Create 169_majority_element.md

Author: Joseph Luce

leetcode/easy/169_majority_element.md

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+# 169. Majority Element
+
+## Solution with dictionary
+- Runtime: O(N)
+- Space: O(N)
+- N = Number of elements in array
+
+Fairly straight forward, use a dictionary to count the number of occurrences for each number. Then compare it to the current best.
+
+```
+from collections import defaultdict
+
+class Solution:
+    def majorityElement(self, nums: List[int]) -> int:
+        counter = defaultdict(lambda: 0)
+        curr_major, curr_count = None, 0
+        for n in nums:
+            counter[n] += 1
+            if counter[n] > curr_count:
+                curr_major = n
+                curr_count = counter[n]
+        return curr_major
+```
+
+## Solution with constant space
+- Runtime: O(N)
+- Space: O(1)
+- N = Number of elements in array
+
+Keeping a dictionary of occurrences when you just want to find one major element is performing extra work.
+You can instead pass through the array once while keeping track of your current major number and any other number is a reason your current major number should be decremented, else increment it.
+Once your current major number's count reaches zero, then it is a reason a new major number takes its place.
+
+```
+class Solution:
+    def majorityElement(self, nums: List[int]) -> int:
+        curr_major, curr_count = None, 0 
+        for n in nums:
+            if curr_major != n:
+                curr_count -= 1
+            else:
+                curr_count += 1
+            if curr_count <= 0:
+                curr_major, curr_count = n, 1
+        return curr_major
+```