Jump Game IV

Problem Statement

In this challenge, your task is to determine the minimum number of moves required to traverse from the first index to the last index of a given integer array, arr. You start at the first element of the array and you can jump under certain conditions:

• To the adjacent index in either direction (left or right).
• Directly to any index that contains the same value as the current index.

The objective is to find out the least number of jumps required to reach the very last element of the array. It's crucial to plan jumps intelligently as you can leap over multiple indices in one move if a value repeats itself in the array, potentially reducing the total number of steps required drastically.

Examples

Example 1

Input:

arr = [100,-23,-23,404,100,23,23,23,3,404]

Output:

3

Explanation:

You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2

Input:

arr = [7]

Output:

0

Explanation:

Start index is the last index. You do not need to jump.

Example 3

Input:

arr = [7,6,9,6,9,6,9,7]

Output:

1

Explanation:

You can jump directly from index 0 to index 7 which is last index of the array.

Constraints

• 1 <= arr.length <= 5 * 104
• -108 <= arr[i] <= 108

Approach and Intuition

The problem at hand essentially deals with finding the shortest path in a special kind of graph where each element of the array represents a node and the possible jumps are the edges. Here's a breakdown of the approach:

1. Visualize each element of the array arr as a node in a graph.
2. Construct edges based on the possible moves:
   • From node i to i+1, if i+1 is within array bounds.
   • From node i to i-1, if i-1 is within array bounds.
   • From node i to any node j (where j != i) if arr[i] == arr[j].
3. The task now is to find the shortest path from the first node (index 0) to the last node (index arr.length - 1).
4. A Breadth-First Search (BFS) is suitable for this scenario:
   • Start with the first index and explore all reachable indices using the aforementioned jump rules.
   • For each vertex, record the minimum number of steps taken to reach it.
   • Use a queue to manage the exploration and a set or boolean array to keep track of already visited nodes to avoid cyclic paths and redundant work.
   • For values that occur multiple times in the array, utilize a hashmap to quickly access all indices of the same value, allowing for rapid jumps across distant parts of the array.

Example insights based on given examples:

• In certain configurations where the array has only one element or where direct jumps can be made to the last index (because of repeated values), the number of steps required can be minimized effectively to zero or one respectively.
• In more complex arrays with multiple repeated values or longer lengths, strategic use of jumps to indices with the same values can cut down numerous incremental steps.

This method should be efficient, given the constraints, although the need to keep track of visited nodes and the dynamic construction of connections (especially where values are repeated frequently) can affect the speed of execution. Thus, the BFS algorithm, supplemented with appropriate data structures for tracking and retrieving connections, stands out as a potent approach to solve the problem.

Solutions

class Solution {
    public static int minimumJumps(int[] array) {
        int length = array.length;
        if (length <= 1) {
            return 0;
        }
    
        Map<Integer, List<Integer>> indexMap = new HashMap<>();
        for (int i = 0; i < length; i++) {
            indexMap.computeIfAbsent(array[i], v -> new LinkedList<>()).add(i);
        }
    
        HashSet<Integer> currentSet = new HashSet<>(); // store layers from start
        currentSet.add(0);
        Set<Integer> visitedIndices = new HashSet<>();
        visitedIndices.add(0);
        visitedIndices.add(length - 1);
        int jumps = 0;
    
        HashSet<Integer> targetSet = new HashSet<>(); // store layers from end
        targetSet.add(length - 1);
    
        // when current layer exists
        while (!currentSet.isEmpty()) {
            // search from the side with fewer nodes
            if (currentSet.size() > targetSet.size()) {
                HashSet<Integer> temp = currentSet;
                currentSet = targetSet;
                targetSet = temp;
            }
    
            HashSet<Integer> nextSet = new HashSet<>();
    
            // iterate the layer
            for (int current : currentSet) {
    
                // check same value
                for (int connected : indexMap.get(array[current])) {
                    if (targetSet.contains(connected)) {
                        return jumps + 1;
                    }
                    if (!visitedIndices.contains(connected)) {
                        visitedIndices.add(connected);
                        nextSet.add(connected);
                    }
                }
    
                // clear the list to prevent redundant search
                indexMap.get(array[current]).clear();
    
                // check neighbors
                if (targetSet.contains(current + 1) || targetSet.contains(current - 1)) {
                    return jumps + 1;
                }
    
                if (current + 1 < length && !visitedIndices.contains(current + 1)) {
                    visitedIndices.add(current + 1);
                    nextSet.add(current + 1);
                }
                if (current - 1 >= 0 && !visitedIndices.contains(current - 1)) {
                    visitedIndices.add(current - 1);
                    nextSet.add(current - 1);
                }
            }
    
            currentSet = nextSet;
            jumps++;
        }
    
        return -1;
    }
}