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Golang program to turn off the k’th bit in a number.
Example
Consider n = 20(00010100), k = 3 The result after turning off the 3rd bit => 00010000 & ^(1<<(3-1)) => 00010000 & ^(1 << 2) => 00010000 => 16s
Approach to solve this problem
Step 1 − Define a method, where n and k would be the arguments, return type is int.
Step 2 − Perform AND operation with n & ^(1<<(k-1)).
Step 3 − Return the obtained number.
Example
package main
import (
"fmt"
"strconv"
)
func TurnOffKthBit(n, k int) int {
return n & ^(1 << (k-1))
}
func main(){
var n = 20
var k = 3
fmt.Printf("Binary of %d is: %s.\n", n, strconv.FormatInt(int64(n), 2))
newNumber := TurnOffKthBit(n, k)
fmt.Printf("After turning off %d rd bit of %d is: %d.\n", k, n, newNumber)
fmt.Printf("Binary of %d is: %s.\n", newNumber, strconv.FormatInt(int64(newNumber), 2))
}Output
Binary of 20 is: 10100. After turning off 3 rd bit of 20 is 16. Binary of 16 is: 10000.
Updated on: 17-Mar-2021
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