**Longest Consecutive Sublist - Google Top Interview Questions**

### Problem Statement :

You are given a list of unique integers nums. Return the length of the longest sublist containing consecutive elements. Constraints 0 ≤ n ≤ 1,000 where n is the length of nums Example 1 Input nums = [1, 4, 5, 3, 2, 9] Output 5 Explanation The sublist [1, 4, 5, 3, 2] contains consecutive elements from 1 to 5.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
int n = nums.size(), ret = 1;
if (n <= 1) return n;
vector<int> pre(n, 0);
for (int i = 0; i < n; i++) {
pre[i] = (i == 0 ? 0 : pre[i - 1]) + nums[i];
}
for (int i = 0; i < n; i++) {
int mini = nums[i] - 1, maxi = nums[i];
for (int j = i + 1; j < n; j++) {
mini = max(0, min(mini, nums[j] - 1));
maxi = max(maxi, nums[j]);
long cs = maxi * (maxi + 1) / 2 - (mini) * (mini + 1) / 2;
int sublistSum = pre[j] - (i == 0 ? 0 : pre[i - 1]);
if (cs == sublistSum) {
ret = max(ret, j - i + 1);
}
}
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
int n = nums.length;
if (n <= 1)
return n;
int res = 1;
for (int i = 0; i < n; i++) {
int min = nums[i];
int max = nums[i];
for (int r = i; r < n; r++) {
min = Math.min(min, nums[r]);
max = Math.max(max, nums[r]);
if (max - min + 1 == r - i + 1) {
res = Math.max(res, r - i + 1);
}
}
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
ret = 0
for i in range(len(nums)):
lhs = nums[i]
rhs = nums[i]
for j in range(i, len(nums)):
lhs = min(lhs, nums[j])
rhs = max(rhs, nums[j])
if rhs - lhs == j - i:
ret = max(ret, j - i + 1)
return ret
```

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