2399. Check Distances Between Same Letters LeetCode Solution

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Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Check Distances Between Same Letters solution in C++
  4. Check Distances Between Same Letters solution in Java
  5. Check Distances Between Same Letters solution in Python
  6. Additional Resources
2399. Check Distances Between Same Letters LeetCode Solution image

Problem Statement of Check Distances Between Same Letters

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.

See also  3201. Find the Maximum Length of Valid Subsequence I LeetCode Solution

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

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(26) = O(1)

2399. Check Distances Between Same Letters LeetCode Solution in C++

class Solution {
 public:
  bool checkDistances(string s, vector<int>& distance) {
    vector<int> firstSeenIndex(26, -1);

    for (int i = 0; i < s.length(); ++i) {
      const int j = s[i] - 'a';
      int& prevIndex = firstSeenIndex[j];
      if (prevIndex != -1 && i - prevIndex - 1 != distance[j])
        return false;
      prevIndex = i;
    }

    return true;
  }
};
/* code provided by PROGIEZ */

2399. Check Distances Between Same Letters LeetCode Solution in Java

class Solution {
  public boolean checkDistances(String s, int[] distance) {
    int[] firstSeenIndex = new int[26];
    Arrays.fill(firstSeenIndex, -1);

    for (int i = 0; i < s.length(); ++i) {
      final int j = s.charAt(i) - 'a';
      final int prevIndex = firstSeenIndex[j];
      if (prevIndex != -1 && i - prevIndex - 1 != distance[j])
        return false;
      firstSeenIndex[j] = i;
    }

    return true;
  }
}
// code provided by PROGIEZ

2399. Check Distances Between Same Letters LeetCode Solution in Python

class Solution:
  def checkDistances(self, s: str, distance: list[int]) -> bool:
    firstSeenIndex = [-1] * 26

    for i, c in enumerate(s):
      j = ord(c) - ord('a')
      prevIndex = firstSeenIndex[j]
      if prevIndex != -1 and i - prevIndex - 1 != distance[j]:
        return False
      firstSeenIndex[j] = i

    return True
 # code by PROGIEZ

Additional Resources

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