Consider The Following Dynamic Programming Implementation Of #2073

Consider the following dynamic programming implementation of the longest common subsequence problem:</p> <pre><code class="language-c"> #include<stdio.h> #include<string.h> int max_num(int a, int b) { if(a > b) return a; return b; } int lcs(char *str1, char *str2) { int i,j,len1,len2; len1 = strlen(str1); len2 = strlen(str2); int arr[len1 + 1][len2 + 1]; for(i = 0; i <= len1; i++) arr[i][0] = 0; for(i = 0; i <= len2; i++) arr[0][i] = 0; for(i = 1; i <= len1; i++) { for(j = 1; j <= len2; j++) { if(str1[i-1] == str2[j - 1]) ______________; else arr[i][j] = max_num(arr[i - 1][j], arr[i][j - 1]); } } return arr[len1][len2]; } int main() { char str1[] = " abcedfg", str2[] = "bcdfh"; int ans = lcs(str1,str2); printf("%d",ans); return 0; }</code></pre> <p>Which of the following lines completes the above code?

This multiple choice question (MCQ) is related to the book/course gs gs122 Data Communication and Computer Network. It can also be found in gs gs122 Dynamic Programming - Longest Common Subsequence - Quiz No.1.

 #include<stdio.h> #include<string.h> int max_num(int a, int b) { if(a > b) return a; return b; } int lcs(char *str1, char *str2) { int i,j,len1,len2; len1 = strlen(str1); len2 = strlen(str2); int arr[len1 + 1][len2 + 1]; for(i = 0; i <= len1; i++) arr[i][0] = 0; for(i = 0; i <= len2; i++) arr[0][i] = 0; for(i = 1; i <= len1; i++) { for(j = 1; j <= len2; j++) { if(str1[i-1] == str2[j - 1]) ______________; else arr[i][j] = max_num(arr[i - 1][j], arr[i][j - 1]); } } return arr[len1][len2]; } int main() { char str1[] = " abcedfg", str2[] = "bcdfh"; int ans = lcs(str1,str2); printf("%d",ans); return 0; }