From 7264dcd8020e7974f0e198a2aea9e827674a6ac7 Mon Sep 17 00:00:00 2001
From: Abdul Shaik <starabdul100@gmail.com>
Date: Thu, 6 Jul 2023 13:17:38 +0530
Subject: [PATCH] Implement Binary Search algorithm for finding Longest
 Increasing Subsequence

---
 ...creasingSubsequenceBinarySearchApproach.js | 37 +++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 src/algorithms/sets/longest-increasing-subsequence/dpLongestIncreasingSubsequenceBinarySearchApproach.js

diff --git a/src/algorithms/sets/longest-increasing-subsequence/dpLongestIncreasingSubsequenceBinarySearchApproach.js b/src/algorithms/sets/longest-increasing-subsequence/dpLongestIncreasingSubsequenceBinarySearchApproach.js
new file mode 100644
index 00000000..7111d8ec
--- /dev/null
+++ b/src/algorithms/sets/longest-increasing-subsequence/dpLongestIncreasingSubsequenceBinarySearchApproach.js
@@ -0,0 +1,37 @@
+/**
+ * Binary search-based approach to find longest increasing subsequence.
+ * Complexity: O(n log n)
+ *
+ * @param {number[]} sequence
+ * @return {number}
+ */
+export default function binarySearchLongestIncreasingSubsequence(sequence) {
+  const tails = []; // Array to store the smallest tail elements of increasing subsequences
+
+  // Iterate over the input sequence
+  for (let i = 0; i < sequence.length; i++) {
+    let left = 0;
+    let right = tails.length - 1;
+
+    // Perform binary search to find the largest index `j` where tails[j] <= sequence[i]
+    while (left <= right) {
+      const mid = Math.floor((left + right) / 2);
+      if (tails[mid] <= sequence[i]) {
+        left = mid + 1;
+      } else {
+        right = mid - 1;
+      }
+    }
+
+    // Update tails[j + 1] to sequence[i]
+    if (left === tails.length) {
+      tails.push(sequence[i]);
+    } else {
+      tails[left] = sequence[i];
+    }
+  }
+
+  // The length of the longest increasing subsequence is the index of the rightmost non-zero element in tails
+  return tails.length;
+}
+