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 :



title-img 




                        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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