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Jump Game VII

Medium

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ArraysStringsDynamic Programming

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled:

i + minJump <= j <= min(i + maxJump, s.length - 1), and
s[j] == '0'.

Return true if you can reach index s.length - 1 in s, or false otherwise.

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3
Output: false

Constraints:

2 <= s.length <= 10⁵
s[i] is either '0' or '1'.
s[0] == '0'
1 <= minJump <= maxJump < s.length

Try coding it in LeetCode

Jump Game VII

Medium

Asked by:

17 views

Topics:

ArraysStringsDynamic Programming

i + minJump <= j <= min(i + maxJump, s.length - 1), and
s[j] == '0'.

Return true if you can reach index s.length - 1 in s, or false otherwise.

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3
Output: false

Constraints:

2 <= s.length <= 10⁵
s[i] is either '0' or '1'.
s[0] == '0'
1 <= minJump <= maxJump < s.length

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for `minJump` and `maxJump`? Can they be zero, negative, or larger than the size of the string?
Is the input string `s` guaranteed to contain only '0' and '1' characters? What should I return if it contains other characters?
If it's impossible to reach the end of the string, what should I return (e.g., `false`, `0`, or throw an exception)?
Is the first character of the string `s` always '0'? What if it's '1'?
Can the input string `s` be empty or null? If so, what should I return?

Brute Force Solution

Approach

Let's say we're trying to reach the end of a path by making jumps of certain lengths. The brute force method tries every possible combination of jumps to see if we can get there. We explore each path until we either reach the end or get stuck.

Here's how the algorithm would work step-by-step:

Start at the beginning of the path.
See if we can reach the end of the path with just one jump.
If that doesn't work, try two jumps in a row.
Keep trying more and more jumps until we reach the end, or run out of possible jumps.
Every time we can make a valid jump, remember that path.
If we explore a path that doesn't lead to the end, then we abandon that path and try another one.
Eventually, if there is a way to reach the end, we will find it by trying every possibility.

Code Implementation

def jump_game_vii_brute_force(path, minimum_jump, maximum_jump):
    path_length = len(path)
    # Start at the beginning of the path
    def can_reach_end(current_position):
        if current_position == path_length - 1:
            return True

        for jump_length in range(minimum_jump, maximum_jump + 1):
            next_position = current_position + jump_length

            # Check if the next position is within the bounds of the path
            if 0 <= next_position < path_length:

                # Only continue if the next spot is a valid jump
                if path[next_position] == '0':

                    # If this jump leads to the end, return True
                    if can_reach_end(next_position):
                        return True

        return False

    # The first spot must be zero
    if path[0] == '0':
        return can_reach_end(0)

    # If the first spot is not zero then you are stuck at the start
    else:
        return False

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible jump combination. At each position, we might have multiple valid jump options based on the min and max jump range. In the worst case, we might need to explore almost all possible paths, resulting in exponential growth. Therefore, the time complexity is approximately O(2^n), as each position potentially branches into multiple possibilities, creating a decision tree explored by the recursive or iterative method.

Space Complexity

O(N) – The brute force approach, as described, explores all possible jump combinations which can be visualized as a tree. In the worst-case scenario, the depth of the call stack will be proportional to the length of the input array. Because each call frame occupies memory, the maximum depth of N potentially contributes to O(N) space. The exploration of paths can be viewed as an implicit stack, where the maximum number of stored intermediate paths directly correlates to the length of the input array. Therefore, the space complexity is O(N).

Optimal Solution

Approach

The goal is to find if you can reach the end, starting from the beginning, given specific jump ranges. We can efficiently track how far we can reach so far and incrementally see if we can extend our reach to the final destination.

Here's how the algorithm would work step-by-step:

Start at the beginning.
Keep track of the furthest you can currently jump.
As you move from each position to the next, update how far you can jump to based on the jump range from that position.
If the place you can jump to goes beyond your current furthest jump, update the furthest jump.
If the furthest you can jump to is at or beyond the end, you can reach the end. If you reach a spot where you can no longer extend your jump, you cannot reach the end.

Code Implementation

def can_reach_end(series_of_jumps, minimum_jump, maximum_jump):
    number_of_jumps = len(series_of_jumps)
    reachable = [False] * number_of_jumps
    reachable[0] = True
    farthest_jump = 0
    
    for i in range(number_of_jumps):
        if not reachable[i]:
            return False

        # Only consider cells with '0'
        if series_of_jumps[i] == '1':
            continue

        # Update farthest jump index we can reach
        start_jump = max(i + minimum_jump, farthest_jump + 1)
        end_jump = min(i + maximum_jump, number_of_jumps - 1)

        for j in range(start_jump, end_jump + 1):
            reachable[j] = True

        farthest_jump = max(farthest_jump, i + maximum_jump)
        
        # We have reached the end.
        if farthest_jump >= number_of_jumps - 1:
            return True

    return reachable[number_of_jumps - 1]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string s of length n (where n represents the number of positions). In each iteration, it updates the reachable range based on the current position's jump range. The crucial aspect is that the algorithm maintains and extends the furthest reachable position in a single pass. Therefore, the time complexity is directly proportional to the number of positions, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm primarily tracks the current reachable position and the furthest reachable position. These positions are stored as scalar variables, and their memory footprint does not depend on the input string's length N. No auxiliary data structures like lists, arrays, or hash maps are created or used to store intermediate results. Therefore, the auxiliary space used remains constant, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Null or empty string s	Return false immediately as there are no jumps possible.
start > end	Return false immediately as the jump range is invalid.
String s with length 1, where s[0] is '0'	Return true, as the starting position is also the ending position.
String s with length 1, where s[0] is '1'	Return false, as it's impossible to reach the end.
String s where s[n-1] == '1' (n is the string length)	Return false, as the last index cannot be reached.
start == 0 and end == 0	This scenario requires checking if the second element of the string is '0' to determine whether it's possible to jump to index 1.
String s with all '1's except s[0] which is '0'	Return false, as no jump is possible from index 0.
String s with length approaching the maximum allowed by the language, and a large start and end range.	Ensure the algorithm uses an efficient approach (e.g., BFS with a sliding window) to avoid timeouts.

Null or empty string s

How to Handle:

Return false immediately as there are no jumps possible.

start > end

How to Handle:

Return false immediately as the jump range is invalid.

String s with length 1, where s[0] is '0'

How to Handle:

Return true, as the starting position is also the ending position.

String s with length 1, where s[0] is '1'

How to Handle:

Return false, as it's impossible to reach the end.

String s where s[n-1] == '1' (n is the string length)

How to Handle:

Return false, as the last index cannot be reached.

start == 0 and end == 0

How to Handle:

This scenario requires checking if the second element of the string is '0' to determine whether it's possible to jump to index 1.

String s with all '1's except s[0] which is '0'

How to Handle:

Return false, as no jump is possible from index 0.

String s with length approaching the maximum allowed by the language, and a large start and end range.

How to Handle:

Ensure the algorithm uses an efficient approach (e.g., BFS with a sliding window) to avoid timeouts.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for `minJump` and `maxJump`? Can they be zero, negative, or larger than the size of the string?
Is the input string `s` guaranteed to contain only '0' and '1' characters? What should I return if it contains other characters?
If it's impossible to reach the end of the string, what should I return (e.g., `false`, `0`, or throw an exception)?
Is the first character of the string `s` always '0'? What if it's '1'?
Can the input string `s` be empty or null? If so, what should I return?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the path.
See if we can reach the end of the path with just one jump.
If that doesn't work, try two jumps in a row.
Keep trying more and more jumps until we reach the end, or run out of possible jumps.
Every time we can make a valid jump, remember that path.
If we explore a path that doesn't lead to the end, then we abandon that path and try another one.
Eventually, if there is a way to reach the end, we will find it by trying every possibility.

Code Implementation

def jump_game_vii_brute_force(path, minimum_jump, maximum_jump):
    path_length = len(path)
    # Start at the beginning of the path
    def can_reach_end(current_position):
        if current_position == path_length - 1:
            return True

        for jump_length in range(minimum_jump, maximum_jump + 1):
            next_position = current_position + jump_length

            # Check if the next position is within the bounds of the path
            if 0 <= next_position < path_length:

                # Only continue if the next spot is a valid jump
                if path[next_position] == '0':

                    # If this jump leads to the end, return True
                    if can_reach_end(next_position):
                        return True

        return False

    # The first spot must be zero
    if path[0] == '0':
        return can_reach_end(0)

    # If the first spot is not zero then you are stuck at the start
    else:
        return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning.
Keep track of the furthest you can currently jump.
As you move from each position to the next, update how far you can jump to based on the jump range from that position.
If the place you can jump to goes beyond your current furthest jump, update the furthest jump.
If the furthest you can jump to is at or beyond the end, you can reach the end. If you reach a spot where you can no longer extend your jump, you cannot reach the end.

Code Implementation

def can_reach_end(series_of_jumps, minimum_jump, maximum_jump):
    number_of_jumps = len(series_of_jumps)
    reachable = [False] * number_of_jumps
    reachable[0] = True
    farthest_jump = 0
    
    for i in range(number_of_jumps):
        if not reachable[i]:
            return False

        # Only consider cells with '0'
        if series_of_jumps[i] == '1':
            continue

        # Update farthest jump index we can reach
        start_jump = max(i + minimum_jump, farthest_jump + 1)
        end_jump = min(i + maximum_jump, number_of_jumps - 1)

        for j in range(start_jump, end_jump + 1):
            reachable[j] = True

        farthest_jump = max(farthest_jump, i + maximum_jump)
        
        # We have reached the end.
        if farthest_jump >= number_of_jumps - 1:
            return True

    return reachable[number_of_jumps - 1]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty string s	Return false immediately as there are no jumps possible.
start > end	Return false immediately as the jump range is invalid.
String s with length 1, where s[0] is '0'	Return true, as the starting position is also the ending position.
String s with length 1, where s[0] is '1'	Return false, as it's impossible to reach the end.
String s where s[n-1] == '1' (n is the string length)	Return false, as the last index cannot be reached.
start == 0 and end == 0	This scenario requires checking if the second element of the string is '0' to determine whether it's possible to jump to index 1.
String s with all '1's except s[0] which is '0'	Return false, as no jump is possible from index 0.
String s with length approaching the maximum allowed by the language, and a large start and end range.	Ensure the algorithm uses an efficient approach (e.g., BFS with a sliding window) to avoid timeouts.

Null or empty string s

How to Handle:

Return false immediately as there are no jumps possible.

start > end

How to Handle:

Return false immediately as the jump range is invalid.

String s with length 1, where s[0] is '0'

How to Handle:

Return true, as the starting position is also the ending position.

String s with length 1, where s[0] is '1'

How to Handle:

Return false, as it's impossible to reach the end.

String s where s[n-1] == '1' (n is the string length)

How to Handle:

Return false, as the last index cannot be reached.

start == 0 and end == 0

How to Handle:

This scenario requires checking if the second element of the string is '0' to determine whether it's possible to jump to index 1.

String s with all '1's except s[0] which is '0'

How to Handle:

Return false, as no jump is possible from index 0.

String s with length approaching the maximum allowed by the language, and a large start and end range.

How to Handle:

Ensure the algorithm uses an efficient approach (e.g., BFS with a sliding window) to avoid timeouts.