## GUPTA MECHANICAL

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# [Solution] Single Operation Part 1 CodeChef Solution

## Problem

Chef has the binary representation $S$ of a number $X$ with him. He can modify the number by applying the following operation exactly once:

• Make $X := X \oplus \lfloor \frac{X}{2^{Y}} \rfloor$, where $(1 \leq Y \leq |S|)$ and $\oplus$ denotes the bitwise XOR operation.

Chef wants to maximize the value of $X$ after performing the operation. Help Chef in determining the value of $Y$ which will maximize the value of $X$ after the operation.

### Input Format

• The first line of input will contain a single integer $T$, denoting the number of test cases.
• Each test case consists of two lines of inputs - the first containing the length of binary string $S$.
• The second line of input contains the binary string $S$.

### Output Format

For each test case, output on a new line, the value of $Y$ which will maximize the value of $X$ after the operation.

Solution Click Below:-  👉
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### Explanation:

Test case $1$: Since $S = 10$ is the binary representation of $2$, the current value of $X = 2$. On choosing $Y = 1$$X$ becomes $2 \oplus \lfloor \frac{2}{2^{1}} \rfloor = 3$. We can show that this is the maximum value of $X$ we can achieve after one operation.

Test case $2$: Since $S = 11$ is the binary representation of $3$, the current value of $X = 3$. On choosing $Y = 2$$X$ becomes $3 \oplus \lfloor \frac{3}{2^{2}} \rfloor = 3$. We can show that this is the maximum value of $X$ we can achieve after one operation.

Test case $3$: Since $S = 101$ is the binary representation of $5$, the current value of $X = 5$. On choosing $Y = 1$$X$ becomes $5 \oplus \lfloor \frac{5}{2^{1}} \rfloor = 7$. We can show that this is the maximum value of $X$ we can achieve after one operation.

Test case $4$: Since $S = 110$ is the binary representation of $6$, the current value of $X = 6$. On choosing $Y = 2$$X$ becomes $6 \oplus \lfloor \frac{6}{2^{2}} \rfloor = 7$. We can show that this is the maximum value of $X$ we can achieve after one operation.