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2\n repeating nos

Find the two repeating elements in a given array

June 9, 2010

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.

For example, array = {4, 2, 4, 5, 2, 3, 1} and n = 5

The above array has n + 2 = 7 elements with all elements occurring once except 2 and 4 which occur twice. So the output should be 4 2.

Method 1 (Basic)
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.

#include<stdio.h>
#include<stdlib.h>
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]);
}    
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

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

Method 2 (Use Count array)
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.

#include<stdio.h>
#include<stdlib.h>
 
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]]++;
  }   
}    
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

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

Method 3 (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

Thanks to geek4u for suggesting this method. As pointed by Beginer , there can be addition and multiplication overflow problem with this approach.
The methods 3 and 4 use all useful information given in the question :)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
 
/* function to get factorial of n */
int fact(int 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);
}   
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

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

Method 4 (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);
}    
 
 
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]); 
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Method 5 (Use array elements as index)
Thanks to Manish K. Aasawat for suggesting this method.

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.

#include <stdio.h>
#include <stdlib.h>
 
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]));
  }        
}    
 
int main()
{
  int arr[] = {1, 3, 2, 2, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]);
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}



Find duplicates in O(n) time and O(1) extra space

January 4, 2011

Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times. Find these repeating numbers in O(n) and using only constant memory space.

For example, let n be 7 and array be {1, 2, 3, 1, 3, 0, 6}, the answer should be 1 & 3.

This problem is an extended version of following problem.

Find the two repeating elements in a given array

Method 1 and Method 2 of the above link are not applicable as the question says O(n) time complexity and O(1) constant space. Also, Method 3 and Method 4 cannot be applied here because there can be more than 2 repeating elements in this problem. Method 5 can be extended to work for this problem. Below is the solution that is similar to the Method 5.

Algorithm:

traverse the list for i= 0 to n-1 elements
{
  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
}

Implementation:

#include <stdio.h>
#include <stdlib.h>
 
void printRepeating(int arr[], int size)
{
  int i;
  printf("The repeating elements are: \n");
  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]));
  }
}
 
int main()
{
  int arr[] = {1, 2, 3, 1, 3, 0, 6};
  int arr_size = sizeof(arr)/sizeof(arr[0]);
  printRepeating(arr, arr_size);
  getchar();
  return 0;
}

Note: The above program doesn’t handle 0 case (If 0 is repeated). The program can be easily modified to handle that also. It is not handled to keep the code simple.

Output:
The repeating elements are:
1 3

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

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