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1498. Number of Subsequences That Satisfy the Given Sum Condition

Description

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

Solutions

  • class Solution {
    public int numSubseq(int[] nums, int target) {
        Arrays.sort(nums);
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int[] f = new int[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = (f[i - 1] * 2) % mod;
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] * 2L > target) {
                break;
            }
            int j = search(nums, target - nums[i], i + 1) - 1;
            ans = (ans + f[j - i]) % mod;
        }
        return ans;
    }

    private int search(int[] nums, int x, int left) {
        int right = nums.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

  • class Solution {
public:
    int numSubseq(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        const int mod = 1e9 + 7;
        int n = nums.size();
        int f[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = (f[i - 1] * 2) % mod;
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] * 2L > target) {
                break;
            }
            int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 1;
            ans = (ans + f[j - i]) % mod;
        }
        return ans;
    }
};

  • class Solution:
    def numSubseq(self, nums: List[int], target: int) -> int:
        mod = 10**9 + 7
        nums.sort()
        n = len(nums)
        f = [1] + [0] * n
        for i in range(1, n + 1):
            f[i] = f[i - 1] * 2 % mod
        ans = 0
        for i, x in enumerate(nums):
            if x * 2 > target:
                break
            j = bisect_right(nums, target - x, i + 1) - 1
            ans = (ans + f[j - i]) % mod
        return ans


  • func numSubseq(nums []int, target int) (ans int) {
	sort.Ints(nums)
	n := len(nums)
	f := make([]int, n+1)
	f[0] = 1
	const mod int = 1e9 + 7
	for i := 1; i <= n; i++ {
		f[i] = f[i-1] * 2 % mod
	}
	for i, x := range nums {
		if x*2 > target {
			break
		}
		j := sort.SearchInts(nums[i+1:], target-x+1) + i
		ans = (ans + f[j-i]) % mod
	}
	return
}

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