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 * 104s consists of lowercase English letters.
Solutions
Solution 1
PythonJavaC++Go
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(code-box)
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 "";
}
}(code-box)
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 "";
}
};(code-box)
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
}(code-box)
