LeetCode #1982 — HARD

Find Array Given Subset Sums

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

The Problem

Problem Statement

You are given an integer n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2ⁿ subset sums of the unknown array (in no particular order).

Return the array ans of length n representing the unknown array. If multiple answers exist, return any of them.

An array sub is a subset of an array arr if sub can be obtained from arr by deleting some (possibly zero or all) elements of arr. The sum of the elements in sub is one possible subset sum of arr. The sum of an empty array is considered to be 0.

Note: Test cases are generated such that there will always be at least one correct answer.

Example 1:

Input: n = 3, sums = [-3,-2,-1,0,0,1,2,3]
Output: [1,2,-3]
Explanation: [1,2,-3] is able to achieve the given subset sums:
- []: sum is 0
- [1]: sum is 1
- [2]: sum is 2
- [1,2]: sum is 3
- [-3]: sum is -3
- [1,-3]: sum is -2
- [2,-3]: sum is -1
- [1,2,-3]: sum is 0
Note that any permutation of [1,2,-3] and also any permutation of [-1,-2,3] will also be accepted.

Example 2:

Input: n = 2, sums = [0,0,0,0]
Output: [0,0]
Explanation: The only correct answer is [0,0].

Example 3:

Input: n = 4, sums = [0,0,5,5,4,-1,4,9,9,-1,4,3,4,8,3,8]
Output: [0,-1,4,5]
Explanation: [0,-1,4,5] is able to achieve the given subset sums.

Constraints:

1 <= n <= 15
sums.length == 2ⁿ
-10⁴ <= sums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2n subset sums of the unknown array (in no particular order). Return the array ans of length n representing the unknown array. If multiple answers exist, return any of them. An array sub is a subset of an array arr if sub can be obtained from arr by deleting some (possibly zero or all) elements of arr. The sum of the elements in sub is one possible subset sum of arr. The sum of an empty array is considered to be 0. Note: Test cases are generated such that there will always be at least one correct answer.

Baseline thinking

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

Pattern signal: Array

Example 1

3
[-3,-2,-1,0,0,1,2,3]

Example 2

2
[0,0,0,0]

Example 3

4
[0,0,5,5,4,-1,4,9,9,-1,4,3,4,8,3,8]

Core Insight

What unlocks the optimal approach

What information do the two largest elements tell us?
Can we use recursion to check all possible states?

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 #1982: Find Array Given Subset Sums
class Solution {
    public int[] recoverArray(int n, int[] sums) {
        int m = 1 << 30;
        for (int x : sums) {
            m = Math.min(m, x);
        }
        m = -m;
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        for (int x : sums) {
            tm.merge(x + m, 1, Integer::sum);
        }
        int[] ans = new int[n];
        if (tm.merge(0, -1, Integer::sum) == 0) {
            tm.remove(0);
        }
        ans[0] = tm.firstKey();
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < 1 << i; ++j) {
                if ((j >> (i - 1) & 1) == 1) {
                    int s = 0;
                    for (int k = 0; k < i; ++k) {
                        if (((j >> k) & 1) == 1) {
                            s += ans[k];
                        }
                    }
                    if (tm.merge(s, -1, Integer::sum) == 0) {
                        tm.remove(s);
                    }
                }
            }
            ans[i] = tm.firstKey();
        }
        for (int i = 0; i < 1 << n; ++i) {
            int s = 0;
            for (int j = 0; j < n; ++j) {
                if (((i >> j) & 1) == 1) {
                    s += ans[j];
                }
            }
            if (s == m) {
                for (int j = 0; j < n; ++j) {
                    if (((i >> j) & 1) == 1) {
                        ans[j] *= -1;
                    }
                }
                break;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1982: Find Array Given Subset Sums
func recoverArray(n int, sums []int) []int {
	m := -slices.Min(sums)
	rbt := redblacktree.NewWithIntComparator()
	merge := func(key int, value int) {
		if v, ok := rbt.Get(key); ok {
			nxt := v.(int) + value
			if nxt == 0 {
				rbt.Remove(key)
			} else {
				rbt.Put(key, nxt)
			}
		} else {
			rbt.Put(key, value)
		}
	}
	for _, x := range sums {
		merge(x+m, 1)
	}
	ans := make([]int, n)
	merge(ans[0], -1)
	ans[0] = rbt.Left().Key.(int)
	for i := 1; i < n; i++ {
		for j := 0; j < 1<<i; j++ {
			if j>>(i-1)&1 == 1 {
				s := 0
				for k := 0; k < i; k++ {
					if j>>k&1 == 1 {
						s += ans[k]
					}
				}
				merge(s, -1)
			}
		}
		ans[i] = rbt.Left().Key.(int)
	}
	for i := 0; i < 1<<n; i++ {
		s := 0
		for j := 0; j < n; j++ {
			if i>>j&1 == 1 {
				s += ans[j]
			}
		}
		if s == m {
			for j := 0; j < n; j++ {
				if i>>j&1 == 1 {
					ans[j] = -ans[j]
				}
			}
			break
		}
	}
	return ans

}

# Accepted solution for LeetCode #1982: Find Array Given Subset Sums
class Solution:
    def recoverArray(self, n: int, sums: List[int]) -> List[int]:
        m = -min(sums)
        sl = SortedList(x + m for x in sums)
        sl.remove(0)
        ans = [sl[0]]
        for i in range(1, n):
            for j in range(1 << i):
                if j >> (i - 1) & 1:
                    s = sum(ans[k] for k in range(i) if j >> k & 1)
                    sl.remove(s)
            ans.append(sl[0])
        for i in range(1 << n):
            s = sum(ans[j] for j in range(n) if i >> j & 1)
            if s == m:
                for j in range(n):
                    if i >> j & 1:
                        ans[j] *= -1
                break
        return ans

// Accepted solution for LeetCode #1982: Find Array Given Subset Sums
/**
 * [1982] Find Array Given Subset Sums
 *
 * You are given an integer n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2^n subset sums of the unknown array (in no particular order).
 * Return the array ans of length n representing the unknown array. If multiple answers exist, return any of them.
 * An array sub is a subset of an array arr if sub can be obtained from arr by deleting some (possibly zero or all) elements of arr. The sum of the elements in sub is one possible subset sum of arr. The sum of an empty array is considered to be 0.
 * Note: Test cases are generated such that there will always be at least one correct answer.
 *  
 * Example 1:
 *
 * Input: n = 3, sums = [-3,-2,-1,0,0,1,2,3]
 * Output: [1,2,-3]
 * Explanation: [1,2,-3] is able to achieve the given subset sums:
 * - []: sum is 0
 * - [1]: sum is 1
 * - [2]: sum is 2
 * - [1,2]: sum is 3
 * - [-3]: sum is -3
 * - [1,-3]: sum is -2
 * - [2,-3]: sum is -1
 * - [1,2,-3]: sum is 0
 * Note that any permutation of [1,2,-3] and also any permutation of [-1,-2,3] will also be accepted.
 *
 * Example 2:
 *
 * Input: n = 2, sums = [0,0,0,0]
 * Output: [0,0]
 * Explanation: The only correct answer is [0,0].
 *
 * Example 3:
 *
 * Input: n = 4, sums = [0,0,5,5,4,-1,4,9,9,-1,4,3,4,8,3,8]
 * Output: [0,-1,4,5]
 * Explanation: [0,-1,4,5] is able to achieve the given subset sums.
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 15
 * 	sums.length == 2^n
 * 	-10^4 <= sums[i] <= 10^4
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/find-array-given-subset-sums/
// discuss: https://leetcode.com/problems/find-array-given-subset-sums/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn recover_array(n: i32, sums: Vec<i32>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

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

    #[test]
    #[ignore]
    fn test_1982_example_1() {
        let n = 3;
        let sums = vec![-3, -2, -1, 0, 0, 1, 2, 3];

        let result = vec![1, 2, -3];

        assert_eq!(Solution::recover_array(n, sums), result);
    }

    #[test]
    #[ignore]
    fn test_1982_example_2() {
        let n = 2;
        let sums = vec![0, 0, 0, 0];

        let result = vec![0, 0];

        assert_eq!(Solution::recover_array(n, sums), result);
    }

    #[test]
    #[ignore]
    fn test_1982_example_3() {
        let n = 4;
        let sums = vec![0, 0, 5, 5, 4, -1, 4, 9, 9, -1, 4, 3, 4, 8, 3, 8];

        let result = vec![0, -1, 4, 5];

        assert_eq!(Solution::recover_array(n, sums), result);
    }
}

// Accepted solution for LeetCode #1982: Find Array Given Subset Sums
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1982: Find Array Given Subset Sums
// class Solution {
//     public int[] recoverArray(int n, int[] sums) {
//         int m = 1 << 30;
//         for (int x : sums) {
//             m = Math.min(m, x);
//         }
//         m = -m;
//         TreeMap<Integer, Integer> tm = new TreeMap<>();
//         for (int x : sums) {
//             tm.merge(x + m, 1, Integer::sum);
//         }
//         int[] ans = new int[n];
//         if (tm.merge(0, -1, Integer::sum) == 0) {
//             tm.remove(0);
//         }
//         ans[0] = tm.firstKey();
//         for (int i = 1; i < n; ++i) {
//             for (int j = 0; j < 1 << i; ++j) {
//                 if ((j >> (i - 1) & 1) == 1) {
//                     int s = 0;
//                     for (int k = 0; k < i; ++k) {
//                         if (((j >> k) & 1) == 1) {
//                             s += ans[k];
//                         }
//                     }
//                     if (tm.merge(s, -1, Integer::sum) == 0) {
//                         tm.remove(s);
//                     }
//                 }
//             }
//             ans[i] = tm.firstKey();
//         }
//         for (int i = 0; i < 1 << n; ++i) {
//             int s = 0;
//             for (int j = 0; j < n; ++j) {
//                 if (((i >> j) & 1) == 1) {
//                     s += ans[j];
//                 }
//             }
//             if (s == m) {
//                 for (int j = 0; j < n; ++j) {
//                     if (((i >> j) & 1) == 1) {
//                         ans[j] *= -1;
//                     }
//                 }
//                 break;
//             }
//         }
//         return ans;
//     }
// }

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 or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.