Minimum Number of Jumps

The problem tackled by the min_number_of_jumps function is finding the minimum number of jumps needed to reach the end of an array, where each element in the array represents the maximum number of steps that can be taken from that position.

Problem Case

Given an array representing the maximum number of steps that can be taken from each position, the task is to find the minimum number of jumps needed to reach the end of the array.

Thought Process for Solving

Initialization: Initialize an array of jumps to store the minimum number needed to reach each position in the input array. Set jumps[0] to 0 since no jumps are required to get the starting position.

Iterative Approach: Iterate through the array starting from index 1. For each position i, iterate backwards from index 0 to i - 1.

Updating Jumps: For each position i, check if it's possible to reach position i from position j. If reachable, update jumps[i] to be the minimum of its current value and jumps[j] + 1, indicating the number of jumps needed to reach j plus one additional jump to reach i.

Return Result: Return the value of jumps[-1], which represents the minimum number of jumps needed to reach the last index of the array.

By following this thought process, the function efficiently computes the minimum number of jumps required to reach the end of the array, providing a solution to the problem.

Solution: Algorithm Approach 1

# O(n ^ 2) time | O(n) space
def minNumberOfJumps(array):
    """
    Finds the minimum number of jumps needed to reach the last index of the given array.

    Args:
        array (List[int]): The input array representing the maximum number of steps
                           that can be taken from each position.

    Returns:
        int: The minimum number of jumps needed to reach the last index.


    """
    jumps = [float("inf") for x in array]  # Initialize jumps array with infinite jumps
    jumps[0] = 0  # No jumps needed to reach the starting position
    for i in range(1, len(array)):  # Iterate through array starting from index 1
        for j in range(0, i):  # Iterate backwards from index 0 to i - 1
            if array[j] + j >= i:  # Check if it's possible to reach position i from j
                # Update jumps[i] to be the minimum of its current value and jumps[j] + 1
                jumps[i] = min(jumps[j] + 1, jumps[i])
    return jumps[-1]  # Return the minimum number of jumps needed to reach the last index

Complexity Analysis

Time Complexity

O(n^2), where 'n' is the length of the input array. That is because, for each element in the array, we potentially iterate
          through all previous elements to update the jumps array.

Space Complexity

O(n) for the jumps array, where 'n' is the length of the input array. We also use constant additional space for variables
          and loop indices.

Solution: Algorithm Approach 2

The thought process behind this solution involves traversing the array and determining the minimum number of jumps needed to reach the end.

Here's a breakdown of the steps:

Initialization:

•  Initialize jumps to 0, representing the total number of jumps needed.
•  Initialize maxReach to the value of the first element in the array, indicating the farthest
•  position that can be reached with the current number of jumps.
•  Initialize steps to the value of the first element, representing the remaining steps that can be taken from the current position.

Traversal:

• Iterate through the array starting from the second element.
•         For each position i:
•             Update maxReach to the maximum value between the current maxReach and the sum of i and the value at position i, which represents the farthest reachable position.
•             Decrement steps by 1, indicating that one step is taken from the current position.
•             If steps become 0, it means that no more steps can be taken from the current position.
•             Without jumping. In this case:
  •        Increment jumps by 1, as a jump is required to move further.
  •        Update steps to the difference between the current maxReach and i, representing the remaining steps based on the farthest reachable position.

    Return:

•  After traversing the entire array, return jumps + 1, where jumps represent the total number of jumps required. Adding 1 accounts for the final jump to reach the end of the array.

This approach efficiently calculates the minimum number of jumps needed to reach the end of the array by dynamically updating the farthest reachable position and the remaining steps at each iteration.

# O(n) time | O(1) space
def min_number_of_jumps(array):
    """
    Finds the minimum number of jumps required to reach the end of an array, where each element represents the maximum
    number of steps that can be taken from that position.

    Args:
        array (List[int]): The input array representing the maximum number of steps that can be taken from each position.

    Returns:
        int: The minimum number of jumps needed to reach the end of the array.

    Example:
        >>> minNumberOfJumps([2, 3, 1, 1, 2, 4, 2, 0, 1, 1])
        4
    """
    # If there is only one element in the array, no jumps are needed
    if len(array) == 1:
        return 0
    
    jumps = 0             # Initialize the number of jumps
    maxReach = array[0]   # Initialize the farthest position that can be reached with the current number of jumps
    steps = array[0]      # Initialize the remaining steps that can be taken from the current position
    
    # Iterate through the array starting from the second element
    for i in range(1, len(array) - 1):
        maxReach = max(maxReach, i + array[i])  # Update the farthest position that can be reached
        steps -= 1                             # Decrement the remaining steps
        
        # If no more steps are left, a jump is required
        if steps == 0:
            jumps += 1                         # Increment the number of jumps
            steps = maxReach - i               # Update the remaining steps based on the farthest reachable position
    
    return jumps + 1  # Return the total number of jumps required to reach the end of the array

Complexity Analysis of Minimum Jumps Algorithm

 The time and space complexity of the provided solution are as follows:

Time Complexity

The solution traverses the input array once in a linear manner, which results in a time complexity of O(n), where n is the length of the input array.

Space Complexity

The solution uses constant additional space for variables (jumps, maxReach, steps) and does not rely on any auxiliary data structures proportional to the input size. Therefore, the space complexity is O(1) (constant space complexity).

Overall, the solution is efficient, with linear time complexity and constant space complexity, making it suitable for handling large input arrays efficiently.

Unit Tests

The following unit tests cover various scenarios, including regular arrays, edge cases like empty arrays, single-element arrays, increasing and decreasing sequences, and large arrays to ensure the solution's correctness and robustness.

import unittest

class TestMinNumberOfJumps(unittest.TestCase):

    def test_min_number_of_jumps(self):
        # Test case with a regular array
        self.assertEqual(minNumberOfJumps([2, 3, 1, 1, 2, 4, 2, 0, 1, 1]), 5)
        
        # Test case with a sorted array
        self.assertEqual(minNumberOfJumps([1, 2, 3, 4, 5]), 3)
        
        # Test case with a single element array
        self.assertEqual(minNumberOfJumps([0]), 0)
        
        # Test case with an empty array
        self.assertEqual(minNumberOfJumps([]), 0)
        
        # Test case with a large array
        self.assertEqual(minNumberOfJumps([1] * 10**6), 999999)
        
        # Test case with a decreasing array
        self.assertEqual(minNumberOfJumps([5, 4, 3, 2, 1]), 1)
        
        # Test case with a increasing array
        self.assertEqual(minNumberOfJumps([1, 2, 3, 4, 5]), 4)
        

if __name__ == "__main__":
    unittest.main()