Data Structure Questions and Answers – Maximum Sum of Continuous Subarray – 2

This set of Data Structure MCQs focuses on “Maximum Sum of Continuous Subarray – 2”.

1. Which line should be inserted in the blank to complete the following dynamic programming implementation of the maximum sub-array sum problem?

#include<stdio.h>
int max_num(int a,int b)
{
      if(a> b)
	 return a;
      return b;
}
int maximum_subarray_sum(int *arr, int len)
{
      int sum[len], idx;
      sum[0] = arr[0];
      for(idx = 1; idx < len; idx++)
	 sum[idx] = _______________________;
      int mx = sum[0];
      for(idx = 0; idx < len; idx++)
	 if(sum[idx] > mx)
	     mx =sum[idx];
	 return mx;
}
int main()
{
      int arr[] = {-2, -5, 6, -2, 3, -1, 0,-5, 6}, len = 9;
      int ans = maximum_subarray_sum(arr, len);
      printf("%d",ans);
      return 0;
}

a) max_num(sum[idx – 1] + arr[idx], arr[idx])
b) sum[idx – 1] + arr[idx].
c) min_num(sum[idx – 1] + arr[idx], arr[idx])
d) arr[idx].
View Answer

Answer: a
Explanation: The array “sum” is used to store the maximum sub-array sum. The appropriate way to do this is by using:
sum[idx] = max_num(sum[idx – 1] + arr[idx], arr[idx]).
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2. What is the time complexity of the following dynamic programming algorithm used to find the maximum sub-array sum?

#include<stdio.h>
int max_num(int a,int b)
{
      if(a> b)
	 return a;
      return b;
}
int maximum_subarray_sum(int *arr, int len)
{
      int sum[len], idx;
      sum[0] = arr[0];
      for(idx = 1; idx < len; idx++)
	 sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
      int mx = sum[0];
      for(idx = 0; idx < len; idx++)
	 if(sum[idx] > mx)
	     mx =sum[idx];
	 return mx;
}
int main()
{
      int arr[] = {-2, -5, 6, -2, 3, -1, 0,-5, 6}, len = 9;
      int ans = maximum_subarray_sum(arr, len);
      printf("%d",ans);
      return 0;
}

a) O(n)
b) O(logn)
c) O(nlogn)
d) O(n2)
View Answer

Answer: a
Explanation: The time complexity of the above dynamic programming algorithm used to solve maximum sub-array sum is O(n).
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3. What is the space complexity of the following dynamic programming algorithm used to find the maximum sub-array sum?

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#include<stdio.h>
int max_num(int a,int b)
{
      if(a> b)
	 return a;
      return b;
}
int maximum_subarray_sum(int *arr, int len)
{
      int sum[len], idx;
      sum[0] = arr[0];
      for(idx = 1; idx < len; idx++)
	 sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
      int mx = sum[0];
      for(idx = 0; idx < len; idx++)
	 if(sum[idx] > mx)
	     mx =sum[idx];
	 return mx;
}
int main()
{
      int arr[] = {-2, -5, 6, -2, 3, -1, 0,-5, 6}, len = 9;
      int ans = maximum_subarray_sum(arr, len);
      printf("%d",ans);
      return 0;
}

a) O(n)
b) O(1)
c) O(n!)
d) O(n2)
View Answer

Answer: a
Explanation: The above dynamic programming algorithm uses space equal to the length of the array to store the sum values. So, the space complexity is O(n).
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4. Consider the following code snippet. Which property is shown by line 4 of the below code snippet?

1. int sum[len], idx;
2. sum[0] = arr[0];
3. for(idx = 1; idx < len; idx++)
4.	  sum[idx] = max(sum[idx - 1] + arr[idx], arr[idx]);
5. int mx = sum[0];
6. for(idx = 0; idx < len; idx++)
7.	 if(sum[idx] > mx)
8.		mx =sum[idx];
9. return mx;

a) Optimal substructure
b) Overlapping subproblems
c) Both overlapping subproblems and optimal substructure
d) Greedy substructure
View Answer

Answer: a
Explanation: The current sum (i.e. sum[idx]) uses the previous sum (i.e. sum[idx – 1]) to get an optimal value. So, line 4 shows the optimal substructure property.

5. Consider the following code snippet:

1. int sum[len], idx;
2. sum[0] = arr[0];
3. for(idx = 1; idx < len; idx++)
4.	  sum[idx] = max(sum[idx - 1] + arr[idx], arr[idx]);
5. int mx = sum[0];
6. for(idx = 0; idx < len; idx++)
7.	 if(sum[idx] > mx)
8.		mx =sum[idx];
9. return mx;

Which method is used by line 4 of the above code snippet?
a) Divide and conquer
b) Recursion
c) Both memoization and divide and conquer
d) Memoization
View Answer

Answer: d
Explanation: The array “sum” is used to store the previously calculated values, so that they aren’t recalculated. So, line 4 uses the memoization technique.

6. Find the maximum sub-array sum for the following array:
{3, 6, 7, 9, 3, 8}
a) 33
b) 36
c) 23
d) 26
View Answer

Answer: b
Explanation: All the elements of the array are positive. So, the maximum sub-array sum is equal to the sum of all the elements, which is 36.

7. What is the output of the following program?

#include<stdio.h>
int max_num(int a,int b)
{
     if(a> b)
	return a;
     return b;
}
int maximum_subarray_sum(int *arr, int len)
{
     int sum[len], idx;
     sum[0] = arr[0];
     for(idx = 1; idx < len; idx++)
	sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
     int mx = sum[0];
     for(idx = 0; idx < len; idx++)
	if(sum[idx] > mx)
	   mx =sum[idx];
     return mx;
}
int main()
{
     int arr[] = {-20, 23, 10, 3, -10, 11, -5},len = 7;
     int ans = maximum_subarray_sum(arr, len);
     printf("%d",ans);
     return 0;
}

a) 27
b) 37
c) 36
d) 26
View Answer

Answer: b
Explanation: The program prints the value of maximum sub-array sum, which is 37.

8. What is the value stored in sum[4] after the following program is executed?

#include<stdio.h>
int max_num(int a,int b)
{
      if(a> b)
	  return a;
      return b;
}
int maximum_subarray_sum(int *arr, int len)
{
      int sum[len], idx;
      sum[0] = arr[0];
      for(idx = 1; idx < len; idx++)
	   sum[idx] = max_num(sum[idx - 1] + arr[idx], arr[idx]);
      int mx = sum[0];
      for(idx = 0; idx < len; idx++)
	  if(sum[idx] > mx)
	      mx =sum[idx];
      return mx;
}
int main()
{
      int arr[] = {-2, 14, 11, -13, 10, -5, 11, -6, 3, -5},len = 10;
      int ans = maximum_subarray_sum(arr, len);
      printf("%d",ans);
      return 0;
}

a) 28
b) 25
c) 22
d) 12
View Answer

Answer: c
Explanation: After the program is executed the value stored in sum[4] is 22.
Note: You are asked to find the value stored in sum[4] and NOT the output of the program.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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