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Python Program for Binary Insertion Sort

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  • Difficulty Level : Easy
  • Last Updated : 23 Jun, 2022
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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.


# Python Program implementation 
# of binary insertion sort
def binary_search(arr, val, start, end):
    # we need to distinguish 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
            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)
        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

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

Please refer complete article on Binary Insertion Sort for more details!

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