Interview Question: Longest Prefix That Contains An Equal Number Of Two Integers - c

Question

I attempted to solve the following question in an online assessment for a technical interview, and failed. I have been thinking about the problem for a while now, and can't seem to find an answer that satisfies me:
You are looking for the longest leading fragment (prefix) of an array A in which there is an equal number of occurrences of X, and Y, where X and Y are integers.
For example, where X=7 and Y=42, the longest prefix where A=[6, 42, 11, 7, 1, 42] would be 4, because A[0]-A[4] contain the same number of X and Y.
Another example, X=6 and Y=13.
A=[13,13,1,6]. The function should return -1, because there is no prefix.
X=100, Y=63, and A=[100,63,1,6,2,13] should return 5.
My attempt at an answer in C:
int solution(int X, int Y, int A[], int N){
int result=-1;
int nX=0; //number of occurrences of X
int nY=0; //number of occurrences of Y
int i;
for(i=0;i<N;i++){//loop through the input array
if(A[i]==X)//occurrence of X
nX += 1;
/*
EDGE CASE BELOW: this should have been an if
statement, because X and Y could be the same
number. I know I missed this in the assessment,
but believe there is another issue, because I
failed almost all the test cases generated by the
assessment.
*/
else if(A[i]==Y)//occurrence of Y
nY += 1;
if((nX!=0)&& (nX==nY))//same number of X and Y
//and there is at least one occurence of each
result=i;//longest prefix is the index
}
return result;
}
Unfortunately, I was not able to generate a test case myself that failed, and the failure test cases are hidden on the assessment. Thus I can't provide much information that would be helpful.
I do know, that every time I failed, my program returned a -1 instead of the correct answer.
If any of you can see something wrong just through thinking through it, I would love to see what I am missing. Thanks!

Answer 1

If you've accurately described the requirements, then they don't specify an equal positive number of occurrences of X and Y. Zero is valid.
So this:
if((nX!=0)&& (nX==nY))//same number of X and Y
//and there is at least one occurence of each
result=i;//longest prefix is the index
}
should be this:
if(nX==nY)//same number of X and Y
result=i;//longest prefix is the index
}
without the check for nX!=0. So if X doesn't appear in the array and/or Y doesn't appear in the array, your code returns −1 unnecessarily.
Additionally, the requirements don't seem to guarantee that X and Y are distinct; if they're not, then your code returns −1, but according to a literal reading of the requirements, the answer would be N−1.

Answer 2

After seeing your requirements and answers there is another change required to your code #Joshua.
The answer from #ruakh is also right but need one more change to handle other test cases.
You have to replace the "else if" condition to "if" condition for the occurrence of Y as shown in below code:
for(i=0;i<N;i++){
if(A[i]==X)
nX += 1;
if(A[i]==Y) //occurrence of Y
nY += 1;
Because, if "X" and "Y" have the same values and only "nX" will be counted, and "nY" is ignored with the "else if" condition. To tackle this situation you have to replace the "else if" condition with the "if" condition.
Sample: X=42, Y=42, and A=[42,63,42,6,2,13]
then with my condition, the above sample is handled perfectly.
I hope my answer solved or added more accuracy to your answer.

Answer 3

For starters never do any assignments in an interview. Interview is not an exam. It is a conversation of two equal participants. Ignore all firms that try to manipulate you and your time such a way.
Secondly, this function declaration
int solution(int X, int Y, int A[], int N);
is a declaration of a beginner.
First of all do not use upper case letters to name parameters.
Secondly the size of the array shall have the type size_t as and the return type of the function.
Thirdly the array should be the first parameter of the function and shall have the qualifier const.
Fourthly, declare variables in the smallest scope where they are used.
The function can be declared as it is shown in the demonstrative program. The function returns 0 if there is no such a prefix. You can change the return value either to the size of the array if there is no the prefix ot to ( size_t ) -1 as you like.
#include <stdio.h>
size_t largest_balanced_seq( const int a[], size_t n, int x, int y )
{
size_t last_index = 0;
for ( size_t i = 0, x_count = 0, y_count = 0; i < n; i++ )
{
x_count += a[i] == x;
y_count += a[i] == y;
if ( x_count != 0 && x_count == y_count )
{
last_index = i;
}
}
return last_index;
}
int main(void)
{
int a[] = { 6, 42, 11, 7, 1, 42 };
printf( "%zu\n", largest_balanced_seq( a, sizeof( a ) / sizeof( *a ), 7, 42 ) );
int b[] = { 100, 63, 1, 6, 2, 13 };
printf( "%zu\n", largest_balanced_seq( b, sizeof( b ) / sizeof( *b ), 100, 63 ) );
return 0;
}
The program output is
4
5
Take into account that it is much better when the function returns the length of the sub-sequence, that is when it specifiers a range like [0, N). For example such an approach is used throughout C++. So who gave you this assignment is not a very high qualified.:)

Answer 4

int solution(int X, int Y, int A[], int N){
int result=-1;
int nX=0; //number of occurrences of X
int nY=0; //number of occurrences of Y
int i;
for(i=0;i<N;i++){//loop through the input array
if(A[i]==X)//occurrence of X
nX += 1;
/*
EDGE CASE BELOW: this should have been an if
statement, because X and Y could be the same
number. I know I missed this in the assessment,
but believe there is another issue, because I
failed almost all the test cases generated by the
assessment.
*/
else if(A[i]==Y)//occurrence of Y
nY += 1;
if((nX!=0)&& (nX==nY))//same number of X and Y
//and there is at least one occurence of each
result=i;//longest prefix is the index
if (X==Y)
result =i;
}---this two lines of code is needed for the code correction
return result;
}

Answer 5

def solution(X, Y, A):
N = len(A)
result = -1
nX = 0
nY = 0
for i in range(N):
if A[i] == X:
nX += 1
if A[i] == Y:
nY += 1
if nX == nY:
result = i
if (X == Y and nX != 0):
break
return result
print( solution(7, 42, [6, 42, 11, 7, 1, 42]) )
print( solution(6, 13, [13, 13, 1, 6]) )
print( solution(100, 63, [100, 63, 1, 6, 2, 13]) )
print( solution(1, 1, [1]) )
print( solution(1, 1, [1, 1, 1]) )
print( solution(1, 1, [1, 2, 1]) )
print( solution(1, 1, [2, 2, 1]) )
print( solution(1, 1, [2, 1, 2]) )

Answer 6

def solution(X, Y, A):
N = len(A)
result = -1
nX = 0
nY = 0
for i in range(N):
if X==Y:
if A[i]==X:
nX += 1
if nX%2==0:
result=i
else:
if A[i] == X:
nX += 1
elif A[i] == Y:
nY += 1
if nX == nY:
result = i
return result
print( solution(7, 42, [6, 42, 11, 7, 1, 42]))
print( solution(6, 13, [13, 13, 1, 6]) )
print( solution(100, 63, [100, 63, 1, 6, 2, 13]) )
print( solution(1, 1, [1]) )
print( solution(1, 1, [1, 1, 1]) )
utput:
4
-1
5
-1
1