If the array elements are positive numbers and also elements are in the range 0 to n-1 , then we can find duplicate elements using negation technique. For each element a[i], go to the array element whose index is a[i], that means we select a[a[i]] and mark this value as -a[a[i]]. That is we negate the value of a[a[i]. Continue this process until we find the element that is already negated.

**Note:**

- The solution does not work if the given array is read only, that means no manipulation is allowed to array elements.
- The solution will work only if the array elements are positive and are in the range 0 to n-1.

**Finding duplicates in an array using negation – O(n)**

```
/**
* <p>Check Duplicates in the array using Math abs</p>
* Time Complexity ::O(n)
* Space Complexity ::O(1)
* @param a
* @return
*/
public static int checkDuplicatesWithMathAbs(int[] a){
int n=a.length;
for(int i=0;i<n;i++){
if(a[Math.abs(a[i])]<0)
return a[i];
else
a[a[i]]=-a[a[i]];
}
return -1;
}
```

Sample **Test Client** for above program

```
public static void main(String[] args) {
/**
* Checking duplicates in the array by placing negative values
* Array elements should be in the range 0 to n-1
* This solution does not work if the given array is read only and all the * elements should be +ve
* Time Complexity :: O(n)
* Space Complexity :: O(1)
*/
int[] arr5={0,1,2,2,4,5,6};
System.out.println(checkDuplicatesWithMathAbs(arr5));
}
```