Skip to main content

Split and sort array

Here was another interesting interview question on arrays

How do you sort the array so that all the odd elements are before all the even elements.

And both these subarrays must be sorted.

We can first split the array so that all odd elements are in first subarray and all even elements come afterwards. Next we can use some sorting method to sort these two sub-arrays separately.

Next comes the question. How do we split the array?
  • For ith element in array, if it is even 
    • remove the element from ith position
    • move it to end of array
  • Repeat this process for all elements from last to 0th
Here is the code to split array into odd and even.

int splitOddEven(int *arr,int n)
{
    int i;
    int evencount=0;
    for(i=n-1;i>=0;i--)
    {
       if(arr[i]%2==0)
           {
           move_extreme_right(arr,n,i);
    evencount++;
           }
    }
    return evencount;
}

The function test whether arr[i] is even. If yes, to moves the element to extreme right and packs the hole. It also maintains the total number of even values in the array.

Here is the function to move ith element to end of array.

void move_extreme_right(int *arr,int n,int position)
{
   int i;
   int temp = arr[position];
   for(i=position+1;i<n;i++)
      arr[i-1]=arr[i];
   arr[n-1]=temp;
}

Now we get all odd elements in first part and even elements in second half. We will sort them using insertion sort. But sorting of second part needs an offset because it starts from n-evencount.


evencount = splitOddEven(arr,n);    
insertion_sort(arr,n-evencount,0);
insertion_sort(arr,evencount,n-evencount);//write these in a separate function 

void insertion_sort(int *arr,int sz,int offset)
{
   int i,j,k;
   for(i = 1;i<sz;i++)
   {
       int temp = arr[i+offset];
       int j = i-1;
       while(j>=0  && arr[j+offset]>temp)
       {          
          arr[j+1+offset] = arr[j+offset];          
          j--;
       }
       arr[j+1+offset] = temp;
   }           
}


Here is the complete program.

#include<stdio.h>
void insertion_sort(int *arr,int sz,int offset)
{
   int i,j,k;
   for(i = 1;i<sz;i++)
   {
       int temp = arr[i+offset];
       int j = i-1;
       while(j>=0  && arr[j+offset]>temp)
       {          
          arr[j+1+offset] = arr[j+offset];          
          j--;
       }
       arr[j+1+offset] = temp;
   }           
}
void move_extreme_right(int *arr,int n,int position)
{
   int i;
   int temp = arr[position];
   for(i=position+1;i<n;i++)
      arr[i-1]=arr[i];
   arr[n-1]=temp;
}
int splitOddEven(int *arr,int n)
{
    int i;
    int evencount=0;
    for(i=n-1;i>=0;i--)
    {
       if(arr[i]%2==0)
           {
           move_extreme_right(arr,n,i);
    evencount++;
           }
    }
    return evencount;
}
       
void read_array(int *arr,int sz)
{
     int i;
     for(i  =0;i<sz;i++)
     {
       printf("arr[%d]=",i);
       scanf("%d",&arr[i]);
     }
}
void print_array(int *arr,int sz)
{
     int i;
     for(i  =0;i<sz;i++)
     {
       printf("arr[%d]=",i);
       printf("%d   ",arr[i]);
     }
     printf("\n");
}
int main()
{
    int arr[40];
    int evencount;
    int n,midpoint;
    printf("Enter size of array:");
    scanf("%d",&n);
    read_array(arr,n);
    evencount = splitOddEven(arr,n);    
    insertion_sort(arr,n-evencount,0);
    insertion_sort(arr,evencount,n-evencount);     
    print_array(arr,n);
}

You can download the program from here

Comments

Popular posts from this blog

Introduction to AVL tree

AVL tree is a balanced binary search tree where the difference between heights of two sub trees is maximum 1. Why balanced tree A binary tree is good data structure because search operation here is of the order of O(logn). But this is true if the tree is balanced - which means the left and right subtrees are almost equal in height. If not balanced, search operation will take longer.  In worst case, if the tree has only one branch, then search is of the order O(n). Look at this example.  Here all nodes have only right children.  To search a value in this tree, we need upto 7 iterations, which is O(n). So this tree is very very inefficient. One way of making the tree efficient is to, balance the tree and make sure that height of two branches of each node are almost equal. Height of a node Height of a node is the distance between the node and its extreme child.  In the above a diagram, height of 37 is 3 and height of left child of 37 is 0 and right child of 37 is 2. Bal...

Reverse a singly linked list

One of the commonly used interview question is - how do you reverse a linked list? If you talk about a recursive function to print the list in reverse order, you are so wrong. The question is to reverse the nodes of list. Not print the nodes in reverse order. So how do you go about reversing the nodes. You need to take each node and link it to previous node. But a singly linked list does not have previous pointer. So if n1 is current node, n2 = n1->next, you should set     n2->next = NULL But doing this would cut off the list at n2. So the solution is recursion. That is to reverse n nodes  n1,n2,n3... of a list, reverse the sub list from n2,n3,n4.... link n2->next to n1 set n1->next to NULL The last step is necessary because, once we reverse the list, first node must become last node and should be pointing to NULL. But now the difficulty is regarding the head? Where is head and how do we set it? Once we reach end of list  viz n1->next ==NULL, th...

Balanced brackets

Have you observed something? When ever you are writing code using any IDE, if you write mismatched brackets, immediately an error is shown by IDE. So how does IDE  know if an expression is having balanced brackets? For that, the expression must have equal number of opening and closing brackets of matching types and also they must be in correct order. Let us look at some examples (a+b)*c+d*{e+(f*g)}   - balanced (p+q*[r+u )] - unbalanced (p+q+r+s) ) - unbalanced (m+n*[p+q]+{g+h}) - balanced So we do we write a program to check if an expression is having balanced brackets? We do need to make use of stack to store the brackets. The algorithm is as follows Scan a character - ch from the expression If the character is opening bracket, push it to stack If the character is closing bracket pop a character from stack If popped opening bracket and ch are not of same type ( ( and ) or [ and ] ) stop the function and return false Repeat steps 2 and 3 till all characters are scanned. On...