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Problem:


https://www.geeksforgeeks.org/problems/find-number-of-elements-in-range-a-b-for-each-query/1

Elements in range [a, b]

Difficulty: Medium Accuracy: 64.1%

Given an unsorted array arr[] and a 2D array queries[][] of size q, where each query is in the form of [a, b]. For each query, count how many elements in arr[] lie within the range [a, b], i.e., elements satisfying a ≤ x ≤ b.

Return the results for all queries as a list of integers, where each integer corresponds to the count of elements in the respective query range.

Examples:

Input: arr[] = [1, 4, 2, 8, 5], queries[][] = [[1, 4], [3, 6], [0, 10]]

Output: [3, 2, 5]

Explanation: Query [1, 4]: Elements in range → [1, 4, 2] → Count = 3

Query [3, 6]: Elements in range → [4, 5] → Count = 2

Query [0, 10]: All elements → [1, 4, 2, 8, 5] → Count = 5

Input: arr[] = [10, 20, 30, 40, 50], queries[][] = [[5, 15], [25, 45], [10, 50]]

Output: [1, 2, 5]

Explanation: Query [5, 15]: Elements in range → [10] → Count = 1

Query [25, 45]: Elements in range → [30, 40] → Count = 2

Query [10, 50]: Elements in range → [10, 20, 30, 40, 50] → Count = 5

Constraints:

1 ≤ arr.size(), q ≤ 105

0 ≤ arr[i] ≤ 106

0 ≤ queries[i][0] ≤ queries[i][1] ≤ 106

Solution:

from bisect import bisect_left, bisect_right

class Solution:

def cntInRange(self, arr, queries):

arr.sort()

res = []

for a, b in queries:

left = bisect_left(arr, a)

right = bisect_right(arr, b)

res.append(right - left)

return res