45. Jump Game II 🏃‍♂️

Difficulty: Hard - Tags: Array, Greedy, Dynamic Programming, Breadth-First Search

Description

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].

Examples

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.

Example 2:

Input:

nums = [2,3,0,1,4]

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.

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Solution 💡

The time complexity is $O(n)$, and the space complexity is $O(1)$, where $n$ is the length of the array.

Java

class Solution {
    public int jump(int[] nums) {
        int jumps = 0, currentEnd = 0, maxReach = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            maxReach = Math.max(maxReach, i + nums[i]);
            if (i == currentEnd) {
                jumps++;
                currentEnd = maxReach;
            }
        }
        return jumps;
    }
}

You can find the full Solution.java file here.