## DEV Community

Aniket Vaishnav

Posted on • Updated on

# Earthquake and the Paint Shop

Geek's Paint Shop is one of the famous shop in Geekland, but 2014 Earthquake caused disarrangement of the items in his shop. Each item in his shop is a 40-digit alpha numeric code .
Now Chunky wants to retain the reputation of his shop, for that he has to arrange all the distinct items in lexicographical order.
Your task is to arrange the all the distinct items in lexicographical ascending order and print them along with their count.

Example 1:

``````Input:
N = 3
A[] =
["2234597891 zmxvvxbcij 8800654113 jihgfedcba",
"1234567891 abcdefghij 9876543219 jihgfedcba",
"2234597891 zmxvvxbcij 8800654113 jihgfedcba"]
Output:
1234567891 abcdefghij 9876543219 jihgfedcba 1
2234597891 zmxvvxbcij 8800654113 jihgfedcba 2
Explanation:
We have 3 items (40 digit alphanumeric codes)
here. We arrange the items based on the
lexicographical order of their alpha-numeric
code. Distinct items are printed only once.
The count of the items describes how many
such items are there, so items that appear
more than once have their count greater than 1.
``````

Example 2:

``````Input:
N = 2
A[] =
["3122049353 onafrnbhtr 9822827304 yzfhdgzcvx",
"2992299540 lpvkgykmlk 6946169466 vdqwvywwgg",
Output:
2992299540 lpvkgykmlk 6946169466 vdqwvywwgg  1
3122049353 onafrnbhtr 9822827304 yzfhdgzcvx  1
Explanation:
Out of the 2 items here no one is repeated.
``````

You don't need to read input or print anything. Your task is to complete the function sortedStrings() which takes an integer N and an array of strings A[ ] and returns the array in sorted order along with the frequency of each distinct string.

Expected Time Complexity:O(N logN)
Expected Auxillary Space:O(N)

Solution:

As per the question, the wants is to,

1. Sort the items which are initially given as the string
2. Remove the distinct, entry and maintain the count

For implementing these we are provided with the user defined class `alphanumeric` (Alas! a very poor naming skills ðŸ˜”).

The approach for these two tasks can be done via:

1. List Collection for quickly sorting is as per the lexicographic order, (here used, ArrayList)
2. For spotting the distinct items we use Map, (here used HashMap). This would map the provided item string item to the converted objects to the alphanumeric.
``````// pre-existing code not touching it
class alphanumeric {
public String name;
public int count;
alphanumeric(String name, int count) {
this.name = name;
this.count = count;
}
};
// pre-existing code not touching it

class Solution {
alphanumeric[] sortedStrings(int N, String A[]) {
ArrayList<alphanumeric> a = new ArrayList<>();
HashMap<String, alphanumeric> map = new HashMap<>();
for(String s : A) {
if(map.containsKey(s)) {
map.get(s).count++;
} else {
alphanumeric obj = new alphanumeric(s, 1);
map.put(s, obj);
}
}

Collections.sort(a, (ii, jj) -> (ii.name.compareTo(jj.name)));
alphanumeric[] res = new alphanumeric[a.size()];
for(int i = 0; i < res.length; i++) {
res[i] = a.get(i);
}
return res;
}
};
``````