# Solution in Python 3

`def separateNumbers(s):    if len(s)==1:        print("NO")        return    for i in range(1, len(s)//2+1):        genstr = s[:i]        prev = int(genstr)        while len(genstr) < len(s):             next = prev+1             genstr+=str(next)             prev = next        if genstr == s:            print("YES", s[:i])            return    print("NO")`

# Solution in Python 3

`from collections import defaultdictn, m = map(int, input().split())d = defaultdict(list)for i in range(n):    d[input()].append(i+1)for j in range(m):    s = input()    if s in d:        print(*d[s])    else:        print(-1)`

# Solution in Python 3

`from collections import Counterdef pickingNumbers(a):        countNums = Counter(a)        maxnum=0                for i in range(1, 100):             maxnum = max(maxnum, countNums[i]+countNums[i+1])                                      return maxnum`

# Solution in Python 3

`def sockMerchant(n, ar):    countOfSocks = []    for sock in set(ar):        countOfSocks.append(ar.count(sock))     return sum([i//2 for i in countOfSocks])`

# Solution in Python 3

`def dayOfProgrammer(year):    if year < 1700 or year > 2700:        return                elif year==1918:        return '26.09.1918'        elif year < 1918:        if year%4==0:            return '12.09.'+str(year)                           else:            return '13.09.'+str(year)                    elif year > 1918:        if year%400==0 or year%4==0 and year%100!=0:            return '12.09.'+str(year) …`

# Solution in Python 3

`def beautifulDays(i, j, k):    beautifulDays = 0    for day in range(i, j+1):        rev = int(str(day)[::-1])        if abs(rev-day)%k==0:            beautifulDays+=1              return beautifulDays`

# Solution in Python 3

`def acmTeam(topic):    ans = []    for i in range(n):        for j in range(i+1, n):            know=0                        for a,b in zip(topic[i], topic[j]):                if a=='1' or b=='1':                   know+=1            ans.append(know)                     return [max(ans), ans.count(max(ans))]`

# Solution in Python 3

`def kangaroo(x1, v1, x2, v2):    if x1 < x2 and v1 < v2:        return 'NO'    else:        if v1!=v2 and (x2-x1)%(v2-v1)==0:            return 'YES'         else:            return 'NO'`

# The Minion Game : HackerRank Solution in Python ## Shounak Lohokare

pretending to teach is one of the best way to learn.