Python Program for 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.
Python Program for Binary Insertion Sort
# 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
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
# Python Program implementation # of binary insertion sortdef 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 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 middef 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 arrprint("Sorted array:")print insertion_sort([37, 23, 0, 17, 12, 72, 31, 46, 100, 88, 54])# Code contributed by Mohit Gupta_OMGOutput
Sorted array: [0, 12, 17, 23, 31, 37, 46, 54, 72, 88, 100]
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(long)
Python Program for Binary Insertion Sort Implementation using bisect module
In this method, we are using bisect.bisect_left() function that returns the index at which the val should be inserted in the sorted array arr[start:end+1], so we just need to add the start index to get the correct index in the original array. The insertion_sort function is the same as in the original code.
import bisect
def binary_search(arr, val, start, end):
idx = bisect.bisect_left(arr[start:end+1], val)
return start + idx
def insertion_sort(arr):
for i in range(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]))
import bisectdef binary_search(arr, val, start, end): idx = bisect.bisect_left(arr[start:end+1], val) return start + idxdef insertion_sort(arr): for i in range(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 arrprint("Sorted array:")print(insertion_sort([37, 23, 0, 17, 12, 72, 31, 46, 100, 88, 54]))Output
Sorted array: [0, 12, 17, 23, 31, 37, 46, 54, 72, 88, 100]
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Please refer complete article on Binary Insertion Sort for more details!
