# Get the number of integers between each given range – without external packages

## Issue

Write a function that receives 3 lists and returns an array. The first list contains n integers, their values range between 0 and 10^9. "numbers".

The second list is a low-range list, which contains the lower end of a range, it contains q integers. "low".

The third list is a high-range list, which contains the higher end of a range, it contains q integers. "high".

The function should return a list that contains the number of integers in the first list, that fall in its range, given by the low-range and high-range lists.

In the returned list, at index i, there should be the number of integers in "numbers" which are bigger or equal to low[i] and smaller or equal to high[i].

You can only import math, no other imports are allowed

the list may not be sorted

Examples:
count_range([12,13,14,15,17],[14],[14]) should return [1]
count_range([12,13,14,15,17],[14,15],[14,18]) should return [1,2]
count_range([12,13,14,15,17],[12],[17]) should return [5]

This is my solution but it’s not efficient enough, I need ways to optimize it or solve it differently without having to import any external packages.

``````def binarySearch(data, val):
highIndex = len(data) - 1
lowIndex = 0
while highIndex > lowIndex:
index = math.ceil((highIndex + lowIndex) / 2)
sub = data[index]
if sub > val:
if highIndex == index:
return sorted([highIndex, lowIndex])
highIndex = index
else:
if lowIndex == index:
return sorted([highIndex, lowIndex])
lowIndex = index
return sorted([highIndex, lowIndex])

def count_range(numbers, low, high):
numbers.sort()
result = []
low_range_dict = {}
high_range_dict = {}
for i in range(len(numbers)):
if numbers[i] not in low_range_dict:
low_range_dict[numbers[i]] = i
high_range_dict[numbers[i]] = i
for i in range(len(low)):
low_r = low[i]
high_r = high[i]
if low_r not in low_range_dict:
low_range_dict[low_r] = binarySearch(numbers, low_r)[0]
high_range_dict[low_r] = low_range_dict[low_r]
low_index = low_range_dict.get(low_r)
if high_r not in high_range_dict:
high_range_dict[high_r] = binarySearch(numbers, high_r)[0]
low_range_dict[high_r] = high_range_dict[high_r]
high_index = high_range_dict.get(high_r)
if low_r in numbers or low_r < numbers[0]:
low_index -= 1
result.append(high_index - low_index)
return result
``````

## Solution

If we could use any module from the standard library, we could do write a very simple solution.

``````from bisect import bisect_left
from functools import lru_cache, partial

def count_range(numbers, lows, highs):
index = lru_cache()(partial(bisect_left, sorted(numbers)))
return [index(hi + 1) - index(lo) for (lo, hi) in zip(lows, highs)]
``````

But we can write our own (simplified) equivalent of `partial`, `lru_cache` and `bisect_left`, so the imports are not needed.

It is less complicated than your original code, and should probably run faster, but I don’t know how big the difference is.

We’ll use a simpler bisect function for the binary search. And we don’t need two different memoization dictionaries for high and low range.

``````# This bisect is based on the reference implementation in the standard library.
# in cpython this is actually implemented in C, and is faster.
def bisect_left(a, x):
"""Return the index where to insert item x in list a, assuming a is sorted."""
lo, hi = 0, len(a)
while lo < hi:
mid = (lo + hi) // 2
if a[mid] < x:
lo = mid + 1
else:
hi = mid
return lo

def count_range(numbers, lows, highs):
numbers.sort()
# instead of both low_range_dict and high_range_dict
# we only need a single memoization dictionary.
# We could also use @functools.cache from the standard library
memo = {}
def index(val):
"""Memoized bisect"""
if not val in memo:
memo[val] = bisect_left(numbers, val)
return memo[val]

return [index(hi + 1) - index(lo) for (lo, hi) in zip(lows, highs)]
``````