LeetCode 1345: Jump Game IV Solution

Master LeetCode problem 1345 (Jump Game IV), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1345. Jump Game IV

Hard

Array Hash Table Breadth-First Search

Problem Explanation

Explanation

To solve this problem, we can use a Breadth-First Search (BFS) approach. We start from the first index and explore all possible jumps to other indices based on the given conditions. We keep track of the minimum number of steps required to reach each index. We continue this process until we reach the last index. The minimum number of steps to reach the last index will be our answer.

Time complexity: O(n) where n is the number of elements in the input array. Space complexity: O(n) for the queue used in BFS.

Solution Code

import java.util.*;

class Solution {
    public int minJumps(int[] arr) {
        int n = arr.length;
        Map<Integer, List<Integer>> indices = new HashMap<>();
        for (int i = 0; i < n; i++) {
            indices.computeIfAbsent(arr[i], key -> new ArrayList<>()).add(i);
        }
        
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(0);
        int[] steps = new int[n];
        Arrays.fill(steps, -1);
        steps[0] = 0;
        
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            if (curr == n - 1) {
                return steps[curr];
            }
            
            List<Integer> neighbors = new ArrayList<>(indices.get(arr[curr]));
            neighbors.add(curr - 1);
            neighbors.add(curr + 1);
            indices.get(arr[curr]).clear();
            
            for (int next : neighbors) {
                if (next >= 0 && next < n && steps[next] == -1) {
                    steps[next] = steps[curr] + 1;
                    queue.offer(next);
                }
            }
        }
        
        return -1;
    }
}

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