Jump game in Python

Questions 1 to 5 of 7

1. 1. In jump game I, what does the variable farthest hold?

   The index that holds the largest valueThe number of jumps used so farThe last index of the arrayThe largest index you can reach so far
2. 2. What does this code print? ``python def can_reach_end(nums): last = len(nums) - 1 farthest = 0 for i in range(last + 1): if i > farthest: return False farthest = max(farthest, i + nums[i]) if farthest >= last: return True return True print(can_reach_end([1, 0, 2, 0, 0])) print(can_reach_end([3, 0, 0, 1])) ``

   False then TrueTrue then TrueTrue then FalseFalse then False
3. 3. What does this code print? ``python def min_jumps_off(nums): last = len(nums) - 1 jumps = 0 cur_end = 0 farthest = 0 for i in range(last + 1): farthest = max(farthest, i + nums[i]) if i == cur_end: jumps += 1 cur_end = farthest return jumps print(min_jumps_off([1, 1, 1, 1])) ``

   3524
4. 4. In jump game II, the sweep runs from index 0 to last - 1, not to last. Why?

   The last index always holds 0, so the loop has nothing to read thereThe last index is the destination and no jump starts from it, so a loop that reaches it would count one jump too manyPython raises an IndexError when the loop reads the last index, so it must stop earlyThe function returns early at index 1, so later indexes are never read
5. 5. True or False: in jump game I, the test if i >= farthest: return False gives the right answer, and i > farthest is only a style choice.

   False, because farthest starts at 0, so i >= farthest returns False at index 0 for every array, even when the end is reachableTrue, because a reachable index is always strictly below farthest, so >= catches the first unreachable index