18/12/2020

Author: jameszu

Daily-challenge/Dec/17/README.md

@@ -25,9 +25,22 @@ The two tuples are:
 ## Code
 ```python
 class Solution:
-    def mirrorReflection(self, p: int, q: int) -> int:
-        while p % 2 == 0 and q % 2 == 0: p, q = p / 2, q / 2
-        return int(1 - p % 2 + q % 2)
-
+    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
+        count = 0
+        num_dict = {}
+        for i in C:
+            for j in D:
+                s = i + j
+                if s in num_dict:
+                    num_dict[s] += 1
+                else:
+                    num_dict[s] = 1
+
+        for i in range(len(A)):
+            for j in range(len(B)):
+                target = 0 - (A[i]+B[j])
+                if target in num_dict:
+                    count += num_dict[target]
+        return count
 ```
 

Daily-challenge/Dec/17/sol.py

@@ -1,4 +1,18 @@
 class Solution:
-    def mirrorReflection(self, p: int, q: int) -> int:
-        while p % 2 == 0 and q % 2 == 0: p, q = p / 2, q / 2
-        return int(1 - p % 2 + q % 2)
+    def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
+        count = 0
+        num_dict = {}
+        for i in C:
+            for j in D:
+                s = i + j
+                if s in num_dict:
+                    num_dict[s] += 1
+                else:
+                    num_dict[s] = 1
+
+        for i in range(len(A)):
+            for j in range(len(B)):
+                target = 0 - (A[i]+B[j])
+                if target in num_dict:
+                    count += num_dict[target]
+        return count

Daily-challenge/Dec/18/README.md

@@ -1,25 +1,32 @@
-# Merge Intervals
-Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
+# Increasing Triplet Subsequence
+Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
 
 
 
 Example 1:
 
-Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
-Output: [[1,6],[8,10],[15,18]]
-Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
+Input: nums = [1,2,3,4,5]
+Output: true
+Explanation: Any triplet where i < j < k is valid.
 Example 2:
 
-Input: intervals = [[1,4],[4,5]]
-Output: [[1,5]]
-Explanation: Intervals [1,4] and [4,5] are considered overlapping.
+Input: nums = [5,4,3,2,1]
+Output: false
+Explanation: No triplet exists.
+Example 3:
+
+Input: nums = [2,1,5,0,4,6]
+Output: true
+Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
 
 
 Constraints:
 
-1 <= intervals.length <= 104
-intervals[i].length == 2
-0 <= starti <= endi <= 104<br>
+1 <= nums.length <= 105
+-231 <= nums[i] <= 231 - 1
+ 
+
+Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?<br>
 
 ## Idea
 Just do while loop and convert to decimal