HackMD
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
