Find Duplicate Elements in a Given Range using Go

We can solve this problem in two different ways. Let’s check the first method.

Method 1: 

Examples

Input Array = [1, 2, 3, 4, 4] => Range is from 1 to 5 but 4 is a duplicate element in this range.

Approach to solve this problem

• Step 1: Define a method that accepts an array.
• Step 2: Declare a visited map.
• Step 3: Iterate the given array. If the element exists in the visited map, then return that element.
• Step 4: Else, return -1.

Program

Live Demo

package main
import "fmt"

func duplicateInArray(arr []int) int{
   visited := make(map[int]bool, 0)
   for i:=0; i<len(arr); i++{
      if visited[arr[i]] == true{
         return arr[i]
      } else {
         visited[arr[i]] = true
      }
   }
   return -1
}

func main(){
   fmt.Println(duplicateInArray([]int{1, 2, 3, 4, 4}))
   fmt.Println(duplicateInArray([]int{4, 5, 6, 7, 7}))
   fmt.Println(duplicateInArray([]int{1, 2, 3, 4, 5}))
}

Output

4
7
-1

Now, let’s check the second method to solve this problem.

Method 2: Using XOR operation

Examples

Input Array = [1, 2, 3, 4, 4] => Range is from 1 to 5 but 4 is duplicate in that range.

Range is from 1 to 5. => XOR => 0^1^2^3^4^4^0^1^2^3^4 => 4 (since 0^1=1).

Approach to solve this problem

• Step 1: Define a method that accepts an array.
• Step 2: Find the range value from the given array and define a variable xor, initialize with 0.
• Step 3: Iterate the given array and do a xor operation with the array’s elements.
• Step 4: Also perform xor operation from the lower range value to the higher range value.
• Step 5: At the end, return the xor variable, non-zero value for duplicate element.

Program

Live Demo

package main
import "fmt"

func duplicateInArray(arr []int, r int) int{
   xor := 0
   for i:=0; i<len(arr); i++{
      xor ^= arr[i]
   }
   for j:=1; j<=r-1; j++{
      xor ^= j
   }
   return xor
}

Output

4
3
1
0