Skip to content# Counting Sort 1 code challenge

### Comparison sorting

### Alternative sorting

### Example

### Note

### Input format

### Returns

### Sample input

### Sample output

### Solution

### Appendix

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— 1 min read

My approach to solving HackerRank’s Counting Sort 1 code challenge.

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).

nother 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.

`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:

`1i arr[i] result20 1 [0, 1, 0, 0]31 1 [0, 2, 0, 0]42 3 [0, 2, 0, 1]53 2 [0, 2, 1, 1]64 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]`

.

For this exercise, always return a frequency array with 100 elements. The example above shows only the first 4 elements, the remainder being zeros.

The first line contains an integer `n`

, the number of items in `arr`

.

Each of the next `n`

lines contains an integer `arr[i]`

where `0 <= i < n`

.

`int[100]`

: a frequency array.

`1100263 25 73 1 98 73 56 84 86 57 16 83 8 25 81 56 9 53 98 67 99 12 83 89 80 91 39 86 76 85 74 39 25 90 59 10 94 32 44 3 89 30 27 79 46 96 27 32 18 21 92 69 81 40 40 34 68 78 24 87 42 69 23 41 78 22 6 90 99 89 50 30 20 1 43 3 70 95 33 46 44 9 69 48 33 60 65 16 82 67 61 32 21 79 75 75 13 87 70 33`

`10 2 0 2 0 0 1 0 1 2 1 0 1 1 0 0 2 0 1 0 1 2 1 1 1 3 0 2 0 0 2 0 3 3 1 0 0 0 0 2 2 1 1 1 2 0 2 0 1 0 1 0 0 1 0 0 2 1 0 1 1 1 0 1 0 1 0 2 1 3 2 0 0 2 1 2 1 0 2 2 1 2 1 2 1 1 2 2 0 3 2 1 1 0 1 1 1 0 2 2`

We create a frequency array (`freqArr`

) variable and begin initialising (`new Array()`

) a new `100`

elements array where each value is `0`

(`Array.fill()`

).

Then we use a`for in`

loop to go through the given array. During each loop, we use value of the given array (`i`

) to access `freqArr`

through bracket notation, where we increment its value by `1`

.

```
1function countingSort(arr) {2 let freqArr = new Array(100).fill(0);3
4 for (let i of arr) {5 freqArr[i] += 1;6 }7
8 return freqArr;9}
```

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