# 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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