Added solution 311

Problems/311.txt

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+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.

Solutions/311.py

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+"""
+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)
+"""
+