1498.Number of Subsequences That Satisfy the Given Sum Condition

tags: Medium,Array,Two Pointers,Binary Search,Sorting

1498. Number of Subsequences That Satisfy the Given Sum Condition

題目描述

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.

範例

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

解答

Python

class Solution:
    def numSubseq(self, nums: List[int], target: int) -> int:
        n = len(nums)
        MOD = 10 ** 9 + 7
        nums.sort()

        ans = 0
        l, r = 0, n - 1
        while l <= r:
            if nums[l] + nums[r] <= target:
                ans = (ans + pow(2, r - l, MOD)) % MOD
                l += 1
            else:
                r -= 1
        return ans

Yen-Chi ChenSun, May 7, 2023

class Solution:
    def numSubseq(self, nums: List[int], target: int) -> int:
        nums.sort()

        ans = 0
        l, r = 0, len(nums) - 1
        while l <= r:
            ans <<= 1
            if nums[l] + nums[r] <= target:
                ans += 1
                l += 1
            else:
                r -= 1
        return ans % (10 ** 9 + 7)

Yen-Chi ChenSun, May 7, 2023

Javascript

function numSubseq(nums, target) {
  nums.sort((a, b) => a - b);
  let count = 0;
  let left = 0;
  let right = nums.length - 1;
  const mod = 1e9 + 7;

  const pow2 = new Array(nums.length).fill(0);
  pow2[0] = 1;
  for (let i = 1; i < nums.length; i++) {
    pow2[i] = (pow2[i - 1] * 2) % mod;
  }

  while (left <= right) {
    if (nums[left] + nums[right] > target) {
      right--;
    } else {
      count = (count + pow2[right - left]) % mod;
      left++;
    }
  }

  return count;
}

MarsgoatSun, May 7, 2023

Reference

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