In this article, we are going to learn to find a number which occurs odd number of times in an array of integers in go golang.
Here, only one number will have odd number of count in array.
Examples:
Input: arr = [1, 2, 3, 1, 2, 3, 2]
Output: 2
Explanation: 2 occurs 3 times in an array.
Input: arr = [1, 2, 3, 1, 2, 3]
Output: 0
Explanation: None number occurs odd time.
Note: XOR (^) of two same number becomes zero i.e 2 ^ 2 = 0.
Algorithm:
- Get the size of array from user
- Insert the number into array
- Apply XOR (^) operator to all numbers
- All even time occurred numbers will become zero. Only odd time occurred number will remain same.
package main
import "fmt"
func check_odd_element(arr []int) {
length := len(arr)
var odd_number int = arr[0]
for i := 1; i < length; i++ {
// Perform XOR (^) operator to all elements in array
odd_number = odd_number ^ arr[i]
}
fmt.Printf("\nOdd number occurance is: %d", odd_number)
}
func main() {
var size, number int
fmt.Print("Enter the size of array: ")
// Get the array size from user
fmt.Scan(&size)
arr := make([]int, size)
for i := 0; i < size; i++ {
fmt.Print("\nEnter the number: ")
fmt.Scan(&number)
// Fill the elements into array
arr[i] = number
}
check_odd_element(arr)
}
Output:
Enter the size of array: 5
Enter the number: 1
Enter the number: 2
Enter the number: 3
Enter the number: 1
Enter the number: 2
Odd number occurance is: 3
Enter the size of array: 6
Enter the number: 1
Enter the number: 2
Enter the number: 3
Enter the number: 1
Enter the number: 2
Enter the number: 3
0
To learn more about golang, Please refer given below link:
https://techieindoor.com/go-lang-tutorial/
References:
https://golang.org/doc/
https://golang.org/pkg/
https://golang.org/pkg/fmt/
https://golang.org/pkg/fmt/#Println