Extending the above soln:
NUM_PER_DIGIT = 3
char c[][NUM_PER_DIGIT] = {"abc","def",...};
char n[] = 2156169 (number is pressed);
int k=7;
for i <-- 0 to NUM_PER_DIGIT ^ k
String s="";
for j <-- 0 to k
int index = getJthDigitinItotheBaseNumPerDigit(NUM_PER_DIGIT,i,j);
// ie get 1st digit in (022)3 returns 2
s += c[n[j]][index];
print(s);
Time Complexity: O((NUM_PER_DIGIT^k)*k^2);
On Mon, May 23, 2011 at 7:32 PM, anshu mishra <anshumishra6...@gmail.com>wrote:
> the same question i have asked in microsoft interview. (if it is the same
> :P)
>
> for 12 perutation are (ad, ae, af, bd, be, bf, cd, ce ,cf);
> i have given them 3 solution(recusrsive, stack based) and the last one what
> they wanted.
>
> take a tertiary number(n) = 3^(number of digits) in case of 12 it is equals
> to 2;
>
> i m solving the save problem. assuming on every key there are only 2
> alphabets.
>
> char c[][2] = {ab , cd, ......};
>
> char n[] = 2156169 (number is pressed);
> so k = 7;
> for (i = 0; i < (1<<k); i++){
> string s = "";
> for (j = 0; j < k; j++){
> s += c[n[j]][i & (1<<j)]
> }
> print(s);
> }
>
> same logic u can use for tertiary too.
>
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