Sort a nearly sorted array using STL

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. It may be assumed that k < n.

Example:

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}

Simple Approach: The basic solution is to sort the array using any standard sorting algorithm.



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// 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(n Log n) time.
int sortK(int arr[], int n, int k)
{
    // Sort the array using
    // inbuilt function
    sort(arr, arr + n);
}
  
// An 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;
}

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Output:

Following is sorted arrayn2 3 6 8 12 56

Complexity Analysis:

  • Time complexity: O(n log n), where n is the size of the array.
    The sorting algorithm takes log n time. Since the size of the array is n, the whole program takes O(n log n) time.
  • Space Complxity: O(1).
    As no extra space is required.

Efficient Solution: Sliding Window technique.
Approach: A better solution is to use a priority queue(or heap data structure). Use sliding window technique to keep consecutive k elements of a window in heap. Then remove the top element(smallest element) and replace the first element of the window with it.
As each element will be at most k distance apart, therefore keeping k consecutive elements in a window while replacing the i-th element with the smallest element from i to (i+k) will suffice(first i-1 elements are sorted).

Algorithm :

  1. Build a priority queue pq of first (k+1) elements.
  2. Initialize index = 0 (For result array).
  3. Do following for elements from k+1 to n-1.
    1. Pop an item from pq and put it at index, increment index.
    2. Push arr[i] to pq.
  4. While pq is not empty,
    Pop an item from pq and put it at index, increment index.

We have discussed a simple implementation in Sort a nearly sorted (or K sorted) array. In this post, an STL based implementation is done.

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// 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)
    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 arrayn";
    printArray(arr, n);
  
    return 0;
}

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Output:

Following is sorted arrayn2 3 6 8 12 56

Complexity Analysis:

  • Time Complexity: O(n Log k).
    For every element, it is pushed in priority queue and the insertion and deletion needs O(log k) time as there are k elements in priority queue.
  • Auxiliary Space : O(k).
    To store k elements in priority queue, O(k) space is required.

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