LeetCode Solutions
624. Maximum Distance in Arrays
Time: $O(n)$ Space: $O(1)$
class Solution {
public:
int maxDistance(vector<vector<int>>& arrays) {
int ans = 0;
int min = 10000;
int max = -10000;
for (const vector<int>& 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;
}
};
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;
}
}
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