## Monday, March 31, 2014

### Find the two repeating elements in ranged array

Problem
You are given an array of n+2 elements. All elements of the array are in range 1 to n. And all elements occur once except two numbers which occur twice. Find the two repeating numbers.
Example
Input : array = {4, 2, 4, 5, 2, 3, 1} and n = 5, array.length = n+2 = 7
Output : 4,2

Solution
This can be achieved by various methods.

Method 1 - Brute force
Use two loops. In the outer loop, pick elements one by one and count the number of occurrences of the picked element in the inner loop.
This method doesn’t use the other useful data provided in questions like range of numbers is between 1 to n and there are only two repeating elements.
```void printRepeating(int arr[], int size)
{
int i, j;
printf(" Repeating elements are ");
for(i = 0; i < size; i++)
for(j = i+1; j < size; j++)
if(arr[i] == arr[j])
printf(" %d ", arr[i]);
}
```
Time Complexity: O(n*n)
Auxiliary Space: O(1)

Method 2 - Use Count array (for hashing)
Traverse the array once. While traversing, keep track of count of all elements in the array using a temp array count[] of size n, when you see an element whose count is already set, print it as duplicate.
This method uses the range given in the question to restrict the size of count[], but doesn’t use the data that there are only two repeating elements.
```void printRepeating(int arr[], int size)
{
int *count = (int *)calloc(sizeof(int), (size - 2));
int i;
printf(" Repeating elements are ");
for(i = 0; i < size; i++)
{
if(count[arr[i]] == 1)
printf(" %d ", arr[i]);
else
count[arr[i]]++;
}
} ```

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 3 - Sorting the array and compare adjacent array
```void printRepeating(int arr[], int size)
{
sort(arr);
int i, j;
printf(" Repeating elements are ");
for(i = 0; i < size; i++)
for(j = i+1; j < size; j++)
if(arr[i] == arr[j])
printf(" %d ", arr[i]);
}
```

Time complexity = O(n) Sorting takes O(n log n) time if we use comparison based sort, but as they are integer we can use counting sort.And then comparing takes O(n) times.

Method 4 - Make two equations
Let the numbers which are being repeated are X and Y. We make two equations for X and Y and the simple task left is to solve the two equations.
We know the sum of integers from 1 to n is n(n+1)/2 and product is n!. We calculate the sum of input array, when this sum is subtracted from n(n+1)/2, we get X + Y because X and Y are the two numbers missing from set [1..n]. Similarly calculate product of input array, when this product is divided from n!, we get X*Y.
```Given sum and product of X and Y, we can find easily out X and Y.
Let summation of all numbers in array be S and product be P
X + Y = S – n(n+1)/2
XY = P/n!

Using above two equations, we can find out X and Y.
For array = 4 2 4 5 2 3 1, we get S = 21 and P as 960.
X + Y = 21 – 15 = 6
XY = 960/5! = 8
X – Y = sqrt((X+Y)^2 – 4*XY) = sqrt(4) = 2

Using below two equations, we easily get X = (6 + 2)/2 and Y = (6-2)/2
X + Y = 6
X – Y = 2
```

Note that there can be addition and multiplication overflow problem with this approach.
The methods 3 and 4 use all useful information given in the question
```// function to get factorial of n

void printRepeating(int arr[], int size)
{
int S = 0;  // S is for sum of elements in arr[]
int P = 1;  // P is for product of elements in arr[]
int x,  y;   // x and y are two repeating elements
int D;      // D is for difference of x and y, i.e., x-y
int n = size - 2,  i;

// Calculate Sum and Product of all elements in arr[]
for(i = 0; i < size; i++)
{
S = S + arr[i];
P = P*arr[i];
}

S = S - n*(n+1)/2;  // S is x + y now
P = P/fact(n);         // P is x*y now

D = sqrt(S*S - 4*P); // D is x - y now

x = (D + S)/2;
y = (S - D)/2;

printf("The two Repeating elements are %d & %d", x, y);
}

int fact(int n)
{
return (n == 0)? 1 : n*fact(n-1);
}
```

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 5 - Use XOR
Thanks to neophyte for suggesting this method.
The approach used here is similar to method 2 of this post.
Let the repeating numbers be X and Y, if we xor all the elements in the array and all integers from 1 to n, then the result is X xor Y.

The 1’s in binary representation of X xor Y is corresponding to the different bits between X and Y. Suppose that the kth bit of X xor Y is 1, we can xor all the elements in the array and all integers from 1 to n, whose kth bits are 1. The result will be one of X and Y.

```void printRepeating(int arr[], int size)
{
int xor = arr[0]; // Will hold xor of all elements
int set_bit_no;  // Will have only single set bit of xor
int i;
int n = size - 2;
int x = 0, y = 0;

// Get the xor of all elements in arr[] and {1, 2 .. n}
for(i = 1; i < size; i++)
xor ^= arr[i];
for(i = 1; i <= n; i++)
xor ^= i;

// Get the rightmost set bit in set_bit_no
set_bit_no = xor & ~(xor-1);

// Now divide elements in two sets by comparing rightmost set
bit of xor with bit at same position in each element.
for(i = 0; i < size; i++)
{
if(arr[i] & set_bit_no)
x = x ^ arr[i]; //XOR of first set in arr[]
else
y = y ^ arr[i]; //XOR of second set in arr[]
}
for(i = 1; i <= n; i++)
{
if(i & set_bit_no)
x = x ^ i; //XOR of first set in arr[] and {1, 2, ...n }
else
y = y ^ i; //XOR of second set in arr[] and {1, 2, ...n }
}

printf("\n The two repeating elements are %d & %d ", x, y);
}

```

Method 6 - Use array elements as index and negation

```Traverse the array. Do following for every index i of A[].
{
check for sign of A[abs(A[i])] ;
if positive then
make it negative by   A[abs(A[i])]=-A[abs(A[i])];
else  // i.e., A[abs(A[i])] is negative
this   element (ith element of list) is a repetition
}
```
Example:
``` A[] =  {1, 1, 2, 3, 2}
i=0;
Check sign of A[abs(A[0])] which is A[1].
A[1] is positive, so make it negative.
Array now becomes {1, -1, 2, 3, 2}

i=1;
Check sign of A[abs(A[1])] which is A[1].
A[1] is negative, so A[1] is a repetition.

i=2;
Check sign of A[abs(A[2])] which is A[2].
A[2] is  positive, so make it negative. '
Array now becomes {1, -1, -2, 3, 2}

i=3;
Check sign of A[abs(A[3])] which is A[3].
A[3] is  positive, so make it negative.
Array now becomes {1, -1, -2, -3, 2}

i=4;
Check sign of A[abs(A[4])] which is A[2].
A[2] is negative, so A[4] is a repetition.
```
Note that this method modifies the original array and may not be a recommended method if we are not allowed to modify the array.
```void printRepeating(int arr[], int size)
{
int i;

printf("\n The repeating elements are");

for(i = 0; i < size; i++)
{
if(arr[abs(arr[i])] > 0)
arr[abs(arr[i])] = -arr[abs(arr[i])];
else
printf(" %d ", abs(arr[i]));
}
}
```

Thanks.
References
http://www.geeksforgeeks.org/find-the-two-repeating-elements-in-a-given-array/