LeetCode 45. Jump Game II

Question

Given an array of non-negative integers nums, 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.

You can assume that you can always reach the last index.

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

Constraints:

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 1000

Source: https://leetcode.com/problems/jump-game-ii/

Solution

1D DP. dp[i] is the min number of jumps from i to the end.

// DP, Time Complexity O(n^2)
public int jump(int[] nums) {
    if (nums == null || nums.length == 0) {
        return -1;
    }
    int len = nums.length;
    // dp[i] is the min number of jumps from i to the end
    // if i cannot reach the end, dp[i] should be -1
    int[] dp = new int[len];
    for (int i = 0; i < len; i++) {
        dp[i] = -1;
    }
    dp[len - 1] = 0;
    for (int i = len - 2; i >= 0; i--) {
        int maxStep = nums[i];
        int minJump = Integer.MAX_VALUE;
        boolean isGood = false;
        for (int k = 0; k <= maxStep && i + k < len; k++) {
            // if i+k can reach the end
            if (dp[i + k] != -1) {
                isGood = true;
                minJump = Math.min(minJump, dp[i + k] + 1);
            }
        }
        if (isGood) {
            dp[i] = minJump;
        }
    }
    return dp[0];
}