Explanation :

int binsearch(int a[],int lb,int ub,int x)
{
  int mid;
  while(lb<=ub)
  {
    mid=(lb+ub)/2;
    if(a[mid]==x)
    return mid;
    if(a[mid]>x)
      if(a[mid]>=a[ub] || x<a[ub])
           ub=mid-1;
      else
           lb=mid+1;
    else
      if(a[mid]>=a[lb] || x<a[lb])
          lb=mid+1;
      else
         ub=mid-1;
  }
   return -1;
}

Explanation :

Method 1:

1. Find sum of first n natural numbers and sum of all numbers in the array. Subtract them. Then we will get a+b = x.
2. Find sum of squares of first n natural numbers and sum of squares of all numbers present in the array. Subtract them. Then we will get a2 + b2 = y.
3. Now solve equation of step 1 and step 2.

     z = a*b = (x2 - y)/2.
     w = a-b = sqrt( x2 - 4*z).
     a=(x+w)/2.
     b=(x-w)/2.
 

1. Find sum : a+b = x.
2. Find XOR : a^b = y.
3. Now for each pair which will form the sum x, check a^b whichever will match y.
4. Take that a & b as repeated numbers.

1. count = 0. This variable "count" represents number of repeated elements found.
2. Start traversing the array.
3. If value == index then go ahead to next element.
4. If value > index then
5. If arr[value] == arr[index] then this is repeated value. Store this at the end of the array.

    count++;
         temp = arr[index];
         arr[index]=arr[n+count];
         arr[n+count]=temp;
   

   else swap arr[index] and arr[value]. Don't update the index.
6. If index > value then this value is being repeated. Store this at the end of the array.

     count++;
     temp = arr[index];
     arr[index]=arr[n+count];
     arr[n+count]=temp;

7. again don't change the index and start comparing again from the same point.

Explanation :

void mergesort(int arr[],int n)
 {
   int l1,l2,u1,u2,i,j,k,size,*aux;
   aux=(int *)malloc(sizeof(int)*n);
   size=1;
   while(size < n)
     {
       l1=0,k=0;
       while(l1 + size < n)
        {
          l2=l1+size;
          u1=l2-1;
          u2=(l2+ size-1< n)?l2 + size-1:n-1;
          for(i= l1,j=l2;i<=u1&&j<=u2;k++)
          if(arr[i]<=arr[j])
          aux[k]=arr[i++];
          else
          aux[k]=arr[j++];
          for(;i<=u1;k++)
          aux[k]=arr[i++];
          for(;j<=u2;k++)
          aux[k]=arr[j++];
          l1=u2+1;
        }
       for(i=l1;k < n;i++)
       aux[k++]=arr[i];
       for(i=0;i < n;i++)
       arr[i]=aux[i];
       size=size*2;
     }
}

Explanation :

int binsearch(int a[],int lb,int ub, int x)
{
  int mid;
  mid=(lb+ub)/2;
  while(lb<=ub && x!=a[mid])
   {
     if(a[mid] >x)
       ub=mid-1;
     else
       lb=mid+1;
     mid=(lb+ub)/2;
	}
  if(x==a[mid])
   return mid;
  else
   return -1;
}

Explanation :

int partition(int a[], int lb, int ub, int x)
{
  int up, down,t;
  up=lb;
  down=ub;
  while(up < down)
   {
    while( a[up] == x)
    up++;
    while(a[down] != x)
    down--;
    if(up < down)
      {
       t=a[up];
       a[up]=a[down];
       a[down]=t;
       up++;
       down--;
      }
   }
  return down+1;
}
In main function the function partition() will be used as:
pivot=partition(a,0,n-1,0);
pivot=partition(a,pivot,n-1,1);