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Author: iamnishantchandra

10X/Python/Array/Choclate Bars.py

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+'''
+Chocolate Bars
+There are N chocolates denoted by array A where A[i] is the length of the i-th chocolate. Alice can 
+melt each chocolate and then convert it into a chocolate whose length is any divisor of the number A[i].
+So, a chocolate of length A[i] can be converted into X different types of chocolate where X is the count 
+of divisors of the number A[i]. So you need to count the total unordered pair (i, j) of chocolates such 
+that their X value is same.
+
+Input
+The first line contains an integer N as input denoting the total number of elements in the array A. 
+The next line contains N space-separated integers that denote the elements of the array A.
+
+Output
+In the output, print the total number of ways as mentioned in the statement.
+
+Example
+Input:
+
+3
+
+2 3 4
+
+Output:
+
+1
+
+Explanation
+X values of each number .
+
+A[1] = 2 , divisors of A[1] are 1, 2.Thus X value of A[1] is 2.
+
+similarly, A[2] = 3, divisors of A[2] are 1, 3.Thus X value of A[2] is 2.
+
+A[3] = 4, divisors of A[3] are 1, 2, 4. Thus X value of A[3] is 3.
+
+Thus we have only one pair (1, 2) for which X value is 2.
+
+'''
+import math
+
+def factor(n):
+    count=0
+    for i in range(1,1+int(math.sqrt(n))):
+        if n%i==0:
+            count+=1
+            if n//i!=i:
+                count+=1
+    return count
+
+def result(ar,n):
+    d={}
+    for i in ar:
+        x=factor(i)
+        if x in d:
+            d[x]+=1
+        else:
+            d[x]=1
+    res=0
+    for x in d:
+        res+=(d[x]*(d[x]-1))//2
+    return res
+
+
+n=int(input())
+ar=[int(x) for x in input().split()]
+print(result(ar,n))
+
+# Help.....

10X/Python/Array/Common_Alphabet.py

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+'''
+Common Alphabets
+Given list of strings. The task is to find the frequency of the elements which are common in all strings 
+given in the list of strings.
+
+Input:
+First line contains N, number of strings.
+
+Next N lines contians a string in each line.
+
+Output:
+N linese each line contains an alphabet and it's frequency in sorted order for each of the input string.
+
+alphabet and the frequency are separated by a space
+
+Problem Constraints:
+1<=A.length<=500
+
+1<=A[0].length<=100
+
+Examples:
+Input :
+
+3
+
+gefgek
+
+gfgk
+
+kinfgg
+
+Output :
+
+f 1
+
+g 2
+
+k 1
+
+Explanation :
+f occurs once in all Strings.
+
+g occurs twice in all the strings.
+
+k occurs once in all string.
+
+Output is in ascending order
+
+'''
+from typing import Counter
+n=int(input())
+arr=[]
+for i in range(n):
+    arr.append(input())
+dic=Counter(arr[0])
+for i in range(1,n):
+    d=Counter(arr[i])
+    for i in dic:
+        if d[i]==0:
+            dic[i]=-1
+        else:
+            dic[i]=min(dic[i],d[i])
+for i in sorted(dic):
+    if(dic[i]>0):
+        print(i+" "+str(dic[i]))
+

10X/Python/Array/FINDING HELLO.PY

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+'''
+Find Hello
+You are given a string. You are supposed to find if "hello" exists in the given string as a substring. See explanation for better understanding. If "hello" exists print "Yes" else print "No".
+
+Note: length of string is greater than 0.
+
+Input
+First line contains n number of strings for which you need to check. After this n lines will follow. Each containing a string.
+
+Output
+Print "Yes" or "No".
+
+Input:
+
+2
+
+Academy
+
+ramhellolaxman
+
+Output:
+
+No
+
+Yes
+
+Explanation
+Testcase 1: There is no hello in the word Academy.
+
+Testcase 2: hello starts at index 3 and ends at 7.
+'''
+'''
+def Finding_Hello(s):
+    l=len(s)
+    w="hello"
+    f=False
+    for i in range(0,(len(s)-4)):
+        if w==s[i:i+5]:
+            return True
+    return f
+
+for _ in range(int(input())):
+    print(Finding_Hello(input()))
+
+'''
+def Finding_Hello(s,target):
+    if target in s:
+        return "Yes"
+    else:
+        return "No"
+target="hello"
+for _ in range(int(input())):
+    print(Finding_Hello(input(),target))

10X/Python/Array/IsAngram.py

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+"""
+IS Anagram
+You are given two strings A and B, you have to check whether they are anagrams of one another. An anagram 
+is a word that can be obtained by rearranging the characters of the initial word.
+
+Note- DO NOT use the inbuilt sort function at any stage of the code
+
+Input
+First line contains the string A and second line contains the string B 1<=len(A),len(B)<=2000
+
+Output
+Print "1" if they are anagrams and print "0" if they are not anagrams
+
+Example
+Input:
+
+ABABCD
+
+AABBCD
+
+Output:
+
+1
+
+explanation
+
+Since ABABCD can be rearranged to get AABBCD we print "1" as output
+"""
+'''
+s1=input()
+s2=input()
+d1={}
+d2={}
+if len(s1)==len(s2):
+    for i in range(len(s1)):
+        j=s1[i]
+        if j in d1:
+            d1[j]+=1
+        else:
+            d1[j]=1
+    for i in range(len(s2)):
+        j=s2[i]
+        if j in d2:
+            d2[j]+=1
+        else:
+            d2[j]=1
+    if sorted(d1)==sorted(d2):
+        print(1)
+    else:
+        print(0)
+else:
+    print(0)
+
+'''
+#######################################
+s1=input()
+s2=input()
+d1={}
+d2={}
+if len(s1)==len(s2):
+    for i in range(len(s1)):
+        j=s1[i]
+        if j in d1:
+            d1[j]+=1
+        else:
+            d1[j]=1
+    for i in range(len(s2)):
+        j=s2[i]
+        if j in d2:
+            d2[j]+=1
+        else:
+            d2[j]=1
+    #if sorted(d1)==sorted(d2):
+    if d1==d2:
+        print(d1)
+        print(d2)
+        print(1)
+    else:
+        print(0)
+else:
+        print(0)
+

10X/Python/Array/Sport_sum.py

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+"""
+Sports Stats
+Max likes football very much. In order to check the popuplarity of the game, he decided to talk to random people and ask them about their favourite game and note it down in a list.
+
+Given a list of name of people and their favourite sport, help Max in finding the sport liked by most of the people, and also how many people like football.
+
+He could have met same people more than once, he will only count response of his first meet with any person.
+
+Note : Name of person as well as sport is a single string in lowercase. Length of name of people as well as sport is less than 11.
+
+Input
+First line will contain no of entries in the list. i.e. N Next N lines will contain two strings, person's name and sports he like.
+
+Output
+In first ine, name of sport liked by most no of people (In case of more than one games is liked by highest no of people, output the first one in lexicographical order). In second line, no of people having football as their favourite game.
+
+Example
+Input:
+
+7
+
+abir cricket
+
+aayush cricket
+
+newton kabaddi
+
+abhinash badminton
+
+sanjay tennis
+
+abhinash badminton
+
+govind football
+
+Output:
+
+cricket
+
+1
+
+Explanation
+2 people likes cricket, ans so it liked by maximum people. 1 person has football as his favourite sport
+"""
+'''
+d={}
+max=0
+sports=""
+for _ in range(int(input())):
+    data=input().split(" ")
+    d[data[0]]=data[1]
+for key in d:
+    count=0
+    for j in d:
+        if d[key]==d[j]:
+        #if key==j:
+            count=count+1
+            if count>max:
+                max=count
+                sports=d[key]
+print(sports)        
+find="football"
+count=0
+for key in d:
+    if d[key]==find:
+        count+=1
+print(count)
+
+#80%
+'''
+d={}
+max=0
+# sports=""
+nd={}
+sp=""
+for _ in range(int(input())):
+    data=input().split(" ")
+    d[data[0]]=data[1]
+for i in d:
+    j=d[i]
+    if j in nd:
+        nd[j]+=1
+    else:
+        nd[j]=1
+for i in sorted(nd):
+    if nd[i]>max:
+        max=nd[i]
+        sp=i
+print(sp)
+# for i in d:
+#     if d[i]==sp:
+#         sports=d[i]
+find="football"
+count=0
+for key in d:
+    if d[key]==find:
+        count+=1
+print(count)