Binary Insertion Sort

2.3

We can use binary search to reduce the number of comparisons in normal insertion sort. Binary Insertion Sort find use binary search to find the proper location to insert the selected item at each iteration.
In normal insertion, sort it takes O(i) (at ith iteration) in worst case. we can reduce it to O(logi) by using binary search.

C/C++

// C program for implementation of binary insertion sort
#include <stdio.h>

// A binary search based function to find the position
// where item should be inserted in a[low..high]
int binarySearch(int a[], int item, int low, int high)
{
    if (high <= low)
        return (item > a[low])?  (low + 1): low;

    int mid = (low + high)/2;

    if(item == a[mid])
        return mid+1;

    if(item > a[mid])
        return binarySearch(a, item, mid+1, high);
    return binarySearch(a, item, low, mid-1);
}

// Function to sort an array a[] of size 'n'
void insertionSort(int a[], int n)
{
    int i, loc, j, k, selected;

    for (i = 1; i < n; ++i)
    {
        j = i - 1;
        selected = a[i];

        // find location where selected sould be inseretd
        loc = binarySearch(a, selected, 0, j);

        // Move all elements after location to create space
        while (j >= loc)
        {
            a[j+1] = a[j];
            j--;
        }
        a[j+1] = selected;
    }
}

// Driver program to test above function
int main()
{
    int a[] = {37, 23, 0, 17, 12, 72, 31,
              46, 100, 88, 54};
    int n = sizeof(a)/sizeof(a[0]), i;

    insertionSort(a, n);

    printf("Sorted array: \n");
    for (i = 0; i < n; i++)
        printf("%d ",a[i]);

    return 0;
}

Java

// Java Program implementing
// binary insertion sort

import java.util.Arrays;
class GFG
{
    public static void main(String[] args)
    {
        final int[] arr = {37, 23, 0, 17, 12, 72, 31,
                             46, 100, 88, 54 };

        new GFG().sort(arr);

        for(int i=0; i<arr.length; i++)
            System.out.print(arr[i]+" ");
    }

    public void sort(int array[])
    {
        for (int i = 1; i < array.length; i++)
        {
            int x = array[i];

            // Find location to insert using binary search
            int j = Math.abs(Arrays.binarySearch(array, 0, i, x) + 1);

            //Shifting array to one location right
            System.arraycopy(array, j, array, j+1, i-j);

            //Placing element at its correct location
            array[j] = x;
        }
    }
}

// Code contributed by Mohit Gupta_OMG 

Python

# Python Program implementation  
# of binary insertion sort

def binary_search(arr, val, start, end):
    # we need to distinugish whether we should insert
    # before or after the left boundary.
    # imagine [0] is the last step of the binary search
    # and we need to decide where to insert -1
    if start == end:
        if arr[start] > val:
            return start
        else:
            return start+1

    # this occurs if we are moving beyond left's boundary
    # meaning the left boundary is the least position to
    # find a number greater than val
    if start > end:
        return start

    mid = (start+end)/2
    if arr[mid] < val:
        return binary_search(arr, val, mid+1, end)
    elif arr[mid] > val:
        return binary_search(arr, val, start, mid-1)
    else:
        return mid

def insertion_sort(arr):
    for i in xrange(1, len(arr)):
        val = arr[i]
        j = binary_search(arr, val, 0, i-1)
        arr = arr[:j] + [val] + arr[j:i] + arr[i+1:]
    return arr

print("Sorted array:")
print insertion_sort([37, 23, 0, 17, 12, 72, 31,
                        46, 100, 88, 54])

# Code contributed by Mohit Gupta_OMG 


Output:
Sorted array:
0 12 17 23 31 37 46 54 72 88 100

Time Complexity: The algorithm as a whole still has a running worst case running time of O(n2) because of the series of swaps required for each insertion.

This article is contributed by Amit Auddy. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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