Solution For Jump Game Vi

Problem Statement:

You are given a 0-indexed integer array nums and an integer k.

You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n – 1, i + k)] inclusive.

You want to reach the last index (index n – 1). Your score is the sum of all nums[i] for each index i you visited in the array.

Return the maximum score you can get.

Solution:

Approach:

This problem can be solved using dynamic programming. We can maintain an array dp where dp[i] represents the maximum score we can get if we reach index i. To calculate dp[i], we need to consider all the indices from i-k to i-1 that can reach i, and choose the one that gives the maximum score.

We can use a deque to maintain the indices that can reach i. We start by adding index 0 to the deque. We then iterate from index 1 to index n-1, and for each index i, we check the indices in the deque from front to back, and remove the indices that can’t reach i. We then add i to the back of the deque. We then calculate dp[i] using the front of the deque. We then remove the indices from the back of the deque that can’t reach i-k. Finally, we return dp[n-1].

Algorithm:

1. Initialize a deque dq.
2. Initialize an array dp of size n with all elements as 0.
3. Add index 0 to dq.
4. For i = 1 to n-1, do:
   a. While dq is not empty and dq front is less than i-k, remove the front element from dq.
   b. Calculate dp[i] as dp[dq front] + nums[i].
   c. While dq is not empty and dp[dq back] <= dp[i], remove the back element from dq.
   d. Add i to dq.
5. Return dp[n-1].

Code:

Here’s the Python implementation of the above algorithm:

“`
from collections import deque

class Solution:
def maxResult(self, nums: List[int], k: int) -> int:
n = len(nums)
dp = [0] * n
dp[0] = nums[0] dq = deque([0])

    for i in range(1, n):
        while dq and dq[0] < i - k:
            dq.popleft()
        dp[i] = dp[dq[0]] + nums[i]
        while dq and dp[dq[-1]] <= dp[i]:
            dq.pop()
        dq.append(i)

    return dp[n-1]

“`

Complexity Analysis:

The time complexity of the above algorithm is O(n) because we process each index only once. The space complexity is O(n) because we use an array dp of size n and a deque that can have at most k elements.

Step by Step Implementation For Jump Game Vi

class Solution {
    public int maxJumps(int[] arr, int d) {
        int n = arr.length;
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        for(int i = 0; i < n; i++) {
            for(int j = i + 1; j <= Math.min(i + d, n - 1); j++) {
                if(arr[j] >= arr[i]) break;
                dp[j] = Math.max(dp[j], dp[i] + 1);
            }
            for(int j = i - 1; j >= Math.max(0, i - d); j--) {
                if(arr[j] >= arr[i]) break;
                dp[j] = Math.max(dp[j], dp[i] + 1);
            }
        }
        int res = 0;
        for(int i = 0; i < n; i++) {
            res = Math.max(res, dp[i]);
        }
        return res;
    }
}

class Solution:
    def maxJumps(self, arr: List[int], d: int) -> int:
        
        # create a list to keep track of the maximum number of jumps 
        # that can be made from each index in the array
        jumps = [0] * len(arr)
        
        # sort the array in reverse order so that we can start
        # from the back of the array and work our way forward
        sorted_arr = sorted(arr, reverse=True)
        
        # for each element in the array, starting from the back
        for i in range(len(sorted_arr)):
            # get the current element
            curr = sorted_arr[i]
            
            # keep track of the maximum number of jumps that can be made
            # from the current element
            max_jumps = 1
            
            # starting from the current element, check the next d elements
            # to see if any jumps can be made
            for j in range(i+1, min(i+d+1, len(sorted_arr))):
                # get the next element
                next = sorted_arr[j]
                
                # if the next element is greater than the current element,
                # then a jump can be made
                if next > curr:
                    # update the maximum number of jumps that can be made
                    # from the current element
                    max_jumps = max(max_jumps, jumps[j]+1)
                    
            # update the list with the maximum number of jumps that can be made
            # from the current element
            jumps[i] = max_jumps
            
        # return the maximum number of jumps that can be made from the first element
        return jumps[0]

var jumpGameVI = function(arr, target) {
    // your code goes here
};

class Solution {
public:
    int maxJumps(vector& arr, int d) {
        int n = arr.size();
        vector dp(n, 1);
        for(int i = 0; i < n; ++i) {
            for(int j = i + 1; j <= min(i + d, n - 1) && arr[j] < arr[i]; ++j) {
                dp[j] = max(dp[j], dp[i] + 1);
            }
            for(int j = i - 1; j >= max(0, i - d) && arr[j] < arr[i]; --j) {
                dp[j] = max(dp[j], dp[i] + 1);
            }
        }
        return *max_element(dp.begin(), dp.end());
    }
};

//This is a C# solution for the leetcode problem "Jump Game VI"

//Given an array of integers arr and an integer target.

//You have to jump to the target index target with the fewest jumps.

//Return the number of jumps needed to reach the target.

public int JumpGameVI(int[] arr, int target)
{
    //edge case:
    if (arr == null || arr.Length == 0)
    {
        return -1; //error
    }
    
    //initialize variables:
    int n = arr.Length;
    int[] dp = new int[n]; //dp[i] = minimum number of jumps needed to reach index i
    dp[0] = 0; //base case
    
    //loop through the array:
    for (int i = 1; i < n; i++)
    {
        //initialize min jumps to reach index i as infinity:
        dp[i] = Int32.MaxValue; 
        
        //loop through all previous indices:
        for (int j = 0; j < i; j++)
        {
            //if we can reach index i from index j:
            if (arr[j] >= i - j)
            {
                //update dp[i] with the minimum number of jumps needed:
                dp[i] = Math.Min(dp[i], dp[j] + 1);
            }
        }
    }
    
    //return the minimum number of jumps needed to reach the target index:
    return dp[target];
}