**DOWNLOAD FREE PDF** **<<CLICK HERE>>**

## Data Structure Questions and Answers-Maximum Sum of Continuous Subarray

Congratulations - you have completed *Data Structure Questions and Answers-Maximum Sum of Continuous Subarray*.

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Given a one-dimensional array of integers, you have to find a sub-array with maximum sum. This is the maximum sub-array sum problem. Which of these methods can be used to solve the problem?

Dynamic programming | |

Two for loops (naive method) | |

Divide and conquer | |

All of the mentioned |

Question 1 Explanation:

All of the methods mentioned can be used to solve the maximum sub-array sum problem.

Question 2 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Find the maximum sub-array sum for the given elements.

{2, -1, 3, -4, 1, -2, -1, 5, -4}

3 | |

5 | |

8 | |

6 |

Question 2 Explanation:

The maximum sub-array sum is 5.

Question 3 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Find the maximum sub-array sum for the given elements.

{-2, -1, -3, -4, -1, -2, -1, -5, -4}

-3 | |

5 | |

3 | |

-1 |

Question 3 Explanation:

All the elements are negative. So, the maximum sub-array sum will be equal to the largest element. The largest element is -1 and therefore, the maximum sub-array sum is -1.

Question 4 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Consider the following naive method to find the maximum sub-array sum:

#include<stdio.h> int main() { int arr[1000]={2, -1, 3, -4, 1, -2, -1, 5, -4}, len=9; int cur....max, tmp....max, strt....idx, sub....arr....idx; cur....max = arr[0]; for(strt....idx = 0; strt....idx < len; strt....idx++) { tmp....max=0; for(sub....arr....idx = strt....idx; sub....arr....idx < len; sub....arr....idx++) { tmp....max +=arr[sub....arr....idx]; if(tmp....max > cur....max) .....; } } printf("%d", cur....max); return 0; }

Which line should be inserted to complete the above code?

tmp....max = cur....max | |

break | |

continue | |

cur....max = tmp....max |

Question 4 Explanation:

If the tmp....max element is greater than the cur....max element, we make cur....max equal to tmp....ma, x i.e. cur....max = tmp....max.

Question 5 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

What is the time complexity of the naive method used to find the maximum sub-array sum in an array containing n elements?

O(n ^{2}) | |

O(n) | |

O(n ^{3}) | |

O(1) |

Question 5 Explanation:

The naive method uses two for loops. The outer loop runs from 0 to n,

i.e. i = 0 : n.

The inner loop runs from i to n, i.e. j = i : n.

Therefore, the inner loop runs:

n times when the outer loop runs the first time.

(n-1) times when the outer loop runs the second time.

:

:

:

2 times when the outer loop runs (n-1)th time.

1 time when the outer loop runs nth time.

Therefore, time complexity = n + (n-1) + (n-2) + ..... + 2 + 1 = n * (n + 1) /2 = O(n^{2}).

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.