Maximum Number of Jumps to Reach the Last Index Interview Question for Google

## Maximum Number of Jumps

This problem asks us to find the maximum number of jumps we can make from index 0 to index `n-1` in a given array `nums`, subject to a constraint on the difference between the values at the jump indices. If it's impossible to reach the end, we should return -1.

### 1. Brute Force Solution

A straightforward approach is to use recursion to explore all possible jump sequences. Starting from index 0, we can jump to any subsequent index `j` if the condition `-target <= nums[j] - nums[i] <= target` is met. We recursively call the function for each valid jump and keep track of the maximum number of jumps.

```python
def max_jumps_recursive(nums, target, current_index):
    n = len(nums)
    if current_index == n - 1:
        return 0  # Reached the end

    max_jumps = -1  # Initialize with -1 (no path)

    for next_index in range(current_index + 1, n):
        if -target <= nums[next_index] - nums[current_index] <= target:
            jumps = max_jumps_recursive(nums, target, next_index)
            if jumps != -1:
                max_jumps = max(max_jumps, 1 + jumps)

    return max_jumps


def maxJumps_brute_force(nums, target):
    return max_jumps_recursive(nums, target, 0)

# Example usage:
nums = [1, 3, 6, 4, 1, 2]
target = 2
result = maxJumps_brute_force(nums, target)
print(f"Maximum jumps (brute force): {result}") # Output: 3

This brute-force approach has exponential time complexity because it explores all possible jump combinations. This will lead to Time Limit Exceeded (TLE) for larger inputs.

2. Optimal Solution: Dynamic Programming

A more efficient solution is to use dynamic programming. We can create a dp array where dp[i] stores the maximum number of jumps to reach index i. We initialize dp[0] to 0 because we start at index 0. For all other indices, we initialize dp[i] to -1, indicating that they are initially unreachable.

We iterate through the array from left to right. For each index i, if dp[i] is not -1 (meaning we can reach index i), we iterate through all subsequent indices j and check if we can jump from i to j. If we can, we update dp[j] with max(dp[j], dp[i] + 1). Finally, we return dp[n-1].

def maxJumps(nums, target):
    n = len(nums)
    dp = [-1] * n
    dp[0] = 0  # Base case: 0 jumps to reach index 0

    for i in range(n):
        if dp[i] != -1:
            for j in range(i + 1, n):
                if -target <= nums[j] - nums[i] <= target:
                    dp[j] = max(dp[j], dp[i] + 1)

    return dp[n - 1]


# Example usage:
nums = [1, 3, 6, 4, 1, 2]
target = 2
result = maxJumps(nums, target)
print(f"Maximum jumps (DP): {result}") # Output: 3

nums = [1, 3, 6, 4, 1, 2]
target = 3
result = maxJumps(nums, target)
print(f"Maximum jumps (DP): {result}") # Output: 5

nums = [1, 3, 6, 4, 1, 2]
target = 0
result = maxJumps(nums, target)
print(f"Maximum jumps (DP): {result}") # Output: -1

3. Big(O) Run-time Analysis

The dynamic programming solution has a time complexity of O(n^2), where n is the length of the nums array. This is because we have a nested loop: the outer loop iterates through each index i from 0 to n-1, and the inner loop iterates through all subsequent indices j from i+1 to n-1. In the worst case, the inner loop runs for each element in the outer loop, leading to a quadratic time complexity.

4. Big(O) Space Usage Analysis

The space complexity of the dynamic programming solution is O(n), where n is the length of the nums array. This is because we use a dp array of size n to store the maximum number of jumps to reach each index. The dp array is the dominant factor in terms of space usage.

5. Edge Cases

Empty Array: If the input array nums is empty or has only one element, there is no way to make any jumps to reach the end, so we should return -1.
Target is Zero: If the target is 0, it means nums[j] must be equal to nums[i] for a jump to be valid. If no such jumps are possible, we will return -1.
No Path to the End: It's possible that there's no sequence of jumps that can reach the end of the array. In this case, the DP solution correctly returns -1, as the last element of the dp array will remain -1.
Negative Numbers: The problem statement specifies that array elements can be negative. The solution correctly handles negative numbers due to the use of -target <= nums[j] - nums[i] <= target condition.
Large Target: If the target is very large relative to the differences in numbers in the nums array, then there are many possible paths and the solution should be able to handle that without errors.