Programming Articles - Page 1293 of 3363

ExamplesConsider two numbers m = 65 => 01000001 and n = 80 => 01010000Number of bits flipped is 2.Approach to solve this problemStep 1 − Convert both numbers into bits.Step 2 − Count number of bits are flipped.Example Live Demopackage main import (    "fmt"    "strconv" ) func FindBits(x, y int) int{    n := x ^ y    count := 0    for ;n!=0; count++{       n = n & (n-1)    }    return count } func main(){    x := 65    y := 80    fmt.Printf("Binary of %d is: %s.", x, strconv.FormatInt(int64(x), 2))   ... Read More

In Golang, we can use the binary operators "&" (AND) and "|" (OR) to convert strings from lowercase to uppercase and vice versa. Let's take a simple example and understand how to use these binary operators in Golang. Examplepackage main import "fmt" func main(){    fmt.Printf("Lowercase characters using OR operator: ")        for ch:='A'; ch

Definition − Parity refers to the count of 1s. If count of 1s is even, then it’s even parity; and if the count of 1s is odd, then the parity is Odd.ExamplesConsider n = 20(00010100)Parity of the given number 20 is even.Approach to solve this problemStep 1 − Define a method, where n and is an argument, return type is int.Step 2 − Calculate the count of 1s in the given number’s bit.Example Live Demopackage main import (    "fmt"    "strconv" ) func FindParity(n int) bool {    parity := false    for n != 0 {       ... Read More

ExamplesConsider n = 20(00010100)Now return log2(20 & -20) => 2+1 => 3Approach to solve this problemStep 1 − Define a method, where n and is an argument, return type is int.Step 2 − Return log2(n & -n)+1.Examplepackage main import (    "fmt"    "math"    "strconv" ) func FindRightMostSetBit(n int) int {    if (n & 1) != 0{       return 1    }    return int(math.Log2(float64(n & -n))) + 1 } func main(){    var n = 20    fmt.Printf("Binary of %d is: %s.", n, strconv.FormatInt(int64(n), 2))    fmt.Printf("Position of the rightmost set bit of the given number %d is %d.", n, FindRightMostSetBit(n)) }OutputBinary of 20 is: 10100. Position of the rightmost set bit of the given number 20 is 3.

ExamplesConsider n = 16(00010000)Now find x = n-1 => 15(00001111) => x & n => 0Approach to solve this problemStep 1 − Define a method, where n and is an argument, returns type is int.Step 2 − Perform x = n & n-1.Step 3 − If x is 0, then the given number is power of 2; else not.Example Live Demopackage main import (    "fmt"    "strconv" ) func CheckNumberPowerOfTwo(n int) int {    return n & (n-1) } func main(){    var n = 16    fmt.Printf("Binary of %d is: %s.", n, strconv.FormatInt(int64(n), 2))    flag := CheckNumberPowerOfTwo(n)   ... Read More

ExamplesConsider n = 20(00010100), k = 3.After toggling the kth bit of the given number: 00010000 => 16.Approach to solve this problemStep 1 − Define a method, where n and k would be the arguments, returns type is int.Step 2 − Perform AND operation with n ^ (1

ExamplesConsider n = 20(00010100), k = 3So, the result after turning off the 3rd bit => 00010000 & (1

ExampleFor example consider n = 20(00010100), k = 4. So result after turning on 4th bit => 00010000 | (1

ExampleConsider n = 20(00010100), k = 3 The result after turning off the 3rd bit => 00010000 & ^(1 16sApproach to solve this problemStep 1 − Define a method, where n and k would be the arguments, return type is int.Step 2 − Perform AND operation with n & ^(1

Example − In the given tree, the root node is 1, the root of its left sub tree is 2, and the root of its right sub tree is 3, ... so on.Preorder Tree Traversal Output: 1, 2, 4, 5, 3, 6, 7.Approach to solve this problemStep 1 − First, we’ll define the node structure.Step 2 − In the main method, we would create the above tree.Step 3 − Finally, we will perform the Preorder Tree Traversal.Example Live Demopackage main import "fmt" type Node struct {    data int    left *Node    right *Node } func (root *Node)PreOrderTraversal(){    if ... Read More