**Number of Hops - Amazon Top Interview Questions**

### Problem Statement :

Given an integer list nums where each number represents the maximum number of hops you can make, return the minimum number of hops it would take to reach the last index starting at index 0. You can assume that you can always reach the last index. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [3, 3, 2, 0, 1] Output 2 Explanation We can jump from index 0 to 1 and then jump straight to the last index by jumping 3 steps.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
if (nums.empty()) return 0;
pair<int, int> interval{0, 0};
int n = nums.size();
if (n == 1) {
if (nums[0] != 0)
return 1;
else
return 0;
}
int jumps = 0;
int can_reach;
while (true) {
jumps++;
can_reach = -1;
for (int i = interval.first; i <= interval.second; i++) {
can_reach = max(can_reach, i + nums[i]);
}
if (can_reach >= n - 1) return jumps;
interval = {interval.second + 1, can_reach};
}
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] a) {
int n = a.length;
int maxEnd = 0;
int currentEnd = 0;
int jumps = 0;
for (int i = 0; i < n; i++) {
maxEnd = Math.max(maxEnd, i + a[i]);
if (i != n - 1 && i == currentEnd) {
currentEnd = maxEnd;
jumps++;
}
}
return jumps;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
n = len(nums)
ans = 0
hi = boundary = 0
for i in range(n):
hi = max(hi, i + nums[i])
if boundary == i < n - 1:
ans += 1
boundary = hi
return ans
```

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