Add solution #1044

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,117 LeetCode solutions in JavaScript
+# 1,118 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -851,6 +851,7 @@
 1041|[Robot Bounded In Circle](./solutions/1041-robot-bounded-in-circle.js)|Medium|
 1042|[Flower Planting With No Adjacent](./solutions/1042-flower-planting-with-no-adjacent.js)|Medium|
 1043|[Partition Array for Maximum Sum](./solutions/1043-partition-array-for-maximum-sum.js)|Medium|
+1044|[Longest Duplicate Substring](./solutions/1044-longest-duplicate-substring.js)|Hard|
 1047|[Remove All Adjacent Duplicates In String](./solutions/1047-remove-all-adjacent-duplicates-in-string.js)|Easy|
 1051|[Height Checker](./solutions/1051-height-checker.js)|Easy|
 1071|[Greatest Common Divisor of Strings](./solutions/1071-greatest-common-divisor-of-strings.js)|Easy|

solutions/1044-longest-duplicate-substring.js

@@ -0,0 +1,69 @@
+/**
+ * 1044. Longest Duplicate Substring
+ * https://leetcode.com/problems/longest-duplicate-substring/
+ * Difficulty: 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 "".
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var longestDupSubstring = function(s) {
+  const MOD = 1e9 + 7;
+  const BASE = 26;
+  let left = 1;
+  let right = s.length - 1;
+  let result = '';
+
+  while (left <= right) {
+    const mid = Math.floor((left + right) / 2);
+    const candidate = checkLength(mid);
+    if (candidate) {
+      result = candidate;
+      left = mid + 1;
+    } else {
+      right = mid - 1;
+    }
+  }
+
+  return result;
+
+  function checkLength(len) {
+    let hash = 0;
+    const seen = new Map();
+    let basePow = 1;
+    for (let i = 0; i < len - 1; i++) {
+      basePow = (basePow * BASE) % MOD;
+    }
+
+    for (let i = 0; i < len; i++) {
+      hash = (hash * BASE + (s.charCodeAt(i) - 97)) % MOD;
+    }
+
+    seen.set(hash, [0]);
+
+    for (let i = 1; i <= s.length - len; i++) {
+      hash = (hash - ((s.charCodeAt(i - 1) - 97) * basePow) % MOD + MOD) % MOD;
+      hash = (hash * BASE + (s.charCodeAt(i + len - 1) - 97)) % MOD;
+      if (seen.has(hash)) {
+        const prevIndices = seen.get(hash);
+        const currSubstr = s.slice(i, i + len);
+        for (const prev of prevIndices) {
+          if (s.slice(prev, prev + len) === currSubstr) {
+            return currSubstr;
+          }
+        }
+        prevIndices.push(i);
+      } else {
+        seen.set(hash, [i]);
+      }
+    }
+    return '';
+  }
+};