Min Max Sets - Google Top Interview Questions

Problem Statement :

```Given list of integers nums and integer k, return the number of non-empty subsets S such that min(S) + max(S) ≤ k.

Note: For this problem, the subsets are multisets. That is, there can be duplicate values in the subsets since they refer to specific elements of the list, not values.

Constraints

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

Example 1

Input

nums = [1, 1, 4, 5]

k = 5

Output

6

Explanation

We can make the following subsets: [1], [1], [1, 1], [1, 4], [1, 4], [1, 1, 4].```

Solution :

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

int solve(vector<int>& nums, int k) {
sort(begin(nums), end(nums));
int ans = 0;
for (int left = 0, righ = nums.size() - 1; left <= righ;) {
if (nums[left] + nums[righ] > k)
righ--;
else {
ans += 1 << (righ - left);
left++;
}
}
return ans;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length;
int l = 0;
int res = 0;
int r = n - 1;
while (l <= r) {
while (l <= r && nums[l] + nums[r] > k) {
r--;
}
res += Math.pow(2, (r - l));
l++;
}
return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums, k):
nums.sort()
sl = SortedList()
ans = sum([x * 2 <= k for x in nums])

for x in nums:
if k >= x:
left = sl.bisect_right(k - x)
right = sl.bisect_right(x)
ans += (2 ** left - 1) * (2 ** (right - left))

return ans```
```

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