Sort a nearly sorted (or K sorted) array

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.

Source: Nearly sorted algorithm

We can use Insertion Sort to sort the elements efficiently. Following is the C code for standard Insertion Sort.

/* 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;

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)

using namespace std;

// Prototype of a utility function to swap two integers
void swap(int *x, int *y);

// A class for Min Heap
class MinHeap
    int *harr; // pointer to array of elements in heap
    int heap_size; // size of min heap
    // Constructor
    MinHeap(int a[], int size);

    // to heapify a subtree with root at given index
    void MinHeapify(int );

    // to get index of left child of node at index i
    int left(int i) { return (2*i + 1); }

    // to get index of right child of node at index i
    int right(int i) { return (2*i + 2); }

    // to remove min (or root), add a new value x, and return old root
    int replaceMin(int x);

    // to extract the root which is the minimum element
    int extractMin();

// 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)
    // Create a Min Heap of first (k+1) elements from
    // input array
    int *harr = new int[k+1];
    for (int i = 0; i<=k && i<n; i++) // i < n condition is needed when k > n
        harr[i] = arr[i];
    MinHeap hp(harr, k+1);

    // i is index for remaining elements in arr[] and ti
    // is target index of for cuurent minimum element in
    // Min Heapm 'hp'.
    for(int i = k+1, ti = 0; ti < n; i++, ti++)
        // If there are remaining elements, then place
        // root of heap at target index and add arr[i]
        // to Min Heap
        if (i < n)
            arr[ti] = hp.replaceMin(arr[i]);

        // Otherwise place root at its target index and
        // reduce heap size
            arr[ti] = hp.extractMin();

// Constructor: Builds a heap from a given array a[] of given size
MinHeap::MinHeap(int a[], int size)
    heap_size = size;
    harr = a;  // store address of array
    int i = (heap_size - 1)/2;
    while (i >= 0)

// Method to remove minimum element (or root) from min heap
int MinHeap::extractMin()
    int root = harr[0];
    if (heap_size > 1)
        harr[0] = harr[heap_size-1];
    return root;

// Method to change root with given value x, and return the old root
int MinHeap::replaceMin(int x)
    int root = harr[0];
    harr[0] = x;
    if (root < x)
    return root;

// A recursive method to heapify a subtree with root at given index
// This method assumes that the subtrees are already heapified
void MinHeap::MinHeapify(int i)
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l] < harr[i])
        smallest = l;
    if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
    if (smallest != i)
        swap(&harr[i], &harr[smallest]);

// A utility function to swap two elements
void swap(int *x, int *y)
    int temp = *x;
    *x = *y;
    *y = temp;

// 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\n";
    printArray (arr, n);

    return 0;


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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