From eb5a9516032d06cbb6686e1053019fdad38c6d78 Mon Sep 17 00:00:00 2001
From: Cho Yin Yong <choyiny@users.noreply.github.com>
Date: Wed, 11 Nov 2020 06:17:54 -0500
Subject: [PATCH] Peak of unimodal list DNC algorithm (#3691)

* Peak of unimodal list DNC algorithm

* fix black formatting issues

* add doctest testing

* make file black compliant
---
 divide_and_conquer/peak.py | 53 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 53 insertions(+)
 create mode 100644 divide_and_conquer/peak.py

diff --git a/divide_and_conquer/peak.py b/divide_and_conquer/peak.py
new file mode 100644
index 000000000..f94f83ed3
--- /dev/null
+++ b/divide_and_conquer/peak.py
@@ -0,0 +1,53 @@
+"""
+Finding the peak of a unimodal list using divide and conquer.
+A unimodal array is defined as follows: array is increasing up to index p,
+then decreasing afterwards. (for p >= 1)
+An obvious solution can be performed in O(n),
+to find the maximum of the array.
+(From Kleinberg and Tardos. Algorithm Design.
+Addison Wesley 2006: Chapter 5 Solved Exercise 1)
+"""
+from typing import List
+
+
+def peak(lst: List[int]) -> int:
+    """
+    Return the peak value of `lst`.
+    >>> peak([1, 2, 3, 4, 5, 4, 3, 2, 1])
+    5
+    >>> peak([1, 10, 9, 8, 7, 6, 5, 4])
+    10
+    >>> peak([1, 9, 8, 7])
+    9
+    >>> peak([1, 2, 3, 4, 5, 6, 7, 0])
+    7
+    >>> peak([1, 2, 3, 4, 3, 2, 1, 0, -1, -2])
+    4
+    """
+    # middle index
+    m = len(lst) // 2
+
+    # choose the middle 3 elements
+    three = lst[m - 1 : m + 2]
+
+    # if middle element is peak
+    if three[1] > three[0] and three[1] > three[2]:
+        return three[1]
+
+    # if increasing, recurse on right
+    elif three[0] < three[2]:
+        if len(lst[:m]) == 2:
+            m -= 1
+        return peak(lst[m:])
+
+    # decreasing
+    else:
+        if len(lst[:m]) == 2:
+            m += 1
+        return peak(lst[:m])
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()