LeetCode #1755 — HARD

Closest Subsequence Sum

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array nums and an integer goal.

You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal).

Return the minimum possible value of abs(sum - goal).

Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.

Example 1:

Input: nums = [5,-7,3,5], goal = 6
Output: 0
Explanation: Choose the whole array as a subsequence, with a sum of 6.
This is equal to the goal, so the absolute difference is 0.

Example 2:

Input: nums = [7,-9,15,-2], goal = -5
Output: 1
Explanation: Choose the subsequence [7,-9,-2], with a sum of -4.
The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum.

Example 3:

Input: nums = [1,2,3], goal = -7
Output: 7

Constraints:

1 <= nums.length <= 40
-10⁷ <= nums[i] <= 10⁷
-10⁹ <= goal <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer goal. You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal). Return the minimum possible value of abs(sum - goal). Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Dynamic Programming · Bit Manipulation

Example 1

[5,-7,3,5]
6

Example 2

[7,-9,15,-2]
-5

Example 3

[1,2,3]
-7

Core Insight

What unlocks the optimal approach

The naive solution is to check all possible subsequences. This works in O(2^n).
Divide the array into two parts of nearly is equal size.
Consider all subsets of one part and make a list of all possible subset sums and sort this list.
Consider all subsets of the other part, and for each one, let its sum = x, do binary search to get the nearest possible value to goal - x in the first part.

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

Largest constraint values

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 #1755: Closest Subsequence Sum
class Solution {
    public int minAbsDifference(int[] nums, int goal) {
        int n = nums.length;
        List<Integer> lsum = new ArrayList<>();
        List<Integer> rsum = new ArrayList<>();
        dfs(nums, lsum, 0, n / 2, 0);
        dfs(nums, rsum, n / 2, n, 0);

        rsum.sort(Integer::compareTo);
        int res = Integer.MAX_VALUE;

        for (Integer x : lsum) {
            int target = goal - x;
            int left = 0, right = rsum.size();
            while (left < right) {
                int mid = (left + right) >> 1;
                if (rsum.get(mid) < target) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            if (left < rsum.size()) {
                res = Math.min(res, Math.abs(target - rsum.get(left)));
            }
            if (left > 0) {
                res = Math.min(res, Math.abs(target - rsum.get(left - 1)));
            }
        }

        return res;
    }

    private void dfs(int[] nums, List<Integer> sum, int i, int n, int cur) {
        if (i == n) {
            sum.add(cur);
            return;
        }

        dfs(nums, sum, i + 1, n, cur);
        dfs(nums, sum, i + 1, n, cur + nums[i]);
    }
}

// Accepted solution for LeetCode #1755: Closest Subsequence Sum
func minAbsDifference(nums []int, goal int) int {
	n := len(nums)
	lsum := make([]int, 0)
	rsum := make([]int, 0)

	dfs(nums[:n/2], &lsum, 0, 0)
	dfs(nums[n/2:], &rsum, 0, 0)

	sort.Ints(rsum)
	res := math.MaxInt32

	for _, x := range lsum {
		t := goal - x
		l, r := 0, len(rsum)
		for l < r {
			m := int(uint(l+r) >> 1)
			if rsum[m] < t {
				l = m + 1
			} else {
				r = m
			}
		}
		if l < len(rsum) {
			res = min(res, abs(t-rsum[l]))
		}
		if l > 0 {
			res = min(res, abs(t-rsum[l-1]))
		}
	}

	return res
}

func dfs(nums []int, sum *[]int, i, cur int) {
	if i == len(nums) {
		*sum = append(*sum, cur)
		return
	}

	dfs(nums, sum, i+1, cur)
	dfs(nums, sum, i+1, cur+nums[i])
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1755: Closest Subsequence Sum
class Solution:
    def minAbsDifference(self, nums: List[int], goal: int) -> int:
        n = len(nums)
        left = set()
        right = set()

        self.getSubSeqSum(0, 0, nums[: n // 2], left)
        self.getSubSeqSum(0, 0, nums[n // 2 :], right)

        result = inf
        right = sorted(right)
        rl = len(right)

        for l in left:
            remaining = goal - l
            idx = bisect_left(right, remaining)

            if idx < rl:
                result = min(result, abs(remaining - right[idx]))

            if idx > 0:
                result = min(result, abs(remaining - right[idx - 1]))

        return result

    def getSubSeqSum(self, i: int, curr: int, arr: List[int], result: Set[int]):
        if i == len(arr):
            result.add(curr)
            return

        self.getSubSeqSum(i + 1, curr, arr, result)
        self.getSubSeqSum(i + 1, curr + arr[i], arr, result)

// Accepted solution for LeetCode #1755: Closest Subsequence Sum
/**
 * [1755] Closest Subsequence Sum
 *
 * You are given an integer array nums and an integer goal.
 * You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal).
 * Return the minimum possible value of abs(sum - goal).
 * Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.
 *  
 * Example 1:
 *
 * Input: nums = [5,-7,3,5], goal = 6
 * Output: 0
 * Explanation: Choose the whole array as a subsequence, with a sum of 6.
 * This is equal to the goal, so the absolute difference is 0.
 *
 * Example 2:
 *
 * Input: nums = [7,-9,15,-2], goal = -5
 * Output: 1
 * Explanation: Choose the subsequence [7,-9,-2], with a sum of -4.
 * The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum.
 *
 * Example 3:
 *
 * Input: nums = [1,2,3], goal = -7
 * Output: 7
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 40
 * 	-10^7 <= nums[i] <= 10^7
 * 	-10^9 <= goal <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/closest-subsequence-sum/
// discuss: https://leetcode.com/problems/closest-subsequence-sum/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/closest-subsequence-sum/solutions/5557655/rust-filter-out-out-of-bound-values/
    pub fn min_abs_difference(nums: Vec<i32>, goal: i32) -> i32 {
        let mut nums = nums;

        let mut pos_sum = nums.iter().filter(|&&x| x > 0).sum::<i32>();
        let mut neg_sum = nums.iter().filter(|&&x| x < 0).sum::<i32>();

        nums.sort_unstable_by_key(|&x| -x.abs());

        let mut v = vec![-goal];
        let mut v2 = Vec::<i32>::new();
        let mut result = goal.abs();

        for &x in nums.iter() {
            if x > 0 {
                pos_sum -= x;
            } else {
                neg_sum -= x;
            }
            for y in v.drain(..).flat_map(|val| [val, val + x]) {
                result = y.abs().min(result);
                if y >= -neg_sum {
                    result = result.min(y + neg_sum);
                } else if -y >= pos_sum {
                    result = result.min(-y - pos_sum);
                } else {
                    v2.push(y);
                }
                if result == 0 {
                    return 0;
                }
            }
            std::mem::swap(&mut v, &mut v2);
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1755_example_1() {
        let nums = vec![5, -7, 3, 5];
        let goal = 6;

        let result = 0;

        assert_eq!(Solution::min_abs_difference(nums, goal), result);
    }

    #[test]
    fn test_1755_example_2() {
        let nums = vec![7, -9, 15, -2];
        let goal = -5;

        let result = 1;

        assert_eq!(Solution::min_abs_difference(nums, goal), result);
    }

    #[test]
    fn test_1755_example_3() {
        let nums = vec![1, 2, 3];
        let goal = -7;

        let result = 7;

        assert_eq!(Solution::min_abs_difference(nums, goal), result);
    }
}

// Accepted solution for LeetCode #1755: Closest Subsequence Sum
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1755: Closest Subsequence Sum
// class Solution {
//     public int minAbsDifference(int[] nums, int goal) {
//         int n = nums.length;
//         List<Integer> lsum = new ArrayList<>();
//         List<Integer> rsum = new ArrayList<>();
//         dfs(nums, lsum, 0, n / 2, 0);
//         dfs(nums, rsum, n / 2, n, 0);
// 
//         rsum.sort(Integer::compareTo);
//         int res = Integer.MAX_VALUE;
// 
//         for (Integer x : lsum) {
//             int target = goal - x;
//             int left = 0, right = rsum.size();
//             while (left < right) {
//                 int mid = (left + right) >> 1;
//                 if (rsum.get(mid) < target) {
//                     left = mid + 1;
//                 } else {
//                     right = mid;
//                 }
//             }
//             if (left < rsum.size()) {
//                 res = Math.min(res, Math.abs(target - rsum.get(left)));
//             }
//             if (left > 0) {
//                 res = Math.min(res, Math.abs(target - rsum.get(left - 1)));
//             }
//         }
// 
//         return res;
//     }
// 
//     private void dfs(int[] nums, List<Integer> sum, int i, int n, int cur) {
//         if (i == n) {
//             sum.add(cur);
//             return;
//         }
// 
//         dfs(nums, sum, i + 1, n, cur);
//         dfs(nums, sum, i + 1, n, cur + nums[i]);
//     }
// }

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)

Space

O(1)

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.