**Sublist with Largest Min-Length Product - Google Top Interview Questions**

### Problem Statement :

You are given a list of integers nums and an integer pos. Find a sublist A of nums that includes the index (0-indexed) pos such that min(A) * A.length is maximized and return the value. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [-1, 1, 4, 3] pos = 3 Output 6 Explanation The best sublist is [4, 3]. Since min(4, 3) = 3 and its length is 2 we have 3 * 2 = 6.

### Solution :

` ````
Solution in C++ :
int par[100005];
int sz[100005];
bool active[100005];
int find(int x) {
return par[x] == x ? x : (par[x] = find(par[x]));
}
void merge(int x, int y) {
x = find(x);
y = find(y);
if (x == y) return;
par[x] = y;
sz[y] += sz[x];
}
int solve(vector<int>& nums, int pos) {
if (nums[pos] < 0) return nums[pos];
int n = nums.size();
vector<pair<int, int>> v;
for (int i = 0; i < n; i++) v.emplace_back(nums[i], i);
sort(v.rbegin(), v.rend());
int ret = -1e9;
for (int i = 0; i < n; i++) {
par[i] = i;
sz[i] = 1;
active[i] = false;
}
for (auto [value, index] : v) {
active[index] = true;
if (index > 0 && active[index - 1]) merge(index - 1, index);
if (index + 1 < n && active[index + 1]) merge(index, index + 1);
if (find(pos) == find(index)) ret = max(ret, sz[find(pos)] * value);
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
static int NEG = -100000000;
public int solve(int[] nums, int pos) {
int l = pos;
int r = pos;
int currMin = nums[pos];
int ret = NEG;
for (int i = 1; i <= nums.length; i++) {
ret = Math.max(ret, currMin * i);
int left = l - 1 >= 0 ? nums[l - 1] : NEG;
int right = r + 1 < nums.length ? nums[r + 1] : NEG;
if (left > right) {
currMin = Math.min(currMin, left);
l--;
} else {
currMin = Math.min(currMin, right);
r++;
}
}
return ret;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, A, pos):
NINF = float("-inf")
ans = m = A[pos]
i = j = pos
for _ in range(len(A) - 1):
left = A[i - 1] if i - 1 >= 0 else NINF
right = A[j + 1] if j + 1 < len(A) else NINF
if left >= right:
i -= 1
m = min(m, A[i])
else:
j += 1
m = min(m, A[j])
ans = max(ans, m * (j - i + 1))
return ans
```

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