Count numbers in the range [l, r] having only three set bits

Given two integers l and r, find the count of numbers x such that:

• l ≤ x ≤ r
• The binary representation of x contains exactly 3 set bits.

Return the total count of such numbers in the given range.

Examples:

Input: l = 11, r = 19
Output: 4
Explanation: There are 4 such numbers with 3 set bits in range 11 to 19. 11 -> 1011, 13 -> 1101, 14 -> 1110, 19 -> 10011. So answer for this test case is 4.

Input: l = 25, r = 29
Output: 3
Explanation: There are 3 such numbers with 3 set bits in range 25 to 29. 25 -> 11001, 26 -> 11010, 28 -> 11100. So answer for this test case is 3

[Naive Approach] Using Bit Counting – O((r − l + 1) × log n) Time and O(1) Space

The idea is to traverse every number in the range [l, r] and count the number of set bits in its binary representation. If a number contains exactly 3 set bits, it is counted in the final answer.
The set bits are counted by repeatedly checking the last bit using (n & 1) and then right-shifting the number.

• Traverse all numbers from l to r
• For each number, count set bits in binary representation and If the count equals 3, increment the result

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[Efficient Approach] Using Precomputation + Binary Search – O(1) Query Time

A number having exactly 3 set bits can be formed by choosing any 3 distinct bit positions. Since a long long integer has at most 63 usable bit positions, all such numbers can be generated beforehand using three nested loops.

After generating all valid numbers:

• Sort them
• Use binary search to count how many lie in the range [l, r]

The count is: upperBound(r)−lowerBound(l)

• Generate all numbers with exactly 3 set bits using:(1LL<<i)  ∣  (1LL<<j)  ∣  (1LL<<k)(1LL << i)
• Store all generated numbers in a array and sort the array
• Use: lowerBound() for first index having value >= l and upperBound() for first index having value > r
• Their difference gives the required count

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