**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))
sl.add(x)
return ans
```

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