Find the repeating and the missing | Added 3 new methods | GeeksforGeeks

Find the repeating and the missing | Added 3 new methods | GeeksforGeeks
Given an unsorted array of size n. Array elements are in range from 1 to n. One number from set {1, 2, …n} is missing and one number occurs twice in array. Find these two numbers.
Method 3 (Use elements as Index and mark the visited places)
Traverse the array. While traversing, use absolute value of every element as index and make the value at this index as negative to mark it visited. If something is already marked negative then this is the repeating element. To find missing, traverse the array again and look for a positive value.

void printTwoElements(int arr[], int size)

{

    int i;

    printf("\n The repeating element is");

    for(i = 0; i < size; i++)

    {

        if(arr[abs(arr[i])-1] > 0)

            arr[abs(arr[i])-1] = -arr[abs(arr[i])-1];

        else

            printf(" %d ", abs(arr[i]));

    }

    printf("\nand the missing element is ");

    for(i=0; i<size; i++)

    {

        if(arr[i]>0)

            printf("%d",i+1);

    }

}

Method 4 (Make two equations)
Let x be the missing and y be the repeating element.

1) Get sum of all numbers.
Sum of array computed S = n(n+1)/2 – x + y
2) Get product of all numbers.
Product of array computed P = 1*2*3*…*n * y / x
3) The above two steps give us two equations, we can solve the equations and get the values of x and y.

Time Complexity: O(n)

This method can cause arithmetic overflow as we calculate product and sum of all array elements. 

Method 5 (Use XOR)

Calculate XOR of all the array elements.

     xor1 = arr[0]^arr[1]^arr[2].....arr[n-1]

XOR the result with all numbers from 1 to n

     xor1 = xor1^1^2^.....^n

In the result xor1, all elements would nullify each other except x and y. All the bits that are set inxor1 will be set in either x or y. So if we take any set bit (We have chosen the rightmost set bit in code) of xor1 and divide the elements of the array in two sets – one set of elements with same bit set and other set with same bit not set. By doing so, we will get x in one set and y in another set. Now if we do XOR of all the elements in first set, we will get x, and by doing same in other set we will get y.

void getTwoElements(int arr[], int n, int *x, int *y)

{

  int xor1;   /* Will hold xor of all elements and numbers from 1 to n */

  int set_bit_no;  /* Will have only single set bit of xor1 */

  int i;

  *x = 0;

  *y = 0;

  xor1 = arr[0];

  /* Get the xor of all array elements elements */

  for(i = 1; i < n; i++)

     xor1 = xor1^arr[i];

  /* XOR the previous result with numbers from 1 to n*/

  for(i = 1; i <= n; i++)

     xor1 = xor1^i;

  /* Get the rightmost set bit in set_bit_no */

  set_bit_no = xor1 & ~(xor1-1);

  /* Now divide elements in two sets by comparing rightmost set

   bit of xor1 with bit at same position in each element. Also, get XORs

   of two sets. The two XORs are the output elements.

   The following two for loops serve the purpose */

  for(i = 0; i < n; i++)

  {

    if(arr[i] & set_bit_no)

     *x = *x ^ arr[i]; /* arr[i] belongs to first set */

    else

     *y = *y ^ arr[i]; /* arr[i] belongs to second set*/

  }

  for(i = 1; i <= n; i++)

  {

    if(i & set_bit_no)

     *x = *x ^ i; /* i belongs to first set */

    else

     *y = *y ^ i; /* i belongs to second set*/

  }

  /* Now *x and *y hold the desired output elements */

}

This method doesn’t cause overflow, but it doesn’t tell which one occurs twice and which one is missing. We can add one more step that checks which one is missing and which one is repeating. This can be easily done in O(n) time.

Method 1 (Use Sorting)
Method 2 (Use count array)
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