Bitwise Operators in C - Hacker Rank Solution
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| Bitwise Operators in C - Hacker Rank Solution |
This challenge will let you learn about bitwise operators in C.
Inside the CPU, mathematical operations like addition, subtraction, multiplication and division are done in bit-level. To perform bit-level operations in C programming, bitwise operators are used which are explained below.
- Bitwise AND operator & The output of bitwise AND is 1 if the corresponding bits of two operands is 1. If either bit of an operand is 0, the result of corresponding bit is evaluated to 0. It is denoted by &. Bitwise OR operator | The output of
- bitwise OR is 1 if at least one corresponding bit of two operands is 1. It is denoted by |.
- Bitwise XOR (exclusive OR) operator ^ The result of bitwise XOR operator is 1 if the corresponding bits of two operands are opposite. It is denoted by XOR
For example, for integers 3 and 5,
3 = 00000011 (In Binary) 5 = 00000101 (In Binary) AND operation OR operation XOR operation 00000011 00000011 00000011 & 00000101 | 00000101 ^ 00000101 ________ ________ ________ 00000001 = 1 00000111 = 7 00000110 = 6
Task
Given set, s= {1,2,3,4......n} find:- the maximum value of a&b which is less than a given integer k , where a and b (where a < b) are two integers from set S.
- the maximum value of a|b which is less than a given integer k, where a and b (where a < b) are two integers from set S.
- the maximum value of a XOR b which is less than a given integer k, where a and b (where a < b) are two integers from set S.
Input Format
The only line contains 2 space-separated integers n, and k, respectively.Constraints
- 2<=n<=10^3
- 2<=k<=n
Output Format
The first line of output contains the maximum possible value of a & b.The second line of output contains the maximum possible value of a | b.
The second line of output contains the maximum possible value of a XOR b.
Sample Input :
5 4
Sample Output :
2 3 3
Explanation :
n = 5, k =4S = {1,2,3,4,5}
All possible values of a and b are:
- a = 1, b = 2 ; a&b = 0 a|b = 3 a XOR b = 3
- a = 1, b = 3 ; a&b = 1 a|b = 3 a XOR b = 2
- a = 1, b = 4 ; a&b = 0 a|b = 5 a XOR b = 5
- a = 1, b = 5 ; a&b = 1 a|b = 5 a XOR b = 4
- a = 2, b = 3 ; a&b = 2 a|b = 3 a XOR b = 1
- a = 2, b = 4 ; a&b = 0 a|b = 6 a XOR b = 6
- a = 2, b = 5 ; a&b = 0 a|b = 7 a XOR b = 7
- a = 3, b = 4 ; a&b = 0 a|b = 7 a XOR b = 7
- a = 3, b = 5 ; a&b = 1 a|b = 7 a XOR b = 6
- a = 4, b = 5 ; a&b = 4 a|b = 5 a XOR b = 1
The maximum possible value of a|b that is also <(k=4) is 3, so we print 3 on first line.
The maximum possible value of a XOR b that is also <(k=4) is 3, so we print 3 on first line.
Solution :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | #include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int n, k; scanf("%d %d", &n, &k); int mxAnd = 0, mxOr = 0, mxXor = 0; for(int i = 1; i <= n; i++){ for(int j = i + 1; j <= n; j++){ if(mxAnd < (i & j) && (i & j) < k) mxAnd = i & j; if(mxOr < (i | j) && (i | j) < k) mxOr = i | j; if(mxXor < (i ^ j) && (i ^ j) < k) mxXor = i ^ j; } } printf("%d\n", mxAnd); printf("%d\n", mxOr); printf("%d\n", mxXor); return 0; } |
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Hey, I am Parthe Omkar
first one is not working , timeout error is shown
The solution is Correct and it's work.
correct
It's not working.....