Minimum steps to reach end of array

//Time complexity:O(n)
//Initially the max_index we can reach is arr[0] so max_reach=arr[0] so we have to traverse till max_reach of 0th index so
//steps reamining to reach max_reach=arr[0],our current jump is at 0th index,now after traversing every index,we will reduce
//steps_remaining and update max_reach.
//when steps_reamining==0 then you have reached the maximum position you can reach from the last jump,so increase minimum_jumps.
//if i>=max_reach,it can only happen when 2 1 0,whenmax_reach and ith index equals and max_reach can be updated ahead because
//cureent index element is 0,so return -1
#define lli int
class Solution {
public:
    int jump(vector<int>& arr) {
        int n=arr.size();
        int max_reach=arr[0];
        int steps_remaining=arr[0];
        int minimum_jumps=1;
        if(n==1) return 0;
        else if(arr[0]==0) return -1;
        else{
            for(int i=1;i<n;i++){
                if(i==n-1) return minimum_jumps;
                max_reach=max(max_reach,i+arr[i]);
                steps_remaining--;
                if(steps_remaining==0){
                    minimum_jumps++;
                    if(i==max_reach) return -1;
                    steps_remaining=max_reach-i;
                }
            }
        }
        return minimum_jumps;
    }
};