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Basic Programming > Bit Manipulation > Basics of Bit Manipulation
Given queries of type , find the count of integers in range such that their bit (1-indexed) is from the LSB (least significant bit) side.
Input Format:
- First line contains an integer , denoting the number of queries.
- Next lines contains three space-separated integres .
Output Format:
For every query, print the number of integers in which satisfy the above condition.
Constraints:
- Binary representation of integers in range are as follows:-
- 2 : 10
- 3 : 11
- 4 : 100
- 5 : 101
- 6 : 110
- There are three integers with X-th bit ON i.e. 2, 3 and 6.
- Thus, required answer is 3.
I am learning Data Structures and Algorithms.
I will provide a screenshot of a problem statement.
Your task is NOT to directly give me the final code solution unless I explicitly ask for it.
Instead, please do the following:
1. First, carefully restate the problem in your own words to confirm correct understanding.
- If anything is unclear in the screenshot, ask me clarification questions before proceeding.
2. Identify:
- What category of problem this is (e.g., prefix sum, greedy, DP, two pointers, binary search, graph, etc.)
- What core DSA concepts are being tested
- What mathematical observations (if any) are involved
- What constraints are important and how they affect the approach
3. Explain:
- What naive/brute-force solution would look like
- Why that may or may not work under the given constraints
- What optimization insight is required
4. Clearly explain the key trick or insight needed to solve it.
- Why does that trick work?
- What pattern should I remember for future problems?
5. Provide:
- The time complexity and space complexity of the optimal approach
- Why it satisfies the constraints
6. If there are multiple valid approaches, compare them.
7. Only after all conceptual explanation, provide clean and correct Python code.
8. Avoid guessing unclear details.
- If any part of the screenshot is ambiguous, ask me instead of assuming.
Focus on correctness over speed. Be explicit in reasoning.
Great 👍 Let’s go step by step exactly as you requested.
✅ 1. Restating the Problem (Understanding Check)
You are given Q queries.
Each query contains three integers:
CodeL, R, X
For each query, you must count how many integers in the range [L, R] have their X-th bit (1-indexed from the LSB) turned ON.
Important details:
-
Bits are counted from the right (LSB).
-
X is 1-indexed.
-
X = 1 → check the least significant bit
-
X = 2 → second bit from right
-
-
Constraints:
-
Example from screenshot:
Query:
CodeL = 2, R = 6, X = 2
Binary representations:
| Number | Binary | 2nd bit |
|---|---|---|
| 2 | 10 | ON |
| 3 | 11 | ON |
| 4 | 100 | OFF |
| 5 | 101 | OFF |
| 6 | 110 | ON |
Answer = 3
✅ 2. What Type of Problem Is This?
📌 Category:
-
Bit manipulation
-
Pattern observation
-
Prefix counting
-
Math-based periodicity
📌 Core DSA Concepts Being Tested
-
Binary representation
-
Understanding bit patterns
-
Efficient range counting
-
Turning range queries into prefix queries
📌 Mathematical Observation
Bits follow a repeating pattern.
For a fixed bit position X:
-
That bit repeats every:
-
In every cycle of size :
-
First numbers → bit is OFF
-
Next numbers → bit is ON
-
So it behaves like:
Code0000 1111 0000 1111 ...
(for X=3 example)
This periodicity is the key.
📌 Constraints Impact
-
Q can be 100,000
-
R can be up to 100,000
If we do O(R−L) per query, worst case:
❌ Impossible.
We need O(1) per query.
✅ 3. Naive / Brute Force Approach
For each query:
Codecount = 0
for i in range(L, R+1):
if (i >> (X-1)) & 1:
count += 1
Time complexity:
Worst case per query: O(R−L)
Worst case overall: O(Q × N) = 10¹⁰
❌ Too slow.
✅ 4. Optimization Insight
Instead of checking each number, we count directly using pattern math.
🎯 Core Trick
Define:
= number of integers from 1 to N whose X-th bit is ON.
Then answer for range:
This converts range query into prefix query.
🔎 How to Compute f(N, X)
Let:
Each block:
-
First half → bit OFF
-
Second half → bit ON
Let:
Codehalf = 2^(X-1)
Steps:
-
Count full cycles:
Codefull_cycles = N // blockEach full cycle contributes:
Codehalf numbers -
Handle remaining part:
Coderemainder = N % blockIf remainder < half:
→ contributes 0Else:
→ contributes (remainder - half + 1)
So:
💡 Why This Works
Because binary bits behave like a square wave:
CodeLength = 2^X
Pattern = [0...0, 1...1]
This repeats forever.
Instead of counting numbers,
we count complete patterns + leftover.
🔁 Pattern to Remember for Future
Whenever you see:
-
Count numbers in range
-
Bit at position X
-
Large Q
Think:
Bits follow periodic pattern of size 2^X
And convert:
This pattern appears in:
-
Bit counting problems
-
Parity counting
-
XOR range problems
-
Power-of-two periodic behavior
✅ 5. Complexity of Optimal Approach
For each query:
-
Only arithmetic operations
-
No loops over range
Time Complexity:
Space Complexity:
This easily satisfies constraints.
✅ 6. Multiple Approaches?
Approach 1 (Best): Mathematical periodic counting
Time: O(1) per query
✔ Recommended
Approach 2: Precompute prefix arrays for each bit
You could build a 2D array:
Codeprefix[bit][i]
But:
-
20 bits
-
N = 10^5
Memory is fine, but unnecessary.
Time complexity:
-
Precompute: O(20 × N)
-
Query: O(1)
Also valid.
But math approach is cleaner.
✅ 7. Clean Python Code (After Full Explanation)
Now here is the implementation:
Pythonimport sys
input = sys.stdin.read
data = input().split()
q = int(data[0])
index = 1
def count_upto(N, X):
if N <= 0:
return 0
block = 1 << X # 2^X
half = 1 << (X - 1) # 2^(X-1)
full_cycles = N // block
remainder = N % block
count = full_cycles * half
if remainder >= half:
count += remainder - half + 1
return count
results = []
for _ in range(q):
L = int(data[index])
R = int(data[index + 1])
X = int(data[index + 2])
index += 3
answer = count_upto(R, X) - count_upto(L - 1, X)
results.append(str(answer))
print("\n".join(results))
If you'd like, next we can:
-
Walk through a dry run manually
-
Derive the formula visually
-
Or convert this to C++ for practice
Your choice 👌
