1498. Number of Subsequences That Satisfy the Given Sum Condition

Problem

You are given an array of integers nums and an integer target.
Return the number of non-empty subsequences of nums such that the sum of the minimum and maximum element on it is less or equal to target. Since the answer may be too large, return it modulo 109 + 7.


Example 1:
Input: nums = [3,5,6,7], target = 9
Output: 4
Explanation: There are 4 subsequences that satisfy the condition.
[3] -> Min value + max value <= target (3 + 3 <= 9)
[3,5] -> (3 + 5 <= 9)
[3,5,6] -> (3 + 6 <= 9)
[3,6] -> (3 + 6 <= 9)

Example 2:
Input: nums = [3,3,6,8], target = 10
Output: 6
Explanation: There are 6 subsequences that satisfy the condition. (nums can have repeated numbers).
[3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]

Example 3:
Input: nums = [2,3,3,4,6,7], target = 12
Output: 61
Explanation: There are 63 non-empty subsequences, two of them do not satisfy the condition ([6,7], [7]).
Number of valid subsequences (63 - 2 = 61).

Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 106
1 <= target <= 106

Intuition

Ví dụ nums = [3,3,6,8], target = 10. Với subsequence [3,3,6] là một subsequence hợp lệ có min = 3, max = 6 và min+max = 9 <= target
Thì subsequence trên, ta cố định số 3, mỗi số có 2 lựa chọn là điền hoặc không điền => có 2^2 cách điền

Approach

Implementation

class Solution {
    private static final int MOD = 1_000_000_007;
    private int[] power2;

    public int numSubseq(int[] nums, int target) {
        int n = nums.length;

        Arrays.sort(nums);
        preCalculatePower2(n);

        int L = 0; int R = n-1;
        int counter = 0;

        while (L <= R) {
            int s = nums[L] + nums[R];
            if (s <= target) {
                counter = (counter + power2[R-L]) % MOD;
                // s <= target => should increase L to increase s
                // we can't decrease R as S also decrease (array is sorted in asc order)
                L++;
            } else {
                R--;
            }
        }

        return counter;
    }

    private void preCalculatePower2(int n) {
        power2 = new int[n+1];

        power2[0] = 1;
        for (int i = 1; i <= n; i++) {
            power2[i] = (2*power2[i-1]) % MOD;
        }
    }
}