45. Jump Game II | LeetCode Solution
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
class Solution {
public:
int jump(vector<int>& nums) {
int n=nums.size();
if(n==0 || n==1)return 0;
pair<int,int>p{0,0};
int ans=0,can_reach;
while(true){
ans++;
can_reach=-1;
for(int i=p.first; i<=p.second; i++){
can_reach=max(can_reach,nums[i]+i);
}
if(can_reach>=n-1)return ans;
p={p.second+1,can_reach};
}
}
};
class Solution {
public:
int jump(vector<int>& nums) {
vector<int>dp(nums.size(),0);
for(int i=nums.size()-2; i>=0; i--){
int minn=INT_MAX;
for(int j=1; j<=nums[i] && (i+j)<nums.size(); j++){
minn=min(dp[i+j],minn);
}
if(minn!=INT_MAX)
dp[i]=1+minn;
else dp[i]=INT_MAX;
}
return dp[0];
}
};
0 Comments
If you have any doubts/suggestion/any query or want to improve this article, you can comment down below and let me know. Will reply to you soon.