README_EN.md

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

C

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;
}

TypeScript

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;
}

Rust

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
    }
}

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