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3271. Hash Divided String

Description

You are given a string s of length n and an integer k, where n is a multiple of k. Your task is to hash the string s into a new string called result, which has a length of n / k.

First, divide s into n / k substrings, each with a length of k. Then, initialize result as an empty string.

For each substring in order from the beginning:

  • The hash value of a character is the index of that character in the English alphabet (e.g., 'a' → 0, 'b' → 1, ..., 'z' → 25).
  • Calculate the sum of all the hash values of the characters in the substring.
  • Find the remainder of this sum when divided by 26, which is called hashedChar.
  • Identify the character in the English lowercase alphabet that corresponds to hashedChar.
  • Append that character to the end of result.

Return result.

 

Example 1:

Input: s = "abcd", k = 2

Output: "bf"

Explanation:

First substring: "ab", 0 + 1 = 1, 1 % 26 = 1, result[0] = 'b'.

Second substring: "cd", 2 + 3 = 5, 5 % 26 = 5, result[1] = 'f'.

Example 2:

Input: s = "mxz", k = 3

Output: "i"

Explanation:

The only substring: "mxz", 12 + 23 + 25 = 60, 60 % 26 = 8, result[0] = 'i'.

 

Constraints:

  • 1 <= k <= 100
  • k <= s.length <= 1000
  • s.length is divisible by k.
  • s consists only of lowercase English letters.

Solutions

Solution 1: Simulation

We can simulate the process according to the steps described in the problem.

Traverse the string $s$, and each time take $k$ characters, calculate the sum of their hash values, denoted as $t$. Then, take $t$ modulo $26$ to find the corresponding character and add it to the end of the result string.

Finally, return the result string.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. Ignoring the space consumption of the answer string, the space complexity is $O(1)$.

  • class Solution {
    public String stringHash(String s, int k) {
        StringBuilder ans = new StringBuilder();
        int n = s.length();
        for (int i = 0; i < n; i += k) {
            int t = 0;
            for (int j = i; j < i + k; ++j) {
                t += s.charAt(j) - 'a';
            }
            int hashedChar = t % 26;
            ans.append((char) ('a' + hashedChar));
        }
        return ans.toString();
    }
}


  • class Solution {
public:
    string stringHash(string s, int k) {
        string ans;
        int n = s.length();
        for (int i = 0; i < n; i += k) {
            int t = 0;
            for (int j = i; j < i + k; ++j) {
                t += s[j] - 'a';
            }
            int hashedChar = t % 26;
            ans += ('a' + hashedChar);
        }
        return ans;
    }
};


  • class Solution:
    def stringHash(self, s: str, k: int) -> str:
        ans = []
        for i in range(0, len(s), k):
            t = 0
            for j in range(i, i + k):
                t += ord(s[j]) - ord("a")
            hashedChar = t % 26
            ans.append(chr(ord("a") + hashedChar))
        return "".join(ans)


  • func stringHash(s string, k int) string {
	n := len(s)
	ans := make([]byte, 0, n/k)

	for i := 0; i < n; i += k {
		t := 0
		for j := i; j < i+k; j++ {
			t += int(s[j] - 'a')
		}
		hashedChar := t % 26
		ans = append(ans, 'a'+byte(hashedChar))
	}

	return string(ans)
}


  • function stringHash(s: string, k: number): string {
    const ans: string[] = [];
    const n: number = s.length;

    for (let i = 0; i < n; i += k) {
        let t: number = 0;
        for (let j = i; j < i + k; j++) {
            t += s.charCodeAt(j) - 97;
        }
        const hashedChar: number = t % 26;
        ans.push(String.fromCharCode(97 + hashedChar));
    }

    return ans.join('');
}

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