LeetCode #1498 — MEDIUM

Number of Subsequences That Satisfy the Given Sum Condition

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[3,5,6,7]
9

Example 2

[3,3,6,8]
10

Example 3

[2,3,3,4,6,7]
12

Core Insight

What unlocks the optimal approach

Sort the array nums.
Use two pointers approach: Given an index i (choose it as the minimum in a subsequence) find the maximum j where j ≥ i and nums[i] +nums[j] ≤ target.
Count the number of subsequences.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1498: Number of Subsequences That Satisfy the Given Sum Condition
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;
    }
}

// Accepted solution for LeetCode #1498: Number of Subsequences That Satisfy the Given Sum Condition
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
}

# Accepted solution for LeetCode #1498: Number of Subsequences That Satisfy the Given Sum Condition
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

// Accepted solution for LeetCode #1498: Number of Subsequences That Satisfy the Given Sum Condition
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
    }
}

// Accepted solution for LeetCode #1498: Number of Subsequences That Satisfy the Given Sum Condition
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.