# Sublists Containing Maximum and Minimum - Google Top Interview Questions

### Problem Statement :

```You are given a list of integers nums and you can remove at most one element in the list.

Return the maximum number of sublists that contain both the maximum and minimum elements of the resulting list.

The answer is guaranteed to fit in a 32-bit signed integer.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [2, 1, 5, 1, 3, 9]

Output

8

Explanation

If we remove 9 we'd get [2, 1, 5, 1, 3] and there's eight sublists where it contains both the max and the
min:

[1, 5]

[5, 1]

[1, 5, 1]

[2, 1, 5]

[5, 1, 3]

[1, 5, 1, 3]

[2, 1, 5, 1]

[2, 1, 5, 1, 3]

Example 2

Input

nums = [5, 5]

Output

3

Explanation

In this case, we don't remove any element. There's three sublists which contain both the max and the
min: [5], [5] and [5, 5].```

### Solution :

```                        ```Solution in C++ :

using ll = long long;

/*
number of sublists containing both the max as well as the min elements
*/
int numSublists(const vector<int>& nums) {
ll mi = INT_MAX + 1LL, mx = INT_MIN - 1LL;
int sz = nums.size(), res = 0, posMin, posMax;

for (int i = sz - 1; i >= 0; --i) {
if (mi > nums[i] || mx < nums[i]) res = 0;

mi = min(mi, (long long)nums[i]);
mx = max(mx, (long long)nums[i]);

if (nums[i] == mi) posMin = i;
if (nums[i] == mx) posMax = i;

res += sz - max(posMax, posMin);
}

return res;
}

int solve(vector<int>& nums) {
if (nums.empty()) return 0;

vector<int> minRemoved = nums;
minRemoved.erase(min_element(minRemoved.begin(), minRemoved.end()));

vector<int> maxRemoved = nums;
maxRemoved.erase(max_element(maxRemoved.begin(), maxRemoved.end()));

return max({numSublists(nums), numSublists(minRemoved), numSublists(maxRemoved)});
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums) {
int N = nums.length;
if (N == 0)
return 0;
if (N == 1)
return 1;
// find the max and the min element
int min = nums[0];
int minI = 0;
int cntMin = 1;
int max = nums[0];
int maxI = 0;
int cntMax = 1;
for (int i = 1; i < N; i++) {
if (nums[i] < min) {
min = nums[i];
minI = i;
cntMin = 1;
} else if (nums[i] == min) {
cntMin++;
}
if (nums[i] > max) {
max = nums[i];
maxI = i;
cntMax = 1;
} else if (nums[i] == max) {
cntMax++;
}
}

int ans = helper(nums); // don't remove anything
if (cntMin == 1)
ans = Math.max(ans, helper(newArr(nums, minI)));
if (cntMax == 1)
ans = Math.max(ans, helper(newArr(nums, maxI)));

return ans;
}

public int[] newArr(int[] nums, int index) {
int N = nums.length;
int[] newNums = new int[N - 1];
for (int i = 0; i < index; i++) newNums[i] = nums[i];
for (int i = index + 1; i < N; i++) newNums[i - 1] = nums[i];
return newNums;
}

public int helper(int[] nums) {
int N = nums.length;
int min = nums[0];
int max = nums[0];
for (int n : nums) {
min = Math.min(min, n);
max = Math.max(max, n);
}

if (min == max) {
// every number is the same
return (N * (N + 1) / 2);
}

// store the indices where the max and min element appear in
TreeSet<Integer> mins = new TreeSet<Integer>();
TreeSet<Integer> maxs = new TreeSet<Integer>();
for (int i = 0; i < N; i++) {
if (nums[i] == min)
if (nums[i] == max)
}

int ans = 0;
// find the number of sublists beginning at i that include the max and min.
for (int i = 0; i < N; i++) {
if (mins.isEmpty() || maxs.isEmpty())
break;
int minF = mins.first();
int maxF = maxs.first();
Integer a1 = maxs.ceiling(minF);
Integer a2 = mins.ceiling(maxF);
int ind = Integer.MAX_VALUE;
if (a1 != null)
ind = Math.min(ind, a1);
if (a2 != null)
ind = Math.min(ind, a2);
if (ind == Integer.MAX_VALUE)
break;
ans += (N - ind);
if (minF == i)
mins.remove(i);
if (maxF == i)
maxs.remove(i);
}
return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):
if len(nums) <= 1:
return len(nums)

def num_sublists(lst):
res = 0
max_val, min_val = max(lst), min(lst)
min_idx, max_idx = None, None

for i, val in enumerate(lst):
if val == max_val:
max_idx = i
if val == min_val:
min_idx = i

if min_idx is None or max_idx is None:
continue
res += min(min_idx, max_idx) + 1
return res

res = num_sublists(nums)
n = len(nums)

if nums.count(min(nums)) == 1:
idx = nums.index(min(nums))
res = max(res, num_sublists(nums[:idx] + nums[idx + 1 :]))

if nums.count(max(nums)) == 1:
idx = nums.index(max(nums))
res = max(res, num_sublists(nums[:idx] + nums[idx + 1 :]))
return res```
```

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