Bitwise AND Operators in C Hackerrank Solution are very easy and simple. Visit the hackerrank website and choose your language C and make your you have clear a coding dashboard now copy the following code and paste it into the editor dashboard and Run the program. After successfully Running submit the code.

Given set S = {1, 2, 3, . . . . , N}. Find two integers, A and B (where A<B), from set S such that the value of A&B is the maximum possible and also less than a given integer, K. In this case, & represents the bitwise AND operator.

**Basic Operators**

Here are some commonly used Java operators you should familiarize yourself with & Bitwise AND (^). This binary operation evaluates to 1 (true) if both operands are true, otherwise 0 (false). In other words:

**&**Bitwise

1 & 1 = 1

1 & 0 = 0

0 & 1 = 0

0 & 0 = 0

**| Bitwise Inclusive OR**

**|**Bitwise Inclusive OR (V). This binary operation evaluates to 1 if either operand is true, otherwise 0 (false) if both operands are false. In other words:

1 | 1 = 1

1 | 0 = 1

0 | 1 = 1

0 | 0 = 0

**^ Bitwise Exclusive OR**

^ Bitwise Exclusive OR or XOR (⊕). This binary operation evaluates to 1 (true) if and only if exactly one of the two operands is 1; if both operands are 1 or 0, it evaluates to 0 (false). In other words:

1 ^ 1 = 0

1 ^ 0 = 1

0 ^ 1 = 1

0 ^ 0 = 0

**~ Bitwise Unary**

~ The unary Bitwise Complement operator flips every bit; for example, the bitwise-inverted 8-bit binary number 01111001 becomes 10000110, and the bitwise-inverted signed decimal integer 8 becomes -9.

```
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
int T, n, k, i, j, val;
scanf("%d", & T);
while (T > 0) {
int maximum = 0;
scanf("%d%d", & n, & k);
int max_val = 0;
int a = 0, b = k - 1;
for (a = n; a > 2; a--) {
if (a == b)
continue;
if ((a & b) > max_val) {
max_val = a & b;
}
}
printf("%d\n", max_val);
T--;
}
return 0;
}
```

### Explanation of Bitwise AND Operators

The explanation of this programming challenge is below please find out and read all the steps carefully.

Since K-1 is the highest possible answer, we will take it as one of the 2 numbers. The other number should be > K-1 due to the AND property and it would be >= K. It’s best to take a number whose binary equivalent is similar to K-1’s binary value. So K would be the best choice.

Note: By doing an OR operation between 2 numbers, the answer would be >= HIGHEST NUMBER of the 2 numbers.

To find the other number we perform OR operation between the highest possible answer and the immediate larger number to it.

i.e [(K-1) OR (K-1)+1] which is nothing but [(K-1) OR K] and its result would be >= K.

Now we got the AND pair which are K-1 and K (the minimum possible result of the above OR operation) and our AND result would be <= K-1.

For most cases we will get the final answer as K-1 itself but not for all cases. When K is odd, the final answer will definitely be K-1 because if K is odd, K-1 will be even. Here only the LSB of the binary equivalent will be different. Eg: K=5(0101); K-1=4(0100)

When K is even, K-1 will be odd and both numbers' binary values might not be similar. Eg: K=8(1000); K-1=7(0111).

K-1 will be the answer only when the result of the OR operation is <= N. If it's> N, we would end up using a number which is not in the given number set for the AND operation which might result in a wrong final answer.

So these cases occur when {(K-1 OR K) > N} and when K is even.

For these scenarios, the highest possible answer would not be K-1 and it'll be the next lesser number K-2. The AND pairs for such scenarios would be K-2 and K-1 resulting in a final answer of K-2.

For the above cases, K-1 cannot be the highest possible answer, so we take the lesser number K-2 as the highest possible answer and start again from STEP 1 replacing K-1 with K-2 and K with K-1.

#### STEP 1 to Bitwise Operators in C Hackerrank Solution

Since K-1 is the highest possible answer, we will take it as one of the 2 numbers. The other number should be > K-1 due to the AND property and it would be >= K. It’s best to take a number whose binary equivalent is similar to K-1’s binary value. So K would be the best choice.

Note: By doing an OR operation between 2 numbers, the answer would be >= HIGHEST NUMBER of the 2 numbers.

#### STEP 2: Bitwise Operators in C Hackerrank Solution

To find the other number we perform OR operation between the highest possible answer and the immediate larger number to it.

i.e [(K-1) OR (K-1)+1] which is nothing but [(K-1) OR K] and its result would be >= K.

#### STEP 3 to Bitwise Operators Hackerrank Solution in C

Now we got the AND pair which are K-1 and K (the minimum possible result of the above OR operation) and our AND result would be <= K-1.

For most cases we will get the final answer as K-1 itself but not for all cases. When K is odd, the final answer will definitely be K-1 because if K is odd, K-1 will be even. Here only the LSB of the binary equivalent will be different. Eg: K=5(0101); K-1=4(0100)

When K is even, K-1 will be odd and both numbers' binary values might not be similar. Eg: K=8(1000); K-1=7(0111).

K-1 will be the answer only when the result of the OR operation is <= N. If it's> N, we would end up using a number which is not in the given number set for the AND operation which might result in a wrong final answer.

So these cases occur when {(K-1 OR K) > N} and when K is even.

For these scenarios, the highest possible answer would not be K-1 and it'll be the next lesser number K-2. The AND pairs for such scenarios would be K-2 and K-1 resulting in a final answer of K-2.

For the above cases, K-1 cannot be the highest possible answer, so we take the lesser number K-2 as the highest possible answer and start again from STEP 1 replacing K-1 with K-2 and K with K-1.

**First Scenario**N=5, K=2 ; K-1 = 1

0010 2(K)

0001 1(K-1)

-----OR----

0011 3(OR result)

3 < N

0011 3(OR result)

0001 1(K-1)

-----AND----

0001 1(final answer)

**Second Scenario**

2.N=8, K=5 ; K-1 = 4

0101 5(K)

0100 4(K-1)

-----OR----

0101 5(OR result)

5 < N

0101 5(OR result)

0100 4(K-1)

-----AND----

0100 4(final answer)

**Third Scenario**

3.N=2, K=2 ; K-1 = 1 ; K-2 = 0

0010 2(K)

0001 1(K-1)

-----OR----

0011 3(OR result)

3 > N

0001 1(K-1)

0000 0(K-2)

-----OR----

0001 1(OR result)

0001 1(OR result)

0000 0(K-2)

-----AND----

0000 0(final answer)

**Fourth Scenario**

4.N=21, K=20 ; K-1 = 19 ; K-2 = 18

10100 20(K)

10011 19(K-1)

-----OR----

10111 23(OR result)

23 > N

10011 19(K-1)

10010 18(K-2)

-----OR----

10011 19(OR result)

10011 19(OR result)

10010 18(K-2)

-----AND----

10010 18(final answer)

The above explanation by rospvar. All the credit goes to Rospvar. Thank you for the detailed explanation.

### Bitwise AND Operators

As we can see above in Bitwise(Bitwise Operators Hackerrank Solution in C) AND if 1 and 1 then the only condition is true. In this problem, we are taking two inputs from the user first one in number N and the second in K. Now we have to find all sets of numbers S = {1, 2, 3, . . . . , N}. Let's take an example and try to understand the problem "Bitwise AND" easily. suppose N=5 and K=2 then set S={1, 2, 3, 4, 5}. then the combination is below.

1. A = 1, B = 2; A & B = 0

2. A = 1, B = 3; A & B = 1

3. A = 1, B = 4; A & B = 0

4. A = 1, B = 5; A & B = 1

5. A = 2, B = 3; A & B = 2

6. A = 2, B = 4; A & B = 0

7. A = 2, B = 5; A & B = 0

8. A = 3, B = 4; A & B = 0

9. A = 4, B = 5; A & B = 1

10. A = 4, B = 5; A & B = 4

The maximum possible value of A & B (LINE NUMBER 5) that is also < (K = 2) is 1, so we print 1 on a new line. We continue to check A & B value is maximum and A & B value is less than or equal to the K.

1. A = 1, B = 2; A & B = 0

2. A = 1, B = 3; A & B = 1

3. A = 1, B = 4; A & B = 0

4. A = 1, B = 5; A & B = 1

5. A = 2, B = 3; A & B = 2

6. A = 2, B = 4; A & B = 0

7. A = 2, B = 5; A & B = 0

8. A = 3, B = 4; A & B = 0

9. A = 4, B = 5; A & B = 1

10. A = 4, B = 5; A & B = 4

The maximum possible value of A & B (LINE NUMBER 5) that is also < (K = 2) is 1, so we print 1 on a new line. We continue to check A & B value is maximum and A & B value is less than or equal to the K.

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