Jump Game III - Iterative Solution in JavaScript (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 represents 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.

This problem is significant in scenarios where you need to optimize the number of steps or moves, such as in game development or pathfinding algorithms. A common pitfall is to assume that the maximum jump length at each position should always be used, which is not necessarily optimal.

Approach

To solve this problem, we can use a greedy approach. However, since the problem requires an iterative solution with O(n^2) time complexity, we will use a dynamic programming-like approach to keep track of the minimum jumps needed to reach each position.

We will maintain an array jumps where jumps[i] represents the minimum number of jumps needed to reach index i. Initially, jumps[0] is 0 because we are already at the first index, and all other elements are set to infinity (or a very large number) because they are initially unreachable.

Algorithm

1. Initialize an array jumps of the same length as the input array, with jumps[0] set to 0 and all other elements set to infinity.

2. Iterate through the array using a nested loop. For each position i, update the jumps array for all reachable positions from i.

3. The value at jumps[n-1] will give the minimum number of jumps needed to reach the last index.

Code Implementation

// Function to find the minimum number of jumps to reach the end
function minJumps(arr) {
    // Length of the array
    const n = arr.length;
    
    // If the array has only one element, no jumps are needed
    if (n === 1) return 0;
    
    // Initialize jumps array with Infinity
    const jumps = new Array(n).fill(Infinity);
    jumps[0] = 0; // Starting point requires 0 jumps
    
    // Iterate through the array
    for (let i = 0; i < n; i++) {
        // Check all reachable positions from i
        for (let j = i + 1; j <= i + arr[i] && j < n; j++) {
            // Update the jumps array with the minimum jumps needed to reach j
            jumps[j] = Math.min(jumps[j], jumps[i] + 1);
        }
    }
    
    // Return the minimum jumps needed to reach the last index
    return jumps[n - 1];
}

// Example usage
const input = [2, 3, 1, 1, 4];
console.log(minJumps(input)); // Output: 2

Complexity Analysis

The time complexity of this approach is O(n^2) because of the nested loops iterating through the array. The space complexity is O(n) due to the additional jumps array used to store the minimum jumps needed for each position.

Edge Cases

Consider the following edge cases:

Array with a single element: The output should be 0 as no jumps are needed.
Array where the last element is unreachable: The output should be Infinity or a large number indicating that the last index cannot be reached.

Example:

Input: [1, 0, 3]
Output: Infinity (or a large number)

Testing

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

Simple cases with small arrays.
Cases where the last index is unreachable.
Large arrays to test the performance and efficiency of the algorithm.

Example test cases:

// Test case 1: Simple case
console.log(minJumps([2, 3, 1, 1, 4])); // Output: 2

// Test case 2: Single element
console.log(minJumps([0])); // Output: 0

// Test case 3: Unreachable last index
console.log(minJumps([1, 0, 3])); // Output: Infinity

// Test case 4: Large array
console.log(minJumps(new Array(100).fill(1))); // Output: 99

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller sub-problems and solve them step-by-step.
Think about edge cases and how your solution handles them.
Practice similar problems to improve your problem-solving skills and understanding of algorithms.

Conclusion

In this blog post, we discussed the problem of finding the minimum number of jumps to reach the last index of an array. We explored an iterative solution with O(n^2) time complexity and O(n) space complexity. By understanding the problem, breaking it down into smaller steps, and considering edge cases, we can develop an efficient solution.

Understanding and solving such problems is crucial for improving your algorithmic thinking and problem-solving skills. Keep practicing and exploring further to enhance your abilities.