Counting Sort 1 Problem in Java | Hacker Rank Interview Preparation | 1 Month Preparation Kit.

This video is about Step by step explanation of Counting Sort 1 problem on Hacker Rank.
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Comparison Sorting:-

Quicksort usually has a running time of n x log(n), but is there an algorithm that can sort even faster? In general,
this is not possible. Most sorting algorithms are comparison sorts, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beat n x log(n) (worst-case) running time, since n x log(n) represents the minimum number of comparisons needed to know where to place each element. For more details, you can see these notes (PDF).

Alternative Sorting:-
Another sorting method, the counting sort, does not require comparison. Instead, you create an integer array whose index range covers the entire range of values in your array to sort. Each time a value occurs in the original array, you increment the counter at that index. At the end, run through your counting array, printing the value of each non-zero valued index that number of times.

Example
arr = = [1, 1, 3, 2, 1]
All of the values are in the range [0...3], so create an array of zeros, result = [0, 0, 0, 0]. The results of each iteration follow:

i arr[i] result
Θ 1 [0, 1, 0, 0]
1 1 [0, 2, 0, 0]
2 3 [0, 2, 0, 1]
3 2 [0, 2, 1, 1]
4 1 [0, 3, 1, 1]

The frequency array is [0, 3, 1, 1]. These values can be used to create the sorted array as well: sorted = [1, 1, 1, 2, 3].

#countingsort1 #hackerrank #solution #sorting #count #counting
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[raghavi.m-mjgk]

hii, I did not understand the constraints they mentioned, does they mean that the size of array can be from 100 to 10^6 and the elements in frequency array ranges form 0 to 100. If I'm wrong please explain.