README_EN.md

1392. Longest Happy Prefix

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Description

A string is called a happy prefix if is a non-empty prefix which is also a suffix (excluding itself).

Given a string s, return the longest happy prefix of s. Return an empty string "" if no such prefix exists.

 

Example 1:

Input: s = "level"
Output: "l"
Explanation: s contains 4 prefix excluding itself ("l", "le", "lev", "leve"), and suffix ("l", "el", "vel", "evel"). The largest prefix which is also suffix is given by "l".

Example 2:

Input: s = "ababab"
Output: "abab"
Explanation: "abab" is the largest prefix which is also suffix. They can overlap in the original string.

 

Constraints:

• 1 <= s.length <= 105
• s contains only lowercase English letters.

Solutions

Python3

class Solution:
    def longestPrefix(self, s: str) -> str:
        for i in range(1, len(s)):
            if s[:-i] == s[i:]:
                return s[i:]
        return ''

Java

class Solution {
    private long[] p;
    private long[] h;

    public String longestPrefix(String s) {
        int base = 131;
        int n = s.length();
        p = new long[n + 10];
        h = new long[n + 10];
        p[0] = 1;
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * base;
            h[i + 1] = h[i] * base + s.charAt(i);
        }
        for (int l = n - 1; l > 0; --l) {
            if (get(1, l) == get(n - l + 1, n)) {
                return s.substring(0, l);
            }
        }
        return "";
    }

    private long get(int l, int r) {
        return h[r] - h[l - 1] * p[r - l + 1];
    }
}

C++

typedef unsigned long long ULL;

class Solution {
public:
    string longestPrefix(string s) {
        int base = 131;
        int n = s.size();
        ULL p[n + 10];
        ULL h[n + 10];
        p[0] = 1;
        h[0] = 0;
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * base;
            h[i + 1] = h[i] * base + s[i];
        }
        for (int l = n - 1; l > 0; --l) {
            ULL prefix = h[l];
            ULL suffix = h[n] - h[n - l] * p[l];
            if (prefix == suffix) return s.substr(0, l);
        }
        return "";
    }
};

Go

func longestPrefix(s string) string {
	base := 131
	n := len(s)
	p := make([]int, n+10)
	h := make([]int, n+10)
	p[0] = 1
	for i, c := range s {
		p[i+1] = p[i] * base
		h[i+1] = h[i]*base + int(c)
	}
	for l := n - 1; l > 0; l-- {
		prefix, suffix := h[l], h[n]-h[n-l]*p[l]
		if prefix == suffix {
			return s[:l]
		}
	}
	return ""
}

TypeScript

function longestPrefix(s: string): string {
    const n = s.length;
    for (let i = n - 1; i >= 0; i--) {
        if (s.slice(0, i) === s.slice(n - i, n)) {
            return s.slice(0, i);
        }
    }
    return '';
}

Rust

impl Solution {
    pub fn longest_prefix(s: String) -> String {
        let n = s.len();
        for i in (0..n).rev() {
            if s[0..i] == s[n - i..n] {
                return s[0..i].to_string();
            }
        }
        String::new()
    }
}

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