Consider The Following Dynamic Programming Implementation Of #2040

Consider the following dynamic programming implementation of the minimum jumps problem:</p> <pre><code class="language-c"> #include<stdio.h> #include<limits.h> int 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 <= arr[idx] && idx + j < len; j++) { if(jumps[idx + j] + 1 < tmp_min) tmp_min = jumps[idx + j] + 1; } jumps[idx] = tmp_min; } return jumps[0]; } int main() { int arr[] ={1, 1, 1, 1, 1, 1, 1, 1, 1},len = 9; int ans = min_jump(arr,len); printf("%d\n",ans); return 0; }</code></pre> <p>Which of the following "for" loops can be used instead of the inner for loop so that the output doesn't change?

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 - Minimum Number of Jumps - Quiz No.1.

 #include<stdio.h> #include<limits.h> int 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 <= arr[idx] && idx + j < len; j++) { if(jumps[idx + j] + 1 < tmp_min) tmp_min = jumps[idx + j] + 1; } jumps[idx] = tmp_min; } return jumps[0]; } int main() { int arr[] ={1, 1, 1, 1, 1, 1, 1, 1, 1},len = 9; int ans = min_jump(arr,len); printf("%d\n",ans); return 0; }