Some Algo Added

ranamabdullah · ranamabdullah · commit e63c5c8edad7 · 2021-10-01T22:56:31.000+05:00
diff --git a/Other Algorithms/Add_Strings.py b/Other Algorithms/Add_Strings.py
@@ -0,0 +1,35 @@
+# Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
+# You must not use any built-in BigInteger library or convert the inputs to integer directly.
+
+#Notes:
+#Both num1 and num2 contains only digits 0-9.
+#Both num1 and num2 does not contain any leading zero.
+
+num1 = '364'
+num2 = '1836'
+
+# Approach 1: 
+def solution(num1,num2):
+    eval(num1) + eval(num2)
+    return str(eval(num1) + eval(num2))
+            
+print(solution(num1,num2))
+
+#Approach2 
+#Given a string of length one, the ord() function returns an integer representing the Unicode code point of the character 
+#when the argument is a unicode object, or the value of the byte when the argument is an 8-bit string.
+
+def solution(num1, num2):
+    n1, n2 = 0, 0
+    m1, m2 = 10**(len(num1)-1), 10**(len(num2)-1)
+
+    for i in num1:
+        n1 += (ord(i) - ord("0")) * m1 
+        m1 = m1//10        
+
+    for i in num2:
+        n2 += (ord(i) - ord("0")) * m2
+        m2 = m2//10
+
+    return str(n1 + n2)
+print(solution(num1, num2))
diff --git a/Other Algorithms/First_Unique_Character.py b/Other Algorithms/First_Unique_Character.py
@@ -0,0 +1,39 @@
+     
+# Given a string, find the first non-repeating character in it and return its index. 
+# If it doesn't exist, return -1. # Note: all the input strings are already lowercase.
+
+#Approach 1
+def solution(s):
+    frequency = {}
+    for i in s:
+        if i not in frequency:
+            frequency[i] = 1
+        else:
+            frequency[i] +=1
+    for i in range(len(s)):
+        if frequency[s[i]] == 1:
+            return i
+    return -1
+
+print(solution('alphabet'))
+print(solution('barbados'))
+print(solution('crunchy'))
+
+print('###')
+
+#Approach 2
+import collections
+
+def solution(s):
+    # build hash map : character and how often it appears
+    count = collections.Counter(s) # <-- gives back a dictionary with words occurrence count 
+                                         #Counter({'l': 1, 'e': 3, 't': 1, 'c': 1, 'o': 1, 'd': 1})
+    # find the index
+    for idx, ch in enumerate(s):
+        if count[ch] == 1:
+            return idx     
+    return -1
+
+print(solution('alphabet'))
+print(solution('barbados'))
+print(solution('crunchy'))
