Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. The position is 1-based, not 0-based. The time complexity of Counting Sort is easy to determine due to the very simple algorithm. It running time complexity is O(n) with space proportional to the range of data. In this chapter, we determine the space complexity, stability, and parallelizability of Counting Sort. A simplified form of Counting Sort can be used when sorting numbers (e.g., int primitives). In the JDK, for example, for: If you liked the article, feel free to share it using one of the share buttons at the end. My question is, when the difference between k and n is too much, such as when k=O(n 2)or O(n 3), can we say that the complexity is O(n 2) or O(n 3)? Counting sort calculates the number of occurrence of objects and stores its key values. To find this efficiently, we first aggregate the values in the histogram. Time Complexity: O(n+k) where n is the number of elements in input array and k is the range of input. Thus: Counting Sort can be parallelized by dividing the input array into as many partitions as there are processors available. Why Quick Sort preferred for Arrays and Merge Sort for Linked Lists? In phase two, we iterate once over the histogram array. The simplified algorithm requires an additional array of size k; therefore: The space complexity of the simplified counting sort algorithm is: O(k). In Phase 3, the general form of the Counting Sort algorithm iterates from right to left over the input array, copying objects with the same key also from right to left into the output array. Your email address will not be published. Counting sort is efficient if the range of input data, k k k, is not significantly greater than the number of objects to be sorted, n n n. Counting sort is a stable sort with a space complexity of O (k + n) O(k + n) O (k + n). This does not quite achieve an acceleration by factor 16, but at least one by factor nine. Because there are no negative array indices. Therefore, the counting sort algorithm has a running time of O (k + n) O(k+n) O (k + n). (For smaller number ranges like byte and short, you can omit to determine the maximum and directly create an array in the size of the corresponding number range.). If elements are sorted in ascending order, they are not changed and do not have to be written back to RAM. The time complexity of Counting Sort is easy to determine due to the very simple algorithm. The count array also uses k iterations, thus has a running time of O(k). Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. How to determine the time complexity of Counting Sort? We decrement the value again to 9 and copy the object to position 9 in the target array, i.e., to the left of the other object with key 6: We repeat these steps for all elements and finally reach the object with the key 3. Time Complexity Analysis. In contrast, if the array is incremented from front to back (or from back to front), 16 consecutive int values can be loaded from and written to the RAM in a single 64-byte block. We write the respective array index into the array to be sorted as often as the histogram indicates at the corresponding position. We now have this information entirely in the histogram.). (I grayed out the rest of the numbers because they are still in the array, but we don’t need them anymore. Counting Sort is mainly used for small number ranges. Then use the following form to subscribe to my newsletter. You find further information and options to switch off these cookies in our. Below you’ll find a simple form of the Counting Sort source code – it only works for non-negative int primitives (e.g., for the array from the example above). Let’s start at the far right in the input array – with the object with key 8. Space Complexity. 4. The GitHub repository contains the UltimateTest program, which allows us to measure the speed of Counting Sort (and all the other sorting algorithms in this article series). Input sequences sorted in descending order are sorted minimally slower than those pre-sorted in ascending order. In phase 3, each processor copies the elements of “its” partition to the target array. When writing back the sorted numbers into the array, the offset is subtracted again by adding. The second object from the right has the key 2. Both iterations are time. It is often used as a sub-routine to another sorting algorithm like radix sort. The algorithm contains one or more loops that iterate to n and one loop that iterates to k. Constant factors are irrelevant for the time complexity; therefore: The time complexity of Counting Sort is: O(n + k). Accordingly, the time required increases by a little more than a factor of two. Then in this case counting sort is not a wise approach and we can use merge sort. That field contains a 1, so we write the 0 exactly once into the array to be sorted. edit It is because the total time taken also depends on some external factors like the compiler used, processor’s speed, etc. In addition to the auxiliary array of size k, the general algorithm requires a temporary target array of size n; thus: The space complexity of the general counting sort algorithm is: O(n + k). Auxiliary Space: O(n+k). The following diagram shows the measurements graphically: If elements are not actually sorted but counted and entirely rearranged, shouldn’t the initial order do not affect the time needed for sorting!? Why is Counting Sort almost ten times faster for presorted number sequences than for unsorted ones despite the same number of operations? In phase 1, each processor counts the elements of “its” partition in a separate auxiliary array. Field 3 in the auxiliary array now contains a 3. You could also determine the maximum using Arrays.stream(elements).max().getAsInt(). It is because the total time taken also depends on some external factors like the compiler used, processor’s speed, etc. Phase 1, the counting phase, remains more or less unchanged. The following table shows the time needed to sort unsorted and ascending and descending presorted elements for the given number of elements n, which in these measurements also corresponds to the size of the number range k: You can find the complete result in the file Test_Results_Counting_Sort.log. The following code demonstrates the general form of Counting Sort for simplicity’s sake using int primitives. Instead of the objects themselves, their keys (determined by a getKey() method, for example) are now counted. 2. The sorted array B[] also gets computed in n iterations, thus requiring O(n) running time. This sorting technique is efficient when difference between different keys are not so big, otherwise it can increase the space complexity.