Given an array of n elements, where each element is at most k away from its target position, devise an algorithm that sorts in O(n log k) time. For example, let us consider k is 2, an element at index 7 in the sorted array, can be at indexes 5, 6, 7, 8, 9 in the given array.
Examples:
Input : arr[] = {6, 5, 3, 2, 8, 10, 9}
k = 3
Output : arr[] = {2, 3, 5, 6, 8, 9, 10}
Input : arr[] = {10, 9, 8, 7, 4, 70, 60, 50}
k = 4
Output : arr[] = {4, 7, 8, 9, 10, 50, 60, 70}
We can use Insertion Sort to sort the elements efficiently. Following is the C code for standard Insertion Sort.
C
/* Function to sort an array using insertion sort*/void insertionSort(int A[], int size) { int i, key, j; for (i = 1; i < size; i++) { key = A[i]; j = i-1; /* Move elements of A[0..i-1], that are greater than key, to one position ahead of their current position. This loop will run at most k times */ while (j >= 0 && A[j] > key) { A[j+1] = A[j]; j = j-1; } A[j+1] = key; } } |
Java
/* Function to sort an array using insertion sort*/static void insertionSort(int A[], int size) { int i, key, j; for (i = 1; i < size; i++) { key = A[i]; j = i-1; /* Move elements of A[0..i-1], that are greater than key, to one position ahead of their current position. This loop will run at most k times */ while (j >= 0 && A[j] > key) { A[j+1] = A[j]; j = j-1; } A[j+1] = key; } } |
Python3
# Function to sort an array using # insertion sort def insertionSort(A, size): i, key, j = 0, 0, 0for i in range(size): key = A[i] j = i-1 # Move elements of A[0..i-1], that are # greater than key, to one position # ahead of their current position. # This loop will run at most k times while j >= 0 and A[j] > key: A[j + 1] = A[j] j = j - 1 A[j + 1] = key |
C#
/* C# Function to sort an array using insertion sort*/static void insertionSort(int A[], int size) { int i, key, j; for (i = 1; i < size; i++) { key = A[i]; j = i-1; /* Move elements of A[0..i-1], that are greater than key, to one position ahead of their current position. This loop will run at most k times */ while (j >= 0 && A[j] > key) { A[j+1] = A[j]; j = j-1; } A[j+1] = key; } } |
The inner loop will run at most k times. To move every element to its correct place, at most k elements need to be moved. So overall complexity will be O(nk)
We can sort such arrays more efficiently with the help of Heap data structure. Following is the detailed process that uses Heap.
1) Create a Min Heap of size k+1 with first k+1 elements. This will take O(k) time (See this GFact)
2) One by one remove min element from heap, put it in result array, and add a new element to heap from remaining elements.
Removing an element and adding a new element to min heap will take Logk time. So overall complexity will be O(k) + O((n-k)*logK)
C++
// A STL based C++ program to sort a nearly sorted array. #include <bits/stdc++.h> using namespace std; // Given an array of size n, where every element // is k away from its target position, sorts the // array in O(nLogk) time. int sortK(int arr[], int n, int k) { // Insert first k+1 items in a priority queue (or min heap) //(A O(k) operation). We assume, k < n. priority_queue<int, vector<int>, greater<int> > pq(arr, arr + k + 1); // i is index for remaining elements in arr[] and index // is target index of for current minimum element in // Min Heapm 'hp'. int index = 0; for (int i = k + 1; i < n; i++) { arr[index++] = pq.top(); pq.pop(); pq.push(arr[i]); } while (pq.empty() == false) { arr[index++] = pq.top(); pq.pop(); } } // A utility function to print array elements void printArray(int arr[], int size) { for (int i = 0; i < size; i++) cout << arr[i] << " "; cout << endl; } // Driver program to test above functions int main() { int k = 3; int arr[] = { 2, 6, 3, 12, 56, 8 }; int n = sizeof(arr) / sizeof(arr[0]); sortK(arr, n, k); cout << "Following is sorted array" << endl; printArray(arr, n); return 0; } |
Java
// A java program to sort a nearly sorted array import java.util.Iterator; import java.util.PriorityQueue; class GFG { private static void kSort(int[] arr, int n, int k) { // min heap PriorityQueue<Integer> priorityQueue = new PriorityQueue<>(); // add first k + 1 items to the min heap for(int i = 0; i < k + 1; i++) { priorityQueue.add(arr[i]); } int index = 0; for(int i = k + 1; i < n; i++) { arr[index++] = priorityQueue.peek(); priorityQueue.poll(); priorityQueue.add(arr[i]); } Iterator<Integer> itr = priorityQueue.iterator(); while(itr.hasNext()) { arr[index++] = priorityQueue.peek(); priorityQueue.poll(); } } // A utility function to print the array private static void printArray(int[] arr, int n) { for(int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Driver Code public static void main(String[] args) { int k = 3; int arr[] = { 2, 6, 3, 12, 56, 8 }; int n = arr.length; kSort(arr, n, k); System.out.println("Following is sorted array"); printArray(arr, n); } } // This code is contributed by // Manpreet Singh(manpreetsngh294) |
Python3
# A Python3 program to sort a # nearly sorted array. from heapq import heappop, heappush, heapify # A utility function to print # array elements def print_array(arr: list): for elem in arr: print(elem, end = ' ') # Given an array of size n, where every # element is k away from its target # position, sorts the array in O(nLogk) time. def sort_k(arr: list, n: int, k: int): """ :param arr: input array :param n: length of the array :param k: max distance, which every element is away from its target position. :return: None """ # List of first k+1 items heap = arr[:k + 1] # using heapify to convert list # into heap(or min heap) heapify(heap) # "rem_elmnts_index" is index for remaining # elements in arr and "target_index" is # target index of for current minimum element # in Min Heap "heap". target_index = 0 for rem_elmnts_index in range(k + 1, n): arr[target_index] = heappop(heap) heappush(heap, arr[rem_elmnts_index]) target_index += 1 while heap: arr[target_index] = heappop(heap) target_index += 1 # Driver Code k = 3arr = [2, 6, 3, 12, 56, 8] n = len(arr) sort_k(arr, n, k) print('Following is sorted array') print_array(arr) # This code is contributed by # Veerat Beri(viratberi) |
Output:
Following is sorted array 2 3 6 8 12 56
The Min Heap based method takes O(nLogk) time and uses O(k) auxiliary space.
We can also use a Balanced Binary Search Tree instead of Heap to store K+1 elements. The insert and delete operations on Balanced BST also take O(Logk) time. So Balanced BST based method will also take O(nLogk) time, but the Heap bassed method seems to be more efficient as the minimum element will always be at root. Also, Heap doesn’t need extra space for left and right pointers.
Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.
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