Merge pull request #120 from BrianLusina/feat/algorithms-greed-jump-game

refactor(algorithms, greedy): jump game algorithm alternatives

Author: BrianLusina

DIRECTORY.md

@@ -85,6 +85,8 @@
   * Greedy
     * Gas Stations
       * [Test Gas Stations](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/gas_stations/test_gas_stations.py)
+    * Jump Game
+      * [Test Jump Game](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/jump_game/test_jump_game.py)
     * Majority Element
       * [Test Majority Element](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/majority_element/test_majority_element.py)
     * Min Arrows
@@ -497,8 +499,6 @@
       * [Test H Index](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/h_index/test_h_index.py)
     * Increasing Triplet Subsequence
       * [Test Increasing Triplet](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/increasing_triplet_subsequence/test_increasing_triplet.py)
-    * Jump Game
-      * [Test Jump Game](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/jump_game/test_jump_game.py)
     * Kid With Greatest No Of Candies
       * [Test Kids With Greatest Candies](https://github.com/BrianLusina/PythonSnips/blob/master/puzzles/arrays/kid_with_greatest_no_of_candies/test_kids_with_greatest_candies.py)
     * Longest Increasing Subsequence

algorithms/greedy/jump_game/README.md

@@ -0,0 +1,205 @@
+# Jump Game
+
+You are given an integer array nums. You are initially positioned at the array's first index, and each element in the
+array represents your maximum jump length at that position.
+
+Return true if you can reach the last index, or false otherwise.
+
+```text
+Example 1:
+
+Input: nums = [2,3,1,1,4]
+Output: true
+Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+```text
+Example 2:
+
+Input: nums = [3,2,1,0,4]
+Output: false
+Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible
+to reach the last index.
+```
+
+## Solutions
+
+1. [Naive Approach](#naive-approach)
+1. [Optimized Approach using Greedy Pattern](#optimized-approach-using-greedy-pattern)
+
+### Naive Approach
+
+The naive approach explores all possible jump sequences from the starting position to the end of the array. It begins 
+at the first index and attempts to jump to every reachable position from the current index, recursively repeating this 
+process for each new position. If a path successfully reaches the last index, the algorithm returns success. If it 
+reaches a position without further moves, it backtracks to try a different path.
+
+While this method guarantees that all possible paths are considered, it is highly inefficient, as it explores many 
+redundant or dead-end routes. The time complexity of this backtracking approach is exponential, making it impractical 
+for large inputs.
+
+### Optimized Approach using Greedy Pattern
+
+An optimized way to solve this problem is using a greedy approach that works in reverse. Instead of trying every 
+possible forward jump, we flip the logic: start from the end and ask, “Can we reach this position from any earlier index?”
+
+We begin with the last index as our initial target—the position we want to reach. Then, we move backward through the 
+array, checking whether the current index has a jump value large enough to reach or go beyond the current target. If it 
+can, we update the target to that index. This means that reaching this earlier position would eventually allow us to 
+reach the end. By continuously updating the target, we’re effectively identifying the leftmost position from which the 
+end is reachable.
+
+This process continues until we reach the first index or determine that no earlier index can reach the current target. 
+If we finish with the target at index 0, it means the start of the array can lead to the end, so we return TRUE. If the 
+target remains beyond index 0, then no path exists from the start to the end, and we return FALSE.
+
+![Solution 1](./images/solutions/jump_game_1_solution_1.png)
+![Solution 2](./images/solutions/jump_game_1_solution_2.png)
+![Solution 3](./images/solutions/jump_game_1_solution_3.png)
+![Solution 4](./images/solutions/jump_game_1_solution_4.png)
+![Solution 5](./images/solutions/jump_game_1_solution_5.png)
+![Solution 6](./images/solutions/jump_game_1_solution_6.png)
+![Solution 7](./images/solutions/jump_game_1_solution_7.png)
+![Solution 8](./images/solutions/jump_game_1_solution_8.png)
+![Solution 9](./images/solutions/jump_game_1_solution_9.png)
+![Solution 10](./images/solutions/jump_game_1_solution_10.png)
+![Solution 11](./images/solutions/jump_game_1_solution_11.png)
+
+#### Algorithm
+
+1. We begin by setting the last index of the array as our initial target using the variable target = len(nums) - 1. This 
+    target represents the position we are trying to reach, starting from the end and working backward. By initializing the 
+    target this way, we define our goal: to find out if there is any index i from which the target is reachable based on the
+    value at that position, nums[i]. This also sets the stage for updating the target if such an index is found.
+
+2. Next, we loop backward through the array using for i in range(len(nums) - 2, -1, -1). Here, i represents the current 
+   index we are analyzing. At each index i, the value nums[i] tells us how far we can jump forward from that position. 
+   By checking whether i + nums[i] >= target, we determine whether it can reach the current target from index i. This 
+   step allows us to use the jump range at each position to decide if it can potentially lead us to the end.
+
+3. If the condition i + nums[i] >= target is TRUE, the current index i can jump far enough to reach the current target. 
+   In that case, we update target = i, effectively saying, “Now we just need to reach index i instead.” If the condition
+   fails, we move back in the array one step further and try again with the previous index. 
+   We repeat this process until we either:
+   - Successfully moving the target back to index 0 means the start of the array can reach the end. In this case, we 
+     return TRUE.
+   - Or reach the start without ever being able to update the target to 0, which means there is no valid path. In this 
+     case, we return FALSE.
+
+#### Solution Summary
+
+1. Set the last index of the array as the target index.
+2. Traverse the array backward and verify if we can reach the target index from any of the previous indexes. 
+   - If we can reach it, we update the target index with the index that allows us to jump to the target index. 
+   - We repeat this process until we’ve traversed the entire array.
+3. Return TRUE if, through this process, we can reach the first index of the array. Otherwise, return FALSE.
+
+#### Time Complexity
+
+The time complexity of the above solution is O(n), since we traverse the array only once, where n is the number of 
+elements in the array.
+
+#### Space Complexity
+
+The space complexity of the above solution is O(1), because we do not use any extra space.
+
+
+---
+
+# Jump Game II
+
+You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].
+
+Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at
+nums[i], you can jump to any nums[i + j] where:
+
+0 <= j <= nums[i] and
+i + j < n
+Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach
+nums[n - 1].
+
+```text
+Example 1:
+
+Input: nums = [2,3,1,1,4]
+Output: 2
+Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to
+the last index.
+```
+
+```text
+Example 2:
+
+Input: nums = [2,3,0,1,4]
+Output: 2
+```
+
+![Example 1](./images/examples/jump_game_2_example_1.png)
+![Example 2](./images/examples/jump_game_2_example_2.png)
+![Example 3](./images/examples/jump_game_2_example_3.png)
+![Example 4](./images/examples/jump_game_2_example_4.png)
+![Example 5](./images/examples/jump_game_2_example_5.png)
+![Example 6](./images/examples/jump_game_2_example_6.png)
+
+## Solution
+
+We’ll solve this problem using a greedy approach. At each step, we choose the jump that allows us to reach the farthest 
+point. The objective is to minimize the number of jumps required to reach the end of the array. This strategy is 
+considered greedy because it selects the best possible move at each step without considering the impact on future jumps.
+By always jumping as far as possible, we cover more distance and use fewer jumps to reach the end.
+
+To find the minimum number of jumps needed to reach the end of the array, keep track of how far you can go with the 
+current number of jumps. As you progress through the array, update this maximum reach based on your current position. 
+When you reach the limit of your current jump range, increase your jump count and adjust the range to the farthest 
+position you can reach with the next jump. Continue this process until you reach or exceed the end of the array.
+
+The steps of the algorithm are given below:
+
+1. Initialize three variables, all set to 0. 
+   - `jumps`: This variable tracks the minimum number of jumps required to reach the end of the array and will be 
+    returned as the final output.
+   - `current_jump_boundary`: This represents the maximum index we can reach with the current number of jumps. It acts 
+     as the boundary of how far we can go before making another jump.
+   - `farthest_jump_index`: This tracks the farthest index we can reach from the current position by considering all 
+     possible jumps within the current jump’s range.
+
+2. Iterate through the nums array and perform the following steps:
+   - Update `farthest_jump_index`: For each index i, calculate i + nums[i], which represents the farthest index we can
+    reach from i. Update farthest_jump_index to be the maximum of its current value and i + nums[i].
+   - Check if a new jump is needed: If i equals current_jump_boundary, it means we’ve reached the limit of the current 
+     jump. Increment jumps by 1, and update current_jump_boundary to farthest_jump_index to set up for the next jump.
+
+3. After iterating through the array, jumps will contain the minimum number of jumps needed to reach the end. Return 
+   this value as the output.
+
+Let’s look at the following illustration to get a better understanding of the solution:
+
+![Solution 1](./images/solutions/jump_game_2_solution_1.png)
+![Solution 2](./images/solutions/jump_game_2_solution_2.png)
+![Solution 3](./images/solutions/jump_game_2_solution_3.png)
+![Solution 4](./images/solutions/jump_game_2_solution_4.png)
+![Solution 5](./images/solutions/jump_game_2_solution_5.png)
+![Solution 6](./images/solutions/jump_game_2_solution_6.png)
+![Solution 7](./images/solutions/jump_game_2_solution_7.png)
+![Solution 8](./images/solutions/jump_game_2_solution_8.png)
+![Solution 9](./images/solutions/jump_game_2_solution_9.png)
+![Solution 10](./images/solutions/jump_game_2_solution_10.png)
+![Solution 11](./images/solutions/jump_game_2_solution_11.png)
+![Solution 12](./images/solutions/jump_game_2_solution_12.png)
+![Solution 13](./images/solutions/jump_game_2_solution_13.png)
+
+### Time Complexity
+
+The time complexity of the above solution is O(n), where n is the length of nums because we are iterating the array once.
+
+### Space Complexity
+
+The space complexity of the above solution is O(1), because we are not using any extra space.
+
+---
+
+# Topics
+
+- Array
+- Dynamic Programming
+- Greedy

algorithms/greedy/jump_game/__init__.py

@@ -0,0 +1,127 @@
+from typing import List
+
+
+def can_jump(nums: List[int]) -> bool:
+    """
+    This function checks if it is possible to reach the last index of the array from the first index.
+    Args:
+        nums(list): list of integers
+    Returns:
+        bool: True if can jump to the last index, False otherwise
+    """
+
+    current_position = nums[0]
+
+    for idx in range(1, len(nums)):
+        if current_position == 0:
+            return False
+
+        current_position -= 1
+
+        current_position = max(current_position, nums[idx])
+
+    return True
+
+
+def can_jump_2(nums: List[int]) -> bool:
+    """
+    This function checks if it is possible to reach the last index of the array from the first index.
+
+    This variation starts checking from the last element in the input list and tracking back to check if it is possible
+    to reach the end.
+
+    Args:
+        nums(list): list of integers
+    Returns:
+        bool: True if can jump to the last index, False otherwise
+    """
+    target_num_index = len(nums) - 1
+
+    for idx in range(len(nums) - 2, -1, -1):
+        if target_num_index <= idx + nums[idx]:
+            target_num_index = idx
+
+    if target_num_index == 0:
+        return True
+    return False
+
+
+def jump(nums: List[int]) -> int:
+    """
+    This function returns the minimum number of jumps needed to reach the last index of the array from the first index.
+    Args:
+        nums(list): list of integers
+    Returns:
+        int: minimum number of jumps needed to reach the last index
+    """
+    size = len(nums)
+    # if start index == destination index == 0
+    if size == 1:
+        return 0
+
+    # destination is last index
+    destination = size - 1
+
+    # record of current coverage, record of last jump index
+    current_coverage, last_jump_index = 0, 0
+
+    # min number of jumps
+    min_jumps = 0
+
+    # Greedy strategy: extend coverage as long as possible with lazy jump
+    for idx in range(size):
+        # extend coverage as far as possible
+        current_coverage = max(current_coverage, idx + nums[idx])
+
+        # forced to jump (by lazy jump) to extend coverage
+        if idx == last_jump_index:
+            # update last jump index
+            last_jump_index = current_coverage
+
+            # update counter of jump by +1
+            min_jumps += 1
+
+            # check if destination has been reached
+            if current_coverage >= destination:
+                return min_jumps
+
+    return min_jumps
+
+
+def jump_2(nums: List[int]) -> int:
+    """
+    This function returns the minimum number of jumps needed to reach the last index of the array from the first index.
+    Args:
+        nums(list): list of integers
+    Returns:
+        int: minimum number of jumps needed to reach the last index
+    """
+    # Store the length of the input array
+    size = len(nums)
+
+    # if start index == destination index == 0
+    if size == 1:
+        return 0
+
+    # Initialize the variables to track the number of jumps,
+    # the current jump's limit, and the farthest reachable index
+    min_jumps = 0
+    current_jump_boundary = 0
+    furthest_jump_index = 0
+
+    # Iterate through the array, stopping before the last element
+    for idx in range(size - 1):
+        # Update the farthest_jump_index to be the maximum of its current value
+        # and the index we can reach from the current position
+        furthest_jump_index = max(furthest_jump_index, idx + nums[idx])
+
+        # If we have reached the limit of the current jump
+        if idx == current_jump_boundary:
+            # update counter of jump by +1
+            min_jumps += 1
+
+            # Update the current_jump_boundary to the farthest we can reach
+            current_jump_boundary = furthest_jump_index
+
+    # Return the total number of jumps needed to reach the end of the array
+    return min_jumps