Jump Game II
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# LeetCode #45: Jump Game II (C/C++).

medium

source: https://leetcode.com/problems/jump-game-ii/
C/C++ Solution to LeetCode problem 45. Jump Game II.

## Problem

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

## Constraints

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 1000
• It’s guaranteed that you can reach nums[n - 1].

## Solution

We will use a Greedy Algorithm (see more problems)

• For each position, we determine what is the maximum distance we can achive.
• Everytime we are at the element of the max distance achieve, we increase the jumps counting.
• When we can reach the last element (or more) we increase the jump count and return the value.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 class Solution { public: int jump(vector<int>& nums) { if (nums.size() <= 1) return 0; int jumps = 0; int curPos = 0; int maxPos = 0; for (int i=0; i<nums.size(); i++) { maxPos = max(maxPos, i + nums[i]); if (maxPos >= nums.size() - 1) { jumps += 1; break; } if (curPos == i) { jumps += 1; curPos = maxPos; } } return jumps; } };