HackerEarth String Problems

Author: OmkarPathak

CompetitiveProgramming/HackerEarth/Algorithms/String/P11_CaesarsCipher.py

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+# Caesar's Cipher is a very famous encryption technique used in cryptography. It is a type of substitution
+# cipher in which each letter in the plaintext is replaced by a letter some fixed number of positions down
+# the alphabet. For example, with a shift of 3, D would be replaced by G, E would become H, X would become A
+# and so on.
+#
+# Encryption of a letter X by a shift K can be described mathematically as
+# EK(X)=(X+K) % 26.
+#
+# Given a plaintext and it's corresponding ciphertext, output the minimum non-negative value of shift that was
+# used to encrypt the plaintext or else output −1 if it is not possible to obtain the given ciphertext from
+# the given plaintext using Caesar's Cipher technique.
+#
+# Input:
+#
+# The first line of the input contains Q, denoting the number of queries.
+#
+# The next Q lines contain two strings S and T consisting of only upper-case letters.
+#
+# Output:
+#
+# For each test-case, output a single non-negative integer denoting the minimum value of shift that was used
+# to encrypt the plaintext or else print −1 if the answer doesn't exist.
+#
+# Constraints:
+# 1≤Q≤5
+# 1≤|S|≤10^5
+# 1≤|T|≤10^5
+# |S| = |T|
+#
+# SAMPLE INPUT
+# 2
+# ABC
+# DEF
+# AAA
+# PQR
+#
+# SAMPLE OUTPUT
+# 3
+# -1
+
+# My Solution
+for _ in range(int(input())):
+    string_one = input()
+    string_two= input()
+    check_one = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
+    #            ZYXWVUTSRQPONMLKJIHGFEDCBA
+    check_two = check_one[::-1]
+    result = []
+    for i in range(len(string_one)):
+        if(check_one.find(string_one[i]) > check_one.find(string_two[i])):
+            result.append(check_two.find(string_one[i]) + check_one.find(string_two[i]) + 1)
+        else:
+            result.append(check_one.find(string_two[i]) - check_one.find(string_one[i]))
+
+    if result.count(result[0]) == len(string_one):
+        print(result[0])
+    else:
+        print(-1)
+
+# More Efficient Solution:
+tests = int(input().strip())
+for i in range(tests):
+    plain = input().strip()
+    cipher = input().strip()
+    shift = (ord(cipher[0])-ord(plain[0])+26)%26
+    valid = True
+    for j in range(len(plain)):
+        if (ord(cipher[j])-ord(plain[j])+26)%26 != shift:
+            valid = False
+            break
+    print(shift) if valid else print("-1")

CompetitiveProgramming/HackerEarth/Algorithms/String/P12_CompilerVersion.py

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+# You are converting an old code for a new version of the compiler.
+#
+# In the old code we have used "->" for pointers. But now we have to replace each "->" with a ".". But this
+# replacement shouldn't be done inside commments. A comment is a string that starts with "//" and terminates
+# at the end of the line.
+#
+# Input:
+#
+# At max.
+# 2000
+# 2000 lines of code.
+#
+# Each line of code consists of at maximum
+# 60
+# 60 characters.
+#
+# Output:
+#
+# New code with required changes.
+#
+# SAMPLE INPUT
+# int t; //variable t
+# t->a=0;  //t->a does something
+# return 0;
+#
+# SAMPLE OUTPUT
+# int t; //variable t
+# t.a=0;  //t->a does something
+# return 0;
+
+import sys
+import re
+while True:
+    line = sys.stdin.readline()
+    if len(line) >=2 :
+        if '//' in line:
+            line = re.split("//", line)
+            line[0] = re.sub("->", ".", line[0])
+            sys.stdout.write('//'.join(line))
+        else:
+            sys.stdout.write(re.sub("->", ".", line))
+    else:
+        break

CompetitiveProgramming/HackerEarth/Algorithms/String/P13_NameGame.py

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+# John has recently learned about ASCII values. With his knowledge of ASCII values and character he has
+# developed a special word and named it John's Magical word.
+#
+# A word which consists of alphabets whose ASCII values is a prime number is a John's Magical word. An
+# alphabet is john's Magical alphabet if its ASCII value is prime.
+#
+# John's nature is to boast about the things he know or have learnt about. So just to defame his friends he
+# gives few string to his friends and ask them to convert it to John's Magical word. None of his friends would
+# like to get insulted. Help them to convert the given strings to John's Magical Word.
+#
+# Rules for converting:
+#
+# 1.Each character should be replaced by the nearest John's Magical alphabet.
+#
+# 2.If the character is equidistant with 2 Magical alphabets. The one with lower ASCII value will be considered as its replacement.
+#
+# Input format:
+#
+# First line of input contains an integer T number of test cases. Each test case contains an integer N (denoting the length of the string) and a string S.
+#
+# Output Format:
+#
+# For each test case, print John's Magical Word in a new line.
+#
+# Constraints:
+#
+# 1 <= T <= 100
+#
+# 1 <= |S| <= 500
+#
+# SAMPLE INPUT
+# 1
+# 8
+# KINGKONG
+#
+# SAMPLE OUTPUT
+# IIOGIOOG
+
+numl = [97, 101, 103, 107, 109, 113]
+numu = [67, 71, 73, 79, 83, 89]
+num = [67, 89, 97, 113]
+
+for _ in range(int(input())):
+	length = input()
+	string = input()
+	result = ''
+
+	for char in string:
+		if char.islower() and char.isalpha():
+			minimum = 200
+			ascii_char = ord(char)
+			temp = 0
+
+			for j in numl:
+				if minimum > abs(ascii_char - j):
+					minimum = abs(ascii_char - j)
+					temp = j
+
+			result = result + chr(temp)
+
+		elif char.isupper() and char.isalpha():
+			minimum = 200
+			ascii_char = ord(char)
+			temp = 0
+
+			for j in numu:
+				if minimum > abs(ascii_char - j):
+					minimum = abs(ascii_char - j)
+					temp = j
+
+			result = result + chr(temp)
+
+		else:
+			minimum = 200
+			ascii_char = ord(char)
+			temp = 0
+
+			for j in num:
+				if minimum > abs(ascii_char - j):
+					minimum = abs(ascii_char - j)
+					temp = j
+
+			result = result + chr(temp)
+
+	print(result)