README_EN.md

1044. Longest Duplicate Substring

中文文档

Description

Given a string s, consider all duplicated substrings: (contiguous) substrings of s that occur 2 or more times. The occurrences may overlap.

Return any duplicated substring that has the longest possible length. If s does not have a duplicated substring, the answer is "".

 

Example 1:

Input: s = "banana"
Output: "ana"

Example 2:

Input: s = "abcd"
Output: ""

 

Constraints:

• 2 <= s.length <= 3 * 104
• s consists of lowercase English letters.

Solutions

Python3

class Solution:
    def longestDupSubstring(self, s: str) -> str:
        def check(l):
            vis = set()
            for i in range(n - l + 1):
                t = s[i : i + l]
                if t in vis:
                    return t
                vis.add(t)
            return ''

        n = len(s)
        left, right = 0, n
        ans = ''
        while left < right:
            mid = (left + right + 1) >> 1
            t = check(mid)
            ans = t or ans
            if t:
                left = mid
            else:
                right = mid - 1
        return ans

Java

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

    public String longestDupSubstring(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);
        }
        String ans = "";
        int left = 0, right = n;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            String t = check(s, mid);
            if (t.length() > 0) {
                left = mid;
                ans = t;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }

    private String check(String s, int len) {
        int n = s.length();
        Set<Long> vis = new HashSet<>();
        for (int i = 1; i + len - 1 <= n; ++i) {
            int j = i + len - 1;
            long t = h[j] - h[i - 1] * p[j - i + 1];
            if (vis.contains(t)) {
                return s.substring(i - 1, j);
            }
            vis.add(t);
        }
        return "";
    }
}

C++

typedef unsigned long long ULL;

class Solution {
public:
    ULL p[30010];
    ULL h[30010];
    string longestDupSubstring(string s) {
        int base = 131, n = s.size();
        p[0] = 1;
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * base;
            h[i + 1] = h[i] * base + s[i];
        }
        int left = 0, right = n;
        string ans = "";
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            string t = check(s, mid);
            if (t.empty())
                right = mid - 1;
            else {
                left = mid;
                ans = t;
            }
        }
        return ans;
    }

    string check(string& s, int len) {
        int n = s.size();
        unordered_set<ULL> vis;
        for (int i = 1; i + len - 1 <= n; ++i) {
            int j = i + len - 1;
            ULL t = h[j] - h[i - 1] * p[j - i + 1];
            if (vis.count(t)) return s.substr(i - 1, len);
            vis.insert(t);
        }
        return "";
    }
};

Go

func longestDupSubstring(s string) string {
	base, n := 131, len(s)
	p := make([]int64, n+10)
	h := make([]int64, n+10)
	p[0] = 1
	for i := 0; i < n; i++ {
		p[i+1] = p[i] * int64(base)
		h[i+1] = h[i]*int64(base) + int64(s[i])
	}
	check := func(l int) string {
		vis := make(map[int64]bool)
		for i := 1; i+l-1 <= n; i++ {
			j := i + l - 1
			t := h[j] - h[i-1]*p[j-i+1]
			if vis[t] {
				return s[i-1 : j]
			}
			vis[t] = true
		}
		return ""
	}
	left, right := 0, n
	ans := ""
	for left < right {
		mid := (left + right + 1) >> 1
		t := check(mid)
		if len(t) > 0 {
			left = mid
			ans = t
		} else {
			right = mid - 1
		}
	}
	return ans
}

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