CareerCup

Another example: it takes aabbcccddddbbbb and returns a2b2c3d4b4.

this is easy using map since it is sorted with map

{void count_item(int a[],int len){
int item, item_count, present_idx;
int i;
i = 0;
present_idx = 0;
item_count = 1;
item = a[i];
for (i=1; i<len; i++) {
if (item ==a[i]) {
item_count++;
}
else {
a[present_idx++] = item;
a[present_idx++] = item_count;
item_count = 1;
item = a[i];
}
}
a[present_idx++] = item;
a[present_idx++] = item_count;
a[present_idx]=0;
for (i=0; i<present_idx; i++) {
if(a[i]==0)
break;
printf("%d ",a[i]);
}
}
}

{void count_item(int a[],int len){
int item, item_count, present_idx;
int i;
i = 0;
present_idx = 0;
item_count = 1;
item = a[i];
for (i=1; i<len; i++) {
if (item ==a[i]) {
item_count++;
}
else {
a[present_idx++] = item;
a[present_idx++] = item_count;
item_count = 1;
item = a[i];
}
}
a[present_idx++] = item;
a[present_idx++] = item_count;
a[present_idx]=0;
for (i=0; i<present_idx; i++) {
if(a[i]==0)
break;
printf("%d ",a[i]);
}
}
}

Problem:
================
given a stream of integers.. 112233344442222 write a function that would return as 1222334424
it takes aabbcccddddbbbb and returns a2b2c3d4b4.

Example:
================
A[aabbcccddddbbbb]

Result: a2b2c3d4b4

ALGO:
================
Take A[i] compare with all and increase the counter
put Str S += A[i] + Counter
Return the complete String

CODE:
================



private string CharCount(Array A[], Int n) \\N IS THE SIZE OF ARRAY
{
FOR(I=0;I<N;I++)
{
FOR(J=0;J<N;J++)
{
IF(a[I]==a[J])
{
COUNT++;
}
ELSE
{
J=N; \\TO KILL THE ITERATIONS
}
}
STRING S += A[I]+COUNT;
}
RETURN s;

}



ORDER:
================
O(N^2)

Its Sudo Code:
*************
PreDigit=GetDigit();
Count=1;
while(IsMoreDigit())
{
Digit=GetDigit();
if(PreDigit==Digit) Count++;
else
{
cout<<PreDigit<<Count;
PreDigit=Digit;
Count=1;
}
}
cout<<PreDigit<<Count;

guys... this is RLE (Run Length encoding) Used for compression in fax type of images..... can easily be done in O(n)

guys... this is RLE (Run Length encoding) Used for compression in fax type of images..... can easily be done in O(n)

{
private static void getCharsCount(String s){
String result = "";
int count = 1;

for(int i = 0; i < s.length(); ++i){
if(i == s.length()-1 || s.charAt(i) != s.charAt(i+1)){
result += "" + s.charAt(i)+count;
count = 1;
}else{
count++;
}
}

System.out.println(result);
}
}

O(n)

what would happen when a digit exceeds more than 10 times for example , if 1 runs for 12 times then we represent it as 112 , how can we decode it ?Is there a limit on number of times a character can occur ?