|
|
"""Bisection algorithms.""" |
|
|
|
|
|
|
|
|
def insort_right(a, x, lo=0, hi=None, *, key=None): |
|
|
"""Insert item x in list a, and keep it sorted assuming a is sorted. |
|
|
|
|
|
If x is already in a, insert it to the right of the rightmost x. |
|
|
|
|
|
Optional args lo (default 0) and hi (default len(a)) bound the |
|
|
slice of a to be searched. |
|
|
""" |
|
|
if key is None: |
|
|
lo = bisect_right(a, x, lo, hi) |
|
|
else: |
|
|
lo = bisect_right(a, key(x), lo, hi, key=key) |
|
|
a.insert(lo, x) |
|
|
|
|
|
|
|
|
def bisect_right(a, x, lo=0, hi=None, *, key=None): |
|
|
"""Return the index where to insert item x in list a, assuming a is sorted. |
|
|
|
|
|
The return value i is such that all e in a[:i] have e <= x, and all e in |
|
|
a[i:] have e > x. So if x already appears in the list, a.insert(i, x) will |
|
|
insert just after the rightmost x already there. |
|
|
|
|
|
Optional args lo (default 0) and hi (default len(a)) bound the |
|
|
slice of a to be searched. |
|
|
""" |
|
|
|
|
|
if lo < 0: |
|
|
raise ValueError('lo must be non-negative') |
|
|
if hi is None: |
|
|
hi = len(a) |
|
|
|
|
|
|
|
|
if key is None: |
|
|
while lo < hi: |
|
|
mid = (lo + hi) // 2 |
|
|
if x < a[mid]: |
|
|
hi = mid |
|
|
else: |
|
|
lo = mid + 1 |
|
|
else: |
|
|
while lo < hi: |
|
|
mid = (lo + hi) // 2 |
|
|
if x < key(a[mid]): |
|
|
hi = mid |
|
|
else: |
|
|
lo = mid + 1 |
|
|
return lo |
|
|
|
|
|
|
|
|
def insort_left(a, x, lo=0, hi=None, *, key=None): |
|
|
"""Insert item x in list a, and keep it sorted assuming a is sorted. |
|
|
|
|
|
If x is already in a, insert it to the left of the leftmost x. |
|
|
|
|
|
Optional args lo (default 0) and hi (default len(a)) bound the |
|
|
slice of a to be searched. |
|
|
""" |
|
|
|
|
|
if key is None: |
|
|
lo = bisect_left(a, x, lo, hi) |
|
|
else: |
|
|
lo = bisect_left(a, key(x), lo, hi, key=key) |
|
|
a.insert(lo, x) |
|
|
|
|
|
def bisect_left(a, x, lo=0, hi=None, *, key=None): |
|
|
"""Return the index where to insert item x in list a, assuming a is sorted. |
|
|
|
|
|
The return value i is such that all e in a[:i] have e < x, and all e in |
|
|
a[i:] have e >= x. So if x already appears in the list, a.insert(i, x) will |
|
|
insert just before the leftmost x already there. |
|
|
|
|
|
Optional args lo (default 0) and hi (default len(a)) bound the |
|
|
slice of a to be searched. |
|
|
""" |
|
|
|
|
|
if lo < 0: |
|
|
raise ValueError('lo must be non-negative') |
|
|
if hi is None: |
|
|
hi = len(a) |
|
|
|
|
|
|
|
|
if key is None: |
|
|
while lo < hi: |
|
|
mid = (lo + hi) // 2 |
|
|
if a[mid] < x: |
|
|
lo = mid + 1 |
|
|
else: |
|
|
hi = mid |
|
|
else: |
|
|
while lo < hi: |
|
|
mid = (lo + hi) // 2 |
|
|
if key(a[mid]) < x: |
|
|
lo = mid + 1 |
|
|
else: |
|
|
hi = mid |
|
|
return lo |
|
|
|
|
|
|
|
|
|
|
|
try: |
|
|
from _bisect import * |
|
|
except ImportError: |
|
|
pass |
|
|
|
|
|
|
|
|
bisect = bisect_right |
|
|
insort = insort_right |
|
|
|