Jump Game III - Iterative Solution in Java (O(n^2) Time Complexity)

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

Example:

Input: [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.

Note:

Convert the previous solution to an iterative one. Your algorithm should run in O(n^2) time and use O(n) extra space.

Understanding the Problem

The core challenge of this problem is to find the minimum number of jumps needed to reach the last index of the array. Each element in the array indicates the maximum number of steps you can jump forward from that position. The goal is to determine the fewest jumps required to move from the start to the end of the array.

Significance and Applications

This problem is significant in scenarios where you need to find the shortest path or minimum steps in a constrained environment. It has applications in game development, pathfinding algorithms, and optimization problems.

Potential Pitfalls and Misconceptions

A common misconception is to think that you need to explore all possible paths to find the minimum jumps, which can lead to inefficient solutions. Another pitfall is not considering the constraints and edge cases, such as arrays with all elements being zero except the first one.

Approach

To solve this problem, we can use a greedy approach combined with dynamic programming. The idea is to keep track of the minimum jumps needed to reach each position in the array.

Naive Solution

A naive solution would involve exploring all possible paths from the start to the end, which is highly inefficient and can lead to exponential time complexity.

Optimized Solution

An optimized solution involves using an iterative approach with dynamic programming to keep track of the minimum jumps needed to reach each position. This approach ensures that we only compute the minimum jumps once for each position, leading to a time complexity of O(n^2).

Algorithm

The algorithm can be broken down into the following steps:

Initialize an array jumps where jumps[i] represents the minimum number of jumps needed to reach index i.
Set jumps[0] to 0 since no jumps are needed to reach the first index.
Iterate through the array and for each position, update the jumps array for all reachable positions from the current position.
Return the value at jumps[n-1], which represents the minimum jumps needed to reach the last index.

Code Implementation

public class JumpGameIII {
    public int minJumps(int[] nums) {
        int n = nums.length;
        if (n == 1) return 0; // If there's only one element, no jumps are needed

        int[] jumps = new int[n]; // Array to store the minimum jumps needed to reach each index
        for (int i = 1; i < n; i++) {
            jumps[i] = Integer.MAX_VALUE; // Initialize all positions with a large value
        }

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j <= i + nums[i] && j < n; j++) {
                jumps[j] = Math.min(jumps[j], jumps[i] + 1); // Update the minimum jumps for each reachable position
            }
        }

        return jumps[n - 1]; // Return the minimum jumps needed to reach the last index
    }

    public static void main(String[] args) {
        JumpGameIII solution = new JumpGameIII();
        int[] nums = {2, 3, 1, 1, 4};
        System.out.println("Minimum jumps needed: " + solution.minJumps(nums)); // Output: 2
    }
}

Complexity Analysis

The time complexity of this approach is O(n^2) because for each element, we potentially iterate through all reachable positions. The space complexity is O(n) due to the additional array used to store the minimum jumps.

Edge Cases

Consider the following edge cases:

Array with a single element: The output should be 0 since no jumps are needed.
Array with all elements being zero except the first one: The output should be Integer.MAX_VALUE or an indication that it's not possible to reach the end.

Testing

To test the solution comprehensively, consider a variety of test cases:

Simple cases with small arrays.
Edge cases with single elements or all zeros.
Large arrays to test the efficiency of the algorithm.

Thinking and Problem-Solving Tips

When approaching such problems, it's essential to:

Break down the problem into smaller sub-problems.
Consider both naive and optimized solutions.
Analyze the time and space complexity of different approaches.
Practice similar problems to improve problem-solving skills.

Conclusion

Understanding and solving the Jump Game III problem helps in developing skills for pathfinding and optimization problems. By practicing and exploring different approaches, you can improve your problem-solving abilities and apply these techniques to various scenarios.

Additional Resources

For further reading and practice, consider the following resources: