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.

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;
  }
};