adds leetcode 88 merge sorted arrays

Python/best-time-to-buy-sell-stock-i.py

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+'''
+https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
+'''
+#using Two-Ptrs
+
+class Solution:
+ def maxProfit(self, prices: List[int]) -> int:
+ l,r=0,1 #left ptr=buy, right ptr=sell
+ maxP=0
+
+ while r<len(prices):
+ if prices[l]<prices[r]:
+ profit=prices[r]-prices[l]
+ maxP=max(maxP,profit) #only updates if the current profit is higher
+ else:
+ l=r
+ r+=1
+ return maxP

Python/merge-sorted-arrays.py

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+'''
+https://leetcode.com/problems/merge-sorted-array/
+'''
+
+class Solution:
+ def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
+ last = m + n - 1 #this will be the last index of nums1
+
+ # m - 1 = this will be the last of actual elements in nums1 before 0 elems starts
+ # n - 1 = this will be the last of n
+
+ # merge in reverse order
+ while m > 0 and n > 0: #m and n are ptrs
+ if nums1[m - 1] > nums2[n - 1]:
+ nums1[last] = nums1[m - 1]
+ m -= 1 #decrementing as merging in reverse
+ else: 
+ nums1[last] = nums2[n -1]
+ n -= 1
+ last -= 1
+
+ # filling num1 with leftover num2 elements
+ while n > 0:
+ nums1[last] = nums2[n - 1]
+ n, last = n -1, last - 1 

Python/valid-anagram.py

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+'''https://leetcode.com/problems/valid-anagram/'''
+#brute
+'''We keep count of every distinct character in a hashmap for both S & T.
+Then compare if the count for each character in both is equal'''
+from typing import Counter
+
+
+class Solution:
+ def isAnagram(self, s: str, t: str) -> bool:
+ if len(s)!=len(t):
+ return False
+ S,T={},{}
+ for i in range(0,len(s)):
+ if s[i] in S.keys():
+ S[s[i]]+=1 #updating count
+ else:
+ S[s[i]]=1 #recording a distinct element
+
+ for i in range(0,len(t)):
+ if t[i] in T.keys():
+ T[t[i]]+=1
+ else:
+ T[t[i]]=1
+
+ for i in range(0,len(t)):
+ if t[i] in S.keys() and S[t[i]]==T[t[i]]:
+ continue
+ else:
+ return False
+ return True
+
+#cocky way 1
+class Solution:
+ def isAnagram(self, s: str, t: str) -> bool:
+ return Counter(s)==Counter(t) #compares the count of every distinct element
+
+#optimised in cocky way
+'''We sort both the strings and compare them side by side.
+Note that It is only space efficient but the one b4 is time efficient
+also your interviewer may want you to write your own sort function for this'''
+class Solution:
+ def isAnagram(self, s: str, t: str) -> bool:
+ return sorted(s)==sorted(t)