DSA

02_Dynamic-Programming/01. Integer Break.py

@@ -33,7 +33,8 @@ def integerBreak(self, n):
                 i += 1
             dp.append(max_product)
         return dp[n]
-
+# Time: O(n^2)
+# Space: O(n)
 
 # Method 3 ------ Mathematics -------
 '''

08_Heap/02. Median of Running stream of Integers.py

@@ -0,0 +1,8 @@
+# https://www.facebookrecruiting.com/portal/coding_practice_question/?problem_id=547645422524434&ppid=454615229006519&practice_plan=0
+# https://practice.geeksforgeeks.org/problems/find-median-in-a-stream-1587115620/1
+# https://www.interviewbit.com/blog/find-median-in-a-stream/
+
+
+
+
+

19_HashMap/05. Majority Element.py

@@ -0,0 +1,20 @@
+# https://leetcode.com/problems/majority-element/
+
+class Solution:
+    def majorityElement(self, nums: List[int]) -> int:
+        me = -1   # me = Majority Element
+        count = 0      # count = Count of Majority Element
+
+        for i in range(len(nums)):
+            if nums[i] == me:
+                count += 1
+            elif count > 0 and nums[i] != me:
+                count -= 1
+
+            if count == 0:
+                me = nums[i]
+                count = 1
+
+        if nums.count(me) >= len(nums) // 2:
+            return me
+        return -1

19_HashMap/06. Majority Element II.py

@@ -0,0 +1,29 @@
+# https://leetcode.com/problems/majority-element-ii/
+
+class Solution:
+    def majorityElement(self, nums: List[int]) -> List[int]:
+        # Note:  *There will be only two majority element*. suppose there are 3 majority element then count of each element will be n/3 and there will be only there 3 unique elements in the array. But question is elements that appear *more than ⌊ n/3 ⌋ times*.
+        n = len(nums)
+        me1 = -1    # Majority element 1
+        me2 = -2    # Majority element 2
+        count1 = 0  # Count of Majority element 1
+        count2 = 0  # Count of Majority element 2
+
+        for num in nums:
+            if num == me1: count1 += 1
+
+            elif num == me2: count2 += 1
+
+            elif count1 == 0: me1 = num; count1 = 1
+
+            elif count2 == 0: me2 = num; count2 = 1
+
+            else:
+                count1 -= 1
+                count2 -= 1
+
+        res = set()  # taking set instead of list as me1 and me2 can be equal
+        if nums.count(me1) > n // 3: res.add(me1)
+        if nums.count(me2) > n // 3: res.add(me2)
+
+        return res