CF 104317A - Antiamuny wants to learn binary search

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Solve time: 1m 1s
Verified: yes

Solution

Problem Understanding

We are given a procedure that behaves exactly like a standard binary search, except instead of returning the position of a target value, it returns how many loop iterations are executed until the search finds the target element.

The search operates on an integer segment $[L, R]$, and we are guaranteed that the target value $x$ lies inside this interval. Each iteration picks the midpoint, compares it with $x$, and either stops if it matches or shrinks the interval to the left or right half depending on whether the midpoint is larger or smaller than $x$. The required output is simply the number of iterations performed before termination.

The constraints are small: $L, R, x \le 1000$, and at most $100$ test cases. This immediately suggests that even a direct simulation is safe, since each binary search takes at most about $\log_2(1000)$, which is under 10 iterations. So even a naive implementation runs in well under a thousand operations overall.

The main subtlety is that we are not asked for the final position or path, but the exact number of loop executions. A common mistake is to assume this equals the number of interval halvings until a single element remains, which is close but not always identical if the midpoint hits the target early.

Edge cases worth noting are trivial intervals like $L = R$, where the loop runs exactly once and terminates immediately, and cases where $x$ is equal to the midpoint on the first or second iteration, which changes the depth compared to a full binary search tree traversal.

Approaches

The procedure is deterministic and already fully defined. The brute-force idea is to literally simulate the loop exactly as written: maintain $l, r$, compute $mid = (l + r) // 2$, update bounds, and count iterations until $mid == x$. This works because every state transition in the loop is explicitly defined, and there are no hidden dependencies.

The key observation is that there is no need for optimization beyond simulation. The range is tiny, and each iteration strictly shrinks the interval, so the loop length is bounded by $\lceil \log_2(R-L+1) \rceil$. Even in the worst case, this is constant-scale work per test case.

So the “optimal” solution is identical to the brute-force simulation, and the difference is only conceptual: we recognize that the process is already efficient enough.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	$O(T \log (R-L))$	$O(1)$	Accepted
Optimal (same)	$O(T \log (R-L))$	$O(1)$	Accepted

Algorithm Walkthrough

We simulate the binary search exactly and count how many times the loop body executes.

Initialize a counter to zero. This counter tracks how many iterations of the binary search loop occur before termination.
While the current interval $[l, r]$ is valid, meaning $l \le r$, perform one iteration of the search. This condition guarantees we only continue while the search space is non-empty.
Increment the counter, since each loop execution corresponds to one binary search step regardless of whether we terminate immediately after.
Compute the midpoint $mid = (l + r) // 2$. This is the same deterministic choice as the given implementation, and it defines the current probe position.
If $mid == x$, break immediately. This models the successful search termination.
If $mid < x$, shift the left boundary to $mid + 1$, eliminating all values that are too small.
Otherwise, shift the right boundary to $mid - 1$, eliminating all values that are too large.

After these steps, the counter reflects exactly how many iterations were required to reach $x$.

The correctness comes from the fact that the algorithm is a direct execution trace of the given procedure. At every iteration, the state $(l, r)$ matches the original function, and the counter increments exactly once per loop execution. Since the loop terminates only when $mid == x$, the number of increments is precisely the number of iterations executed before termination.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        l, r, x = map(int, input().split())
        cnt = 0
        while l <= r:
            cnt += 1
            mid = (l + r) // 2
            if mid == x:
                break
            if mid < x:
                l = mid + 1
            else:
                r = mid - 1
        print(cnt)

if __name__ == "__main__":
    solve()

The solution is a direct transcription of the given function. The only additional component is the counter, which increments at the start of each loop iteration to match the exact semantics of the original code. The loop condition and updates are unchanged, ensuring identical behavior. Integer division is used to match Python semantics exactly with the provided pseudocode.

A subtle detail is that the counter is incremented before checking whether mid == x. This is necessary because even the terminating iteration should be counted, matching the problem definition.

Worked Examples

Consider the first sample input case 3 7 6.

We track the execution step by step.

Iteration	l	r	mid	Action	cnt
1	3	7	5	mid < x, move right	1
2	6	7	6	mid == x, stop	2

The loop runs twice, so the answer is 2. This demonstrates that termination can happen before the interval fully collapses, depending on midpoint placement.

Now consider 5 8 6.

Iteration	l	r	mid	Action	cnt
1	5	8	6	mid == x, stop	1

Here the target is found immediately, so only one iteration is executed.

This shows that the count depends on the exact midpoint sequence, not just interval size.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(T \log (R-L))$	Each test performs a standard binary search loop, which halves the range each iteration
Space	$O(1)$	Only a few integers are used per test case

The constraints guarantee at most 100 tests and a range up to 1000, so the total number of loop iterations is negligible. The solution easily fits within both time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from math import log2

    data = inp.strip().split()
    t = int(data[0])
    idx = 1
    out = []

    for _ in range(t):
        l = int(data[idx]); r = int(data[idx+1]); x = int(data[idx+2])
        idx += 3

        cnt = 0
        while l <= r:
            cnt += 1
            mid = (l + r) // 2
            if mid == x:
                break
            if mid < x:
                l = mid + 1
            else:
                r = mid - 1
        out.append(str(cnt))

    return "\n".join(out)

# provided samples
assert run("5\n3 7 6\n6 12 7\n2 10 2\n6 14 13\n5 8 6") == "2\n2\n3\n3\n1"
assert run("1\n1 1 1") == "1"

# minimum interval
assert run("1\n1 1 1") == "1"

# already at midpoint early termination
assert run("1\n1 3 2") == "1"

# skewed range
assert run("1\n1 8 1") >= "1"

# symmetric mid-depth check
assert run("1\n1 7 7") == "3"

Test input	Expected output	What it validates
`1\n1 1 1`	`1`	single-element interval
`1\n1 3 2`	`1`	immediate midpoint hit
`1\n1 7 7`	`3`	deeper right-side search path

Edge Cases

When $L = R$, the loop executes exactly once because the condition $l \le r$ is true, $mid$ equals the only value, and the function terminates immediately. The implementation increments the counter before checking equality, so the output is 1, matching the expected behavior.

For a case like $L = 1, R = 3, x = 2$, the first midpoint is 2, so termination happens in the first iteration. The counter becomes 1, and no further updates are applied. This confirms that early termination does not skip counting the successful iteration.

For a case where the target is at an extreme boundary, such as $L = 1, R = 7, x = 7$, the search goes through several midpoints before reaching the right edge. Each iteration shrinks the interval correctly, and the loop count matches the depth of the implicit binary search tree path to the rightmost leaf.