• Time:O(n)
• Space:O(1)

## C++

``````class Solution {
public:
int maxDistance(vector<vector<int>>& arrays) {
int ans = 0;
int min = 10000;
int max = -10000;

for (const auto& A : arrays) {
ans = std::max({ans, A.back() - min, max - A.front()});
min = std::min(min, A.front());
max = std::max(max, A.back());
}

return ans;
}
};
``````

## JAVA

``````class Solution {
public int maxDistance(List<List<Integer>> arrays) {
int ans = 0;
int min = 10000;
int max = -10000;

for (List<Integer> A : arrays) {
ans = Math.max(ans, Math.max(A.get(A.size() - 1) - min, max - A.get(0)));
min = Math.min(min, A.get(0));
max = Math.max(max, A.get(A.size() - 1));
}

return ans;
}
}
``````

## Python

``````class Solution:
def maxDistance(self, arrays: List[List[int]]) -> int:
ans = 0
mini = 10000
maxi = -10000

for A in arrays:
ans = max(ans, A[-1] - mini, maxi - A[0])
mini = min(mini, A[0])
maxi = max(maxi, A[-1])

return ans
``````

• Time:O(n)
• Space:O(1)

## C++

``````class Solution:
def maxDistance(self, arrays: List[List[int]]) -> int:
min1, index_min1 = min((A[0], i) for i, A in enumerate(arrays))
max1, index_max1 = max((A[-1], i) for i, A in enumerate(arrays))
if index_min1 != index_max1:
return max1 - min1

min2, index_min2 = min((A[0], i)
for i, A in enumerate(arrays) if i != index_min1)
max2, index_min2 = max((A[-1], i)
for i, A in enumerate(arrays) if i != index_max1)
return max(max1 - min2, max2 - min1)
``````