Solution file of Umar Farooq . i191289@nu.edu.pk, umar5555f@gmail.com

Q1/solution.py

@@ -1 +1,27 @@
 ## Add code below with answer clearly stated
+def factorial(num):
+    if num < 0: #if given number is negative or zero  its a invalid number
+        print("Not Defined!!!")
+        return None
+    elif num==1 or num== 0:
+        return 1
+    else:
+        var = 1
+        for i in range(1, num):
+            var *= i
+        return var
+## The above function will be used to calculate the factorial of given number.
+def sum_factorial(num):
+    if num < 1: #if given number is negative or zero  its a invalid number
+        print("Invalid Value!!!")
+        return None
+    if num == 1:
+        return 1
+    else:
+        fact_num = factorial(n) # calculate the factorial of tgiven number
+        var = 0 # initiallize a veriable
+        for i in str(fact_num): #First convert the number to string to iteratively choose one digit from the output
+            var += int(i) # add that number to the Var that is initially zero
+        return var
+
+print("The sum of factorial is ", sum_factorial(100))

Q2/solution.py

@@ -5,4 +5,24 @@
 #         self.next = next
 class Solution:
     def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        temp = head
+        c = 0
+        while temp:                             #Find the length of the linked list
+            c+=1
+            temp = temp.next
+        a = c-n+1                               #Calculate the node to be removed
+        i = 1
+        temp2 = head
+        prev = None
+        while i < a:                            #Traverse till the node to be removed
+            i+=1
+            prev = temp2                        #Prev pointer to point the previous node of the deletion node
+            temp2 = temp2.next
+        if temp2==head:
+            return head.next
+        prev.next = temp2.next                  #Link the previous node to the next of the deletion node
+        return head
 
+
+        # I've done this problem already 0n 02/21/2022 17:58 at that time my submission was faster than 97% people at leetcode.
+        # Myprofile link : https://leetcode.com/umikhaan/

Q3/solution.py

@@ -1,3 +1,49 @@
 class Solution:
     def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
+        if len(nums1+nums2)%2!=0: # check the length of the solution 
+            return sorted((nums1+nums2))[len(nums1+nums2)//2]/1 # Simple code to return the median if the length is odd
+        else:
+            return sum(sorted((nums1+nums2))[(len(nums1+nums2)//2)-1:(len(nums1+nums2)//2)+1])/2 # Simple code to return the median if the length is even.
+        #It is not efficient code. but easy to understand and simplest approach to my knowledge.
+
+
+# in this case runtime increases but its memory efficient.
+class Solution:
+    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
+        a = []
+        half = (len(nums1) + len(nums2))
+        idx = 0
+        idy = 0
+
+        while True:
+
+            if idx == len(nums1):
+                a += nums2[idy:]
+                break
+            elif idy == len(nums2):
+                a += nums1[idx:]
+                break
+            if nums1[idx] < nums2[idy]:
+                a.append(nums1[idx])
+                idx += 1
+            else:
+                a.append(nums2[idy])
+                idy += 1
+        return a[half // 2] if half % 2 != 0 else (a[half //2] + a[(half - 1) // 2]) / 2
+    # While this one is more efficient,.
+class Solution:
+    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
+        a = []
+        half = (len(nums1) + len(nums2))
+
+
+        while nums1 and nums2:
+            if nums1[0] < nums2[0]:
+                a.append(nums1.pop(0))
+            else:
+                a.append(nums2.pop(0))
+
+        a += nums1 + nums2
+
+        return a[half // 2] if half % 2 != 0 else (a[half //2] + a[(half - 1) // 2]) / 2