+4 votes

I am looking for a more efficient way than brute-forcing my way through in the following problem in python 3.

Problem statement:

Input:

• An array of n integers, scores, where each score is denoted by scores_j

• An array of q integers, lowerLimits, where each lowerLimits_i denotes the lower limit for score range i.

• An array of q integers, upper limits, where each upperLimits_i denotes the upper limit for score range i.

Output: A function that returns an array of Q integers where the value at each index I denotes the number of integers that are in the inclusive range [lowerLimits_i, upperLimits_i].

Constraints:

• 1 ≤ n ≤ 1e5
• 1 ≤ scores_j ≤ 1e9
• 1 ≤ q ≤ 1e5
• 1 ≤ lowerLimits_i ≤ upperLimits_i ≤ 1e9

Example: Given scores= [5, 8, 7], lowerLimits = [3, 7], and upperLimits = [9, 7] I want to check how many of the integers are contained in each interval (inclusive). In this examples: intervals are [3,9] and [7,7], and the result would be [3, 1].

My code looks like this:

```def check(scores, lowerLimits, upperLimits):
res = []
for l, u in zip(lowerLimits, upperLimits):
res.append(sum([l <= y <= u for y in scores]))
return res
if __name__ == "__main__":
scores= [5, 8, 7]
lowerLimits = [3, 7]
upperLimits = [9, 7]

print(check(scores, lowerLimits, upperLimits))```

`Regards,`
`Riya williams.`
`python developer`
Sep 25, 2018 in Python 1,115 views

## 1 answer to this question.

+1 vote
Am not sure as this would give a better TC but still what you can try is

First, sort the values
Second, iterate over the sorted list and find the lower hand limit and keep a count till the upper limit
That how you get the count of all the values and the remaining values that are not in the range can be discarded.
answered Sep 27, 2018 by
• 15,520 points

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