Amcat Automata Fix Questions with Answers 2018 (Latest)

Amcat Automata Fix is used to test the candidate’s ability to debug an erroneous code and fix it. Companies like Cognizant have recently included Amcat Automata Fix section in their online test. To know about Cognizant Automata Fix pattern and questions, click here.

Amcat Automata Fix Questions Pattern

You need to select the language in which you want to attempt this section. You can choose between C, C++, and Java. The are three types of Amcat Automata Fix questions asked in this section.

amcat automata fix questions

  • Logical: You need to fix all the logical errors in the given code. To attempt these logical type of question, you need to have knowledge of concepts such as looping, conditions etc.
  • Compilation: You need to correct the syntax of the given code without modifying its logic. The logic along with test cases will be given in the question itself. 
  • Code Reuse: The code reuse type of questions are difficult when compared to the logical or compilation type of questions. You need to complete the given code by reusing the existing functions. There is a tab called “helper code tab” on the test screen. This can be used to find out the details of the function/classes provided for reuse.

Amcat Automata Fix Questions – Important instructions

  • You can use the compile and run option any number of times to check the output of your code.
  • You can use printf to debug your code.
  • The submitted code should be syntactically/logically correct and should pass all the test cases.
  • Donot write the main() function as it is not required.
  • You only need to correct the given code. You don’t have to modify the approach or incorporate any additional library methods.

Amcat Automata Fix Questions with Answers

1) Find the syntax error in the below code without modifying the logic.

#include 
int main() 
{
   float x = 1.1;
   switch (x)
   {
       case 1: printf("Choice is 1");
                 break;
       default: printf("Invalid choice");
                break; 
   }
   return 0;
}

Answer:

#include 
int main() 
{
   int x = 1;
   switch (x)
   {
       case 1: printf("Choice is 1");
                 break;
       default: printf("Invalid choice");
                break; 
   }
   return 0;
}

The expression used in the switch must be an integral type (int, char, and enum). Any other type of expression is not allowed.

2) Find the logical error in the below code.

void main () {
  int i, j, n = 5;
  for(i=1; i<n; i++)
  {
for(j=i;j<n;j++);
{
        printf("%d", i);
}
printf("\n");
  }
}

Solution:

void main () {
  int i, j, n = 5;
  for(i=1; i<n; i++)
  {
              for(j=i;j<n;j++)
{
printf("%d", i);
   }
    printf("\n");
  }
}

we use a semicolon in C statement to tell the compiler where’s the end of our statement. Second for loop executes one time.

3) Complete the below code by reusing the existing function.

Find the index of equilibrium element in the given array. In an array equilibrium element is the one where the sum of all the elements to the left side is equal to the sum of all the elements in the right side.

Return Value:

1) Return -1 if no equilibrium element is found

2) In case there is more than one equilibrium element, return the element with least index value.

You are required to complete the given code by reusing the existing function. You can click on Compile & run anytime to check the compilation/execution status of the program you can use printf to debug your code. The submitted code should be logically/syntactically correct and pass all the test cases. 

Code approach For the question:

You will need to correct the given implementation. 
We do not expect you to modify the approach or incorporate any additional library methods. 

Test Case:

a[] = {1,2,3,4,3,3}. 4 is the equilibrium element since its left side sum (1+2+3) is equal to its right side sum (3+3)

#include <stdio.h>
// Return the sum of elements from index 0 to (idx - 1)
int left_side_sum(int a[], int n, int idx)
{
    int sum = 0, i;
    for(i = 0; i < idx; i++)
    {
        sum += a[i];
    }
    return sum;
}

// Return the sum of elements from index (idx + 1) to (n - 1)
int right_side_sum(int a[], int n, int idx)
{
    int sum = 0, i;
    for(i = idx + 1; i < n; i++)
    {
        sum += a[i];
    }

    return sum;
}

// returns -1 if no equilibrium index found
int findEquilibriumIndex(int a[], int n)
{
    // Type your code here
}

int main() {
              //code
   int a[10], n, i;

   // get the elements count
   scanf("%d", &n);

   // get the array elements
   for(i=0; i<n; i++)
   {
       scanf("%d", &a[i]);
   }

   int equiIndex = findEquilibriumIndex(a, n);
   if(equiIndex != -1) {
       printf("%d", a[equiIndex]);
   }
   return 0;

Solution:

// Return the sum of elements from index 0 to (idx - 1)
int left_side_sum(int a[], int n, int idx)
{
    int sum = 0, i;
    for(i = 0; i < idx; i++)
    {
        sum += a[i];
    }

    return sum;
}

// Return the sum of elements from index (idx + 1) to (n - 1)
int right_side_sum(int a[], int n, int idx)
{
    int sum = 0, i;
    for(i = idx + 1; i < n; i++)
    {
        sum += a[i];
    }

    return sum;
}

// returns -1 if no equilibrium index found
int findEquilibriumIndex(int a[], int n)
{
    // Type your code here
    int i;
    for(i = 0; i < n; i++)
    {
        if(left_side_sum(a, n, i) == right_side_sum(a, n, i))
        {
            return i;
        }
    }
   
    return -1;
}

int main() {
              //code
   int a[10], n, i;
   // get the elements count
   scanf("%d", &n);
   // get the array elements
   for(i=0; i<n; i++)
   {
       scanf("%d", &a[i]);
   }

   int equiIndex = findEquilibriumIndex(a, n);
   if(equiIndex != -1) {
       printf("%d", a[equiIndex]);
   }

   return 0;
}

For more such Amcat automata fix questions asked in recent drives, check here.

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