New Strings InterviewBit Solution

Author: rahulcode22

LINKEDLIST/Nth-Node-From-Last.py

@@ -0,0 +1,34 @@
+class Node:
+    def __init__(self,data):
+        self.data = data
+        self.next = None
+
+class LinkedList:
+    def __init__(self):
+        self.head = None
+
+    def push(self, new_data):
+        new_node = Node(new_data)
+        new_node.next = self.head
+        self.head = new_node
+
+    def printNthFromEnd(self,n):
+        main_ptr = self.head
+        ref_ptr = self.head
+
+        count = 0
+        if (self.head is not None):
+            while(count < n):
+                ref_ptr = ref_ptr.next
+                count += 1
+        while (ref_ptr is not None):
+            main_ptr = main_ptr.next
+            ref_ptr = ref_ptr.next
+        return main_ptr.data
+
+ll = LinkedList()
+ll.push(20)
+ll.push(4)
+ll.push(15)
+ll.push(35)
+print ll.printNthFromEnd(4)

Math/Sort-An-array-of-0_1.py

@@ -0,0 +1,18 @@
+def sort012(arr):
+    n = len(arr)
+    hi = n - 1
+    low = 0
+    high = n - 1
+    mid = 0
+    while mid <= high:
+        if arr[mid] == 0:
+            arr[low], arr[mid] = arr[mid], arr[low]
+            low += 1
+            mid += 1
+        elif arr[mid] == 1:
+            arr[mid], arr[high] = arr[high], arr[mid]
+            high -= 1
+    return arr
+
+arr = [0,1,1,0,1,1,0,0,0,1]
+print sort012(arr)

String/Add-binary-strings.py

@@ -0,0 +1,13 @@
+'''
+
+Given two binary strings, return their sum (also a binary string).
+'''
+class Solution:
+    # @param A : string
+    # @param B : string
+    # @return a strings
+    def addBinary(self, A, B):
+        a = int(A,2)
+        b = int(B,2)
+        c = bin(a+b)[2:]
+        return c

String/Implement-StrStr.py

@@ -0,0 +1,41 @@
+class Solution:
+    # @param haystack : string
+    # @param needle : string
+    # @return an integer
+    def strStr(self, txt, pat):
+        def computeLPSArray(pat,m,lps):
+            j = 0 #length of previous lps
+            lps[0] = 0
+            i = 1
+            while i < m:
+                if pat[i] == pat[j]:
+                    j += 1
+                    lps[i] = j
+                    i += 1
+                else:
+                    if j != 0:
+                        j = lps[j-1]
+                    else:
+                        lps[i] = 0
+                        i += 1
+        m = len(pat)
+        n = len(txt)
+        #Now create a lps array that will hold the longest prefix suffix value
+        lps = [0]*m
+        j = 0 #index for pat
+        computeLPSArray(pat,m,lps)
+
+        i = 0 #index For txt[]
+
+        while i < n:
+            if pat[j] == txt[i]:
+                i += 1
+                j += 1
+            if j == m:
+                return (i-j)
+            elif i < n and pat[j] != txt[i]:
+                if j != 0:
+                    j = lps[j-1]
+                else:
+                    i += 1
+        return -1

String/Integer-To-Roman.py

@@ -0,0 +1,13 @@
+def integerToRoman(num):
+    m = ["", "M" ,"MM", "MMM"]
+    c = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
+    x = ["", "X", "XX", "XXX", "XL", "L" ,"LX", "LXX", "LXXX", "XC"]
+    i = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
+    thousands = m[num/1000]
+    hundreds = c[(num%1000)/100]
+    tens = x[(num%100)/10]
+    ones = i[num%10]
+    ans = thousands + hundreds + tens + ones
+    return ans
+
+print integerToRoman(3549)

String/KMP-Algo.py

@@ -39,6 +39,6 @@ def computeLPSArray(pat,m,lps):
                 i += 1
 
 #Now time to test
-txt = "ABABDABACDABABCABAB"
-pat  = "ABABCABAB"
+txt = "abdefgh"
+pat  = "defg"
 KMPSearch(pat,txt)

String/Power-of-2.py

@@ -0,0 +1,16 @@
+'''
+Find if Given number is power of 2 or not.
+More specifically, find if given number can be expressed as 2^k where k >= 1.
+'''
+class Solution:
+    # @param A : string
+    # @return an integer
+    def power(self, num):
+        num = int(num)
+        if num == 1:
+            return 0
+        while num != 1:
+            if num%2 != 0:
+                return 0
+            num = num/2
+        return 1

String/addBinaryString.py

@@ -0,0 +1,46 @@
+'''
+Given two binary strings, return their sum (also a binary string).
+'''
+def binarySum(a,b):
+    n1 = len(a)
+    n2 = len(b)
+    if n1 > n2:
+        b = "0"*(n1-n2) + b
+    elif n2 > n1:
+        a = "0"*(n2-n1) + a
+    carry = 0
+    a = list(a)
+    b = list(b)
+    for i in range(n2-1,-1,-1):
+        if (a[i] == b[i]) and a[i] == '0':
+            a[i] = str(0+carry)
+            carry = 0
+        elif (a[i] == b[i]) and a[i] == '1':
+            a[i] = str(0 + carry)
+            carry = 1
+        else:
+            if carry == 1:
+                a[i] = str(0)
+                carry = 1
+            else:
+                a[i] = str(1)
+                carry = 0
+    if carry == 1:
+        a = ["1"] + a
+
+    return ''.join(a)
+
+a = "1110000000010110111010100100111"
+b = "101001"
+
+print binarySum(a,b)
+
+
+'''
+Another Approach
+'''
+def add(a,b):
+    a = int(a,2)
+    b = int(b,2)
+    c = bin(a+b)[2:]
+    return c