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Formatted question description: https://leetcode.ca/all/1044.html
1044. Longest Duplicate Substring (Hard)
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: "banana" Output: "ana"
Example 2:
Input: "abcd" Output: ""
Note:
2 <= S.length <= 10^5Sconsists of lowercase English letters.
Related Topics:
Hash Table, Binary Search
Solution 1. Binary Search + Rabin Karp
// OJ: https://leetcode.com/problems/longest-duplicate-substring/
// Time: average O(NlogN), worst O(N^2 * logN)
// Space: O(N)
class Solution {
int findDup(string &s, int len) {
unordered_map<unsigned, vector<int>> m;
unsigned h = 0, p = 1, d = 16777619;
for (int i = 0; i < s.size(); ++i) {
h = h * d + s[i] - 'a';
if (i >= len) h -= (s[i - len] - 'a') * p;
else p *= d;
if (i >= len - 1) {
if (m.count(h)) {
for (int k : m[h]) {
int j = 0;
for (; j < len && s[k + j] == s[i - len + 1 + j]; ++j);
if (j == len) return k;
}
}
m[h].push_back(i - len + 1);
}
}
return -1;
}
public:
string longestDupSubstring(string S) {
int L = 1, R = S.size(), start = 0;
while (L <= R) {
int M = (L + R) / 2, i = findDup(S, M);
if (i != -1) {
L = M + 1;
start = i;
} else R = M - 1;
}
return S.substr(start, R);
}
};
Solution 2.
// OJ: https://leetcode.com/problems/longest-duplicate-substring/
// Time: O(N^2 * logN)
// Space: O(N)
// Ref: https://leetcode.com/problems/longest-duplicate-substring/discuss/695101/C%2B%2B-short-O(n-log(n))-solution-with-std%3A%3Aunordered_setlessstd%3A%3Astring_viewgreater
class Solution {
public:
string longestDupSubstring(string S) {
int N = S.size(), L = 0, R = N - 1;
string_view ans;
while (L <= R) {
unordered_set<string_view> s;
int M = (L + R) / 2;
bool found = false;
for (int i = 0; i <= N - M; ++i) {
const auto [it, inserted] = s.emplace(S.data() + i, M);
if (!inserted) {
found = true;
ans = *it;
break;
}
}
if (found) L = M + 1;
else R = M - 1;
}
return {ans.begin(), ans.end()};
}
};
-
class Solution { public String longestDupSubstring(String S) { int length = S.length(); int[] nums = new int[length]; for (int i = 0; i < length; i++) nums[i] = S.charAt(i) - 'a'; int base = 26; final long MODULO = (long) Math.pow(2, 32); int low = 1, high = length; while (low < high) { int subLength = (high - low) / 2 + low; if (search(subLength, base, MODULO, length, nums) >= 0) low = subLength + 1; else high = subLength; } int start = search(low - 1, base, MODULO, length, nums); if (start >= 0) return S.substring(start, start + low - 1); else return ""; } public int search(int subLength, int base, final long MODULO, int length, int[] nums) { long hash = 0; for (int i = 0; i < subLength; i++) hash = (hash * base + nums[i]) % MODULO; Set<Long> set = new HashSet<Long>(); set.add(hash); long baseLong = 1; for (int i = 0; i < subLength; i++) baseLong = (baseLong * base) % MODULO; int maxIndex = length - subLength; for (int i = 1; i <= maxIndex; i++) { hash = (hash * base - nums[i - 1] * baseLong % MODULO + MODULO) % MODULO; hash = (hash + nums[i + subLength - 1]) % MODULO; if (set.contains(hash)) return i; set.add(hash); } return -1; } } ############ 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 ""; } } -
// OJ: https://leetcode.com/problems/longest-duplicate-substring/ // Time: O(NlogN) // Space: O(N) class Solution { int findDup(string &s, int len) { unordered_set<unsigned long long> st; unsigned long long d = 16777619, h = 0, p = 1; for (int i = 0; i < s.size(); ++i) { h = h * d + s[i]; if (i < len) p *= d; else h -= s[i - len] * p; if (i >= len - 1) { if (st.count(h)) return i - len + 1; st.insert(h); } } return -1; } public: string longestDupSubstring(string s) { int L = 0, R = s.size() - 1, start = 0; while (L < R) { int M = (L + R + 1) / 2, i = findDup(s, M); if (i != -1) { L = M; start = i; } else R = M - 1; } return s.substr(start, L); } }; -
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 -
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 }