Binary Insertion Sort

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 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];
        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;


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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