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Formatted question description: https://leetcode.ca/all/2399.html

2399. Check Distances Between Same Letters

• Difficulty: Easy.
• Related Topics: Array, Hash Table, String.
• Similar Questions: Two Sum, Shortest Distance to a Character.

Problem

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

Solution (Java, C++, Python)

• Java
• Python
• C++
• Go
• TypeScript
• RenderScript

• class Solution {
      public boolean checkDistances(String s, int[] distance) {
          
          boolean valid = true;
          
          HashMap<Character,Integer> map = new HashMap<>();
          int n = s.length();
          
          // System.out.println('c'-'a');
          
          for(int i = 0 ; i < n ; i++){
              
              char c = s.charAt(i);
              if(map.containsKey(c)){
                  int value = i - map.get(c) - 1;
                  if(value != distance[c-'a'])return false;
              }else{
                  map.put(c,i);
              }
          }
          
          return valid;
      }
  }
  
  ############
  
  class Solution {
      public boolean checkDistances(String s, int[] distance) {
          int[] d = new int[26];
          for (int i = 0; i < s.length(); ++i) {
              int j = s.charAt(i) - 'a';
              if (d[j] > 0 && i - d[j] != distance[j]) {
                  return false;
              }
              d[j] = i + 1;
          }
          return true;
      }
  }
  

• class Solution:
      def checkDistances(self, s: str, distance: List[int]) -> bool:
          d = [0] * 26
          for i, c in enumerate(s):
              j = ord(c) - ord("a")
              if d[j] and i - d[j] != distance[j]:
                  return False
              d[j] = i + 1
          return True
  
  ############
  
  # 2399. Check Distances Between Same Letters
  # https://leetcode.com/problems/check-distances-between-same-letters/
  
  class Solution:
      def checkDistances(self, s: str, distance: List[int]) -> bool:
          dist = {}
          
          for i, x in enumerate(s):
              if x not in dist:
                  dist[x] = i
              else:
                  k = ord(x) - ord("a")
                  
                  if distance[k] != i - dist[x] - 1:
                      return False
              
          return True
  
  

• class Solution {
  public:
      bool checkDistances(string s, vector<int>& distance) {
          vector<int> d(26);
          for (int i = 0; i < s.size(); ++i) {
              int j = s[i] - 'a';
              if (d[j] && i - d[j] != distance[j]) {
                  return false;
              }
              d[j] = i + 1;
          }
          return true;
      }
  };
  

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

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

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