Answers for "mutateTheArray(n, a)"

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mutateTheArray(n, a)

def concatenationsSum(a):
    t=sum(a)
    t1=t*len(a)
    t2=sum([t*[len(str(x))-1 for x in a].count(j)*10**(j+1) for j in range(7)])
    return t1+t2


def mutateTheArray(n, a):
    return [a[0]+a[1]]+[sum(a[x-1:x+2])for x in range(1,len(a))]if n>1 else a

def countTinyPairs(a, b, k):
    return sum([1 if int(str(x)+str(y))<k else 0for x,y in zip(a,b[::-1])])


def meanGroups(a):
    d,e=[],[]
    for i,j in enumerate(a):
        if sum(j)/len(j)not in e:
            e+=[sum(j)/len(j)]
            d+=[[i]]
        else:
            d[e.index(sum(j)/len(j))]+=[i]
    return d

def alternatingSort(a):
    c,l=0,a[0]
    while c!=(len(a)-1)//2:
        c=-c if c<0 else -c-1
        if a[c]<=l:
            return False
        l=a[c]
    return True

def mergeStrings(s1, s2):
    s,o1,o2='',s1,s2
    while len(s1)*len(s2)!=0:
        if o1.count(s1[0])>o2.count(s2[0]) or (o1.count(s1[0])==o2.count(s2[0]) and s1[0]>s2[0]):
            s+=s2[0]
            s2=s2[1:]
        else:
            s+=s1[0]
            s1=s1[1:]
    return s+s1+s2


def hashMap(queryType, query):
    d,s,c1,c2={},0,0,0
    for i,j in zip(queryType,query):
        if i == 'insert':
            d[j[0]-c1]=j[1]-c2
        elif i == 'addToValue':
            c2+=j[0]
        elif i == 'addToKey':
            c1+=j[0]
        elif i== 'get':
            s+=d[j[0]-c1]+c2
    return s
Posted by: Guest on November-21-2020

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