@Jalaj
if u consider only Space complexity then it can be done in O(n^2) time
and constant space
here i assume string u taken only hv a-z character nothing else
try this
for(i=0;i<length;i++){
if( ! (st[i]>='a' && st[i]<='z') )
continue;
count=1;
for(j=i+1;j<length;j++){
if(st[j]==st[i]){
count++;
st[j]+=30;
}
}
cout<<st[i]<<count;
}
On May 17, 2:46 pm, kaushik sur <kaushik....@gmail.com> wrote:
> Hi Friends
>
> Hash Map takes 2byte [in Java] for holding a character
>
> So in Amazon -
> It takes A - 1
> M - 1
> Z - 1
> O - 1
> N - 1
>
> But it's time effective!
>
> Yes it takes additional space for intergers, for each key 4 byte for an
> integer!!! :-(
>
> ***********
>
> public void checkTheFrequency() {
> for (int i = 0; i < str.length(); i++) {
> char key = str.charAt(i);
>
> if (checkMap.containsKey(key)!=
> false) {
> int value = checkMap.get(key);
> checkMap.put(key, ++value);
> } else {
> checkMap.put(key, 1);
> }
> }
>
> *****************
> Best Regards
> Kaushik
>
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