Questions and answers for exam Preparation

MCQs

Total Questions : 22 | Page 3 of 3 pages

Question 21. Consider the following dynamic programming implementation of the minimum jumps problem:#include#includeint min_jump(int *arr, int len){ int j, idx, jumps[len]; jumps[len - 1] = 0; for(idx = len - 2; idx >= 0; idx--) { int tmp_min = INT_MAX; for(j = 1; j

for(j = 1; j < arr[idx] + len; j++)
for(j = 0; j < arr[idx] "“ len; j++)
for(j = idx + 1; j < len && j
None of the mentioned

Discuss Question

Answer: Option D. -> None of the mentioned

None of the above mentioned "for loops can be used instead of the inner for loop. Note, for(j = idx + 1; j < len && j

Question 22. Consider the following dynamic programming implementation:#include#includeint max(int a, int b){ if(a > b) return a; return b;}int min_ins(char *s){ int len = strlen(s), i, j; int arr[len + 1][len + 1]; char rev[len + 1]; strcpy(rev, s); strrev(rev); for(i = 0;i

arr[len][len].
len + arr[len][len].
len
len "“ arr[len][len].

Discuss Question

Answer: Option C. -> len

arr[len][len] contains the length of the longest palindromic subsequence. So, len “ arr[len][len] gives the minimum number of insertions required to form a palindrome.