From 126aab53c4187782f16b270e68c0af976cdc02e2 Mon Sep 17 00:00:00 2001
From: RamKuppagiri <kramfl5@gmail.com>
Date: Tue, 30 Dec 2025 19:48:43 -0600
Subject: [PATCH 1/3] Find peak element js code

---
 findPeakElement.js | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)
 create mode 100644 findPeakElement.js

diff --git a/findPeakElement.js b/findPeakElement.js
new file mode 100644
index 00000000..3c876509
--- /dev/null
+++ b/findPeakElement.js
@@ -0,0 +1,21 @@
+/**
+ * 
+  Intution is to use binary search as the requirement is to write an algorithm which completes in O(log n)
+  I  usually use the condition left<=right in the while loop, but this scenario i removed the equal to condition as return left covers it.
+  I am moving my pointers towards the peak always, if the peak resides on the left iam moving my right pointer towards left 
+  otherwise left pointer to the right.
+  
+ */
+var findPeakElement = function (nums) {
+    let left = 0, right = nums.length - 1;
+    while(left < right) {
+        const mid = Math.floor(left + (right - left) / 2);
+        if (nums[mid] < nums[mid + 1]) {
+            left = mid + 1;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return left;
+};
+

From b7c42a86abd77db385ca9bfd05c103e451be2a0b Mon Sep 17 00:00:00 2001
From: RamKuppagiri <kramfl5@gmail.com>
Date: Tue, 30 Dec 2025 19:48:57 -0600
Subject: [PATCH 2/3] comments updated

---
 findPeakElement.js | 3 ++-
 1 file changed, 2 insertions(+), 1 deletion(-)

diff --git a/findPeakElement.js b/findPeakElement.js
index 3c876509..e540b452 100644
--- a/findPeakElement.js
+++ b/findPeakElement.js
@@ -4,7 +4,8 @@
   I  usually use the condition left<=right in the while loop, but this scenario i removed the equal to condition as return left covers it.
   I am moving my pointers towards the peak always, if the peak resides on the left iam moving my right pointer towards left 
   otherwise left pointer to the right.
-  
+  Time complexity : O(log n)
+  Space Complexity: O(1) as i am not using any extra space other than left and right pointers.
  */
 var findPeakElement = function (nums) {
     let left = 0, right = nums.length - 1;

From e7cb58ebc91ba7434b1ea90cc7530af63b261cda Mon Sep 17 00:00:00 2001
From: RamKuppagiri <kramfl5@gmail.com>
Date: Tue, 30 Dec 2025 20:17:38 -0600
Subject: [PATCH 3/3] updated as per the suggestions

---
 findPeakElement.js | 5 ++---
 1 file changed, 2 insertions(+), 3 deletions(-)

diff --git a/findPeakElement.js b/findPeakElement.js
index e540b452..6690ded6 100644
--- a/findPeakElement.js
+++ b/findPeakElement.js
@@ -1,8 +1,7 @@
 /**
  * 
   Intution is to use binary search as the requirement is to write an algorithm which completes in O(log n)
-  I  usually use the condition left<=right in the while loop, but this scenario i removed the equal to condition as return left covers it.
-  I am moving my pointers towards the peak always, if the peak resides on the left iam moving my right pointer towards left 
+  I am moving my pointers towards the peak always, if the peak resides on the left iam moving my right pointer towards mid 
   otherwise left pointer to the right.
   Time complexity : O(log n)
   Space Complexity: O(1) as i am not using any extra space other than left and right pointers.
@@ -14,7 +13,7 @@ var findPeakElement = function (nums) {
         if (nums[mid] < nums[mid + 1]) {
             left = mid + 1;
         } else {
-            right = mid - 1;
+            right = mid;
         }
     }
     return left;