From 5dea68d75eac7ab4d8382cc25dd68155e6d8cf2c Mon Sep 17 00:00:00 2001
From: WHAHA-HA <kingsman00127@gmail.com>
Date: Sat, 26 Sep 2020 08:56:07 +0530
Subject: [PATCH] Added solution 311

---
 Problems/311.txt |  5 +++++
 Solutions/311.py | 45 +++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 50 insertions(+)
 create mode 100644 Problems/311.txt
 create mode 100644 Solutions/311.py

diff --git a/Problems/311.txt b/Problems/311.txt
new file mode 100644
index 0000000..ed26f6d
--- /dev/null
+++ b/Problems/311.txt
@@ -0,0 +1,5 @@
+PROBLEM 311:
+
+Given an unsorted array, in which all elements are distinct, find a "peak" element in O(log N) time.
+
+An element is considered a peak if it is greater than both its left and right neighbors. It is guaranteed that the first and last elements are lower than all others.
\ No newline at end of file
diff --git a/Solutions/311.py b/Solutions/311.py
new file mode 100644
index 0000000..435306a
--- /dev/null
+++ b/Solutions/311.py
@@ -0,0 +1,45 @@
+"""
+Problem:
+
+Given an unsorted array, in which all elements are distinct, find a "peak" element in
+O(log N) time.
+
+An element is considered a peak if it is greater than both its left and right
+neighbors. It is guaranteed that the first and last elements are lower than all others.
+"""
+
+from typing import List
+
+
+def get_peak(arr: List[int]) -> int:
+    # implement similar method as binary search [since the element being searched is
+    # not a concrete value (unlike binary search), but any value which is greater than
+    # its neighbours, it can only be found without sorting]
+    mid = len(arr) // 2
+    if (
+        mid > 0
+        and arr[mid - 1] < arr[mid]
+        and mid < len(arr)
+        and arr[mid + 1] < arr[mid]
+    ):
+        return arr[mid]
+    elif mid > 0 and arr[mid - 1] < arr[mid]:
+        return get_peak(arr[mid:])
+    return get_peak(arr[: mid + 1])
+
+
+if __name__ == "__main__":
+    print(get_peak([0, 2, 4, -1, 3, 1]))
+    print(get_peak([0, 2, 4, 5, 3, 1]))
+    print(get_peak([0, 2, 6, 5, 3, 1]))
+    print(get_peak([0, 2, 4, 5, 7, 1]))
+    print(get_peak([0, 8, 7, 5, 16, 1]))
+
+
+"""
+SPECS:
+
+TIME COMPLEXITY: O(log(n))
+SPACE COMPLEXITY: O(1)
+"""
+