## Interview Question

• 0

Country: United States

Comment hidden because of low score. Click to expand.
1
of 3 vote

As you can see, the *left* half has all the small elements while the *right* half has all the large elements. Thus, the problem is partition the array into two halves of equal sizes, left contains all the small elements ( w/o violating the order ) and right has the large ones.
Thus, for odd sized array the middle element would be the median.
For lack of editing capability, posting as comment.

Comment hidden because of low score. Click to expand.
1
of 3 vote

I cannot think of an in-place algorithm right now. but what we can do is:
1. Find the median in O(n) time using quickselect.
2. Using two pointers (one for the elements smaller than the median and one for the larger) copy the elements over to a buffer array preserving the relative order. Smaller elements will be placed in the buffer starting from index = 0 and larger elements will be placed in the array starting from n/2
We also need to handle both of the cases where the original array has even/odd number of elements.

Comment hidden because of low score. Click to expand.
0
of 0 vote

package carrercup;
import java.util.*;

public class array1 {
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int temp,temp1;
int[] arr=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
int[] newarr=new int[n];
for(int r=0;r<n;r++)
{
newarr[r]=arr[r];
}
for(int j=0;j<n;j++)
{
for(int k=j;k<n;k++)
{
if(newarr[k]<newarr[j])
{
temp=newarr[k];
newarr[k]=newarr[j];
newarr[j]=temp;
}
}
}
temp1=newarr[n/2-1];
int count=0;
for(int t=0;t<n;t++)
{
if(arr[t]<=temp1)
{
temp=arr[t];

for(int u=t;u>count;u--)
{
arr[u]=arr[u-1];

}
System.out.println(temp1);
arr[count]=temp;
count++;

}
}
for(int o=0;o<n;o++)
{
System.out.print(arr[o]);
}
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````void PartitionArray(int* array, int cnt)
{
int* pTmp = new int[cnt];
std::copy(array, array + cnt, pTmp);
std::nth_element(pTmp, pTmp + (cnt-1)/2, pTmp + cnt);
int median = pTmp[(cnt-1)/2];

for (int lIdx = 0, hIdx = (cnt + 1)/2, i = 0; i < cnt; i++)
pTmp[(array[i] <= median) ? lIdx++ : hIdx++] = array[i];

std::copy(pTmp, pTmp + cnt, array);
delete[] pTmp;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Edited to remove extraneous copy

``````void PartitionArray(int* array, int cnt)
{
int* pTmp = new int[cnt];
std::copy(array, array + cnt, pTmp);
std::nth_element(array, array + (cnt-1)/2, array + cnt);
int median = array[(cnt-1)/2];

for (int lIdx = 0, hIdx = (cnt + 1)/2, i = 0; i < cnt; i++)
array[(pTmp[i] <= median) ? lIdx++ : hIdx++] = pTmp[i];

delete[] pTmp;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````// ZoomBA
def left_right( a ){
len = #|a|
odd = !(2 /? len)
left_heap = fold( a , heap( len/2 + 1 ) ) -> { \$.partial  += \$.item  }

#(l,r) = fold( a, [list(),list()] ) -> {
if ( \$.item < left_heap.max ){ \$.partial.0 += \$.item }
if ( \$.item > left_heap.max ){ \$.partial.1 += \$.item }
if ( !odd && \$.item == left_heap.max ) { \$.partial.1 += \$.item }
\$.partial
}
println ( l +  (odd ? left_heap.max : []) + r )
}

a = [ 6, 4, 5, 0, 2, 1, 11 , -1 ]
left_right(a)
a = [ 6, 4, 5, 0, 2, 1, 11 , -1, 9  ]
left_right(a)``````

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Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

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