactly twice.
• distance.length == 26
• 0 <= distance[i] <= 50

Solutions

Solution 1

class Solution:
  def checkDistances(self, s: str, distance: List[int]) -> bool:
    d = defaultdict(int)
    for i, c in enumerate(s, 1):
      if d[c] and i - d[c] - 1 != distance[ord(c) - ord('a')]:
        return False
      d[c] = i
    return True

class Solution {
  public boolean checkDistances(String s, int[] distance) {
    int[] d = new int[26];
    for (int i = 1, n = s.length(); i <= n; ++i) {
      int j = s.charAt(i - 1) - 'a';
      if (d[j] > 0 && i - d[j] - 1 != distance[j]) {
        return false;
      }
      d[j] = i;
    }
    return true;
  }
}

class Solution {
public:
  bool checkDistances(string s, vector<int>& distance) {
    int d[26]{};
    for (int i = 1; i <= s.size(); ++i) {
      int j = s[i - 1] - 'a';
      if (d[j] && i - d[j] - 1 != distance[j]) {
        return false;
      }
      d[j] = i;
    }
    return true;
  }
};

func checkDistances(s string, distance []int) bool {
  d := [26]int{}
  for i, c := range s {
    c -= 'a'
    if d[c] > 0 && i-d[c] != distance[c] {
      return false
    }
    d[c] = i + 1
  }
  return true
}

function checkDistances(s: string, distance: number[]): boolean {
  const n = s.length;
  const d: number[] = new Array(26).fill(0);
  for (let i = 1; i <= n; ++i) {
    const j = s.charCodeAt(i - 1) - 97;
    if (d[j] && i - d[j] - 1 !== distance[j]) {
      return false;
    }
    d[j] = i;
  }
  return true;
}

impl Solution {
  pub fn check_distances(s: String, distance: Vec<i32>) -> bool {
    let n = s.len();
    let s = s.as_bytes();
    let mut d = [0; 26];
    for i in 0..n {
      let j = (s[i] - b'a') as usize;
      let i = i as i32;
      if d[j] > 0 && i - d[j] != distance[j] {
        return false;
      }
      d[j] = i + 1;
    }
    true
  }
}

bool checkDistances(char* s, int* distance, int distanceSize) {
  int n = strlen(s);
  int d[26] = {0};
  for (int i = 0; i < n; i++) {
    int j = s[i] - 'a';
    if (d[j] > 0 && i - d[j] != distance[j]) {
      return false;
    }
    d[j] = i + 1;
  }
  return true;
}