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

Solution 1: Sorting + Binary Search

Since the problem is about subsequences and involves the sum of the minimum and maximum elements, we can first sort the array \(\textit{nums}\).

Then we enumerate the minimum element \(\textit{nums}[i]\). For each \(\textit{nums}[i]\), we can find the maximum element \(\textit{nums}[j]\) in \(\textit{nums}[i + 1]\) to \(\textit{nums}[n - 1]\) such that \(\textit{nums}[i] + \textit{nums}[j] \leq \textit{target}\). The number of valid subsequences in this case is \(2^{j - i}\), where \(2^{j - i}\) represents all possible subsequences from \(\textit{nums}[i + 1]\) to \(\textit{nums}[j]\). We sum up the counts of all such subsequences.

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\).

Python3JavaC++GoTypeScriptRust

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

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 && nums[i] * 2 <= target; ++i) {
            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 && nums[i] * 2 <= target; ++i) {
            int j = upper_bound(nums.begin() + i + 1, nums.end(), target - nums[i]) - nums.begin() - 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
}

function numSubseq(nums: number[], target: number): number {
    nums.sort((a, b) => a - b);
    const mod = 1e9 + 7;
    const n = nums.length;
    const f: number[] = Array(n + 1).fill(1);
    for (let i = 1; i <= n; ++i) {
        f[i] = (f[i - 1] * 2) % mod;
    }

    let ans = 0;
    for (let i = 0; i < n && nums[i] * 2 <= target; ++i) {
        const j = search(nums, target - nums[i], i + 1) - 1;
        if (j >= i) {
            ans = (ans + f[j - i]) % mod;
        }
    }
    return ans;
}

function search(nums: number[], x: number, left: number): number {
    let right = nums.length;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] > x) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

impl Solution {
    pub fn num_subseq(mut nums: Vec<i32>, target: i32) -> i32 {
        nums.sort();
        const MOD: i32 = 1_000_000_007;
        let n = nums.len();
        let mut f = vec![1; n + 1];
        for i in 1..=n {
            f[i] = (f[i - 1] * 2) % MOD;
        }
        let mut ans = 0;
        for i in 0..n {
            if nums[i] * 2 > target {
                break;
            }
            let mut l = i + 1;
            let mut r = n;
            while l < r {
                let m = (l + r) / 2;
                if nums[m] > target - nums[i] {
                    r = m;
                } else {
                    l = m + 1;
                }
            }
            let j = l - 1;
            ans = (ans + f[j - i]) % MOD;
        }
        ans
    }
}

1498. Number of Subsequences That Satisfy the Given Sum Condition

Description

Solutions

Solution 1: Sorting + Binary Search

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