Jump Game Viii - Leetcode Solution

Problem Statement:

You are given an integer array nums with length n. You are initially positioned at the start index of the array. At each step, you can move at most k steps to the right in the array.

Return the minimum number of steps required to reach the last index of the array.

Notice that you can not stay in the same index for more than one turn.

Example 1: Input: nums = [10,-5,-2,4,0,3], k = 3, start = 0 Output: 3 Explanation:

• Move from index 0 to index 4. nums[4] = 0
• Move from index 4 to index 5. nums[5] = 3

Example 2: Input: nums = [2,-1,1,1,2], k = 1, start = 0 Output: 4 Explanation:

• Move from index 0 to index 2. nums[2] = 1
• Move from index 2 to index 3. nums[3] = 1
• Move from index 3 to index 4. nums[4] = 2
• Move from index 4 to index 5. nums[5] = end of array

Approach:

The problem can be solved using a modified version of Breadth First Search (BFS) approach. The idea is to explore all the possible paths that can be taken from the given starting index and keep track of the minimum steps required to reach the end index.

For each index i, we can explore all the possible indices that can be reached in at most k steps to the right. We can store these indices in a deque and keep adding them to a set. We also keep track of the smallest index we have visited so far, as we cannot move to the same index twice. We repeat this process until we reach the last index of the array.

Algorithm:

1. Initialize a queue with the starting index and set the minimum steps required to reach the starting index as 0.
2. Initialize a set to keep track of the indices that have been visited so far and add the starting index to the set.
3. Initialize a variable to keep track of the smallest index we have visited so far. Set its initial value as the starting index.
4. Repeat the following steps until the queue is empty: a. Dequeue an index from the queue and set the current steps required to reach this index as its value in the queue. b. Explore all the possible indices that can be reached from this index in at most k steps to the right. c. For each index, check if it has already been visited or if it is smaller than the smallest index visited so far. If not, enqueue it to the queue with the current steps required to reach it as its value and add it to the set of visited indices. d. Update the smallest index visited so far. e. If the last index of the array has been visited, return the steps required to reach it.
5. If the last index of the array has not been visited, return -1.

Code:

Here's the Python code for the above algorithm:

from collections import deque

def minSteps(nums, k, start): n = len(nums) queue = deque([(start, 0)]) visited = {start} smallest = start while queue: idx, steps = queue.popleft() for i in range(max(smallest, idx-k), min(n, idx+k+1)): if i in visited or i <= smallest: continue if i == n-1: return steps+1 visited.add(i) queue.append((i, steps+1)) smallest = idx+1 return -1

Complexity Analysis:

Time Complexity: O(nk), where n is the length of the given array and k is the maximum number of steps we can take to the right from each index. In the worst case, we may need to explore all possible paths to reach the last index of the array. Space Complexity: O(n), for the visited set and the queue. In the worst case, we may need to store all the indices in the queue.

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