# Given a collection of 1 million integers, all ranging between 1 to 9, sort them in Big O(n) time

Upasana | November 22, 2020 | 2 min read | 669 views |

This is a typical Integer Sorting problem with a constraint that the number range to sort is very limited in spite 1 million total entries. Integer Sorting with limited range is achieved efficiently with Bucket Sorting.

Bucket Sorting Algorithm

Bucket sort, counting sort, radix sort, and van Emde Boas tree sorting all work best when the key size is small; for large enough keys, they become slower than comparison sorting algorithm.

## Algorithm Steps

1. Create a array of size 9

2. At each index store the occurrence count of the respective integers. Doing this will achieve this sorting with time complexity of Big O(n) and even the memory requirements are minimized.

3. In Order to print the output just traverse the above created array.

Bucket Sort 0-9
``````public class BucketSort {
public int[] sort(int[] array, int min, int max) {
int range = max - min + 1;
int[] result = new int[range];
for (int i: array) {
result[i]++;    (1)
}
return result;
}
}

public class BucketSortTest {

@Test
public void testBucketSortFor1To9() {
int[] array = {
2, 1, 5, 1, 2, 3, 4, 3, 5, 6, 7, 8, 5, 6, 7, 0
};
int[] sort = new BucketSort().sort(array, 0, 8);

for (int i = 0; i < sort.length; i++) {
for (int j = 0; j < sort[i]; j++) {
System.out.println(i);
}
}
}
}``````
 1 Storing the count of occurrences for each number
Program output:
`0,1,1,2,2,3,3,4,5,5,5,6,6,7,7,8`
Time Complexity
`Big O(n)`

## Notes

In a very similar fashion we can sort millions of people with their age (0-130).

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