LeetCode #2122 — HARD

Recover the Original Array

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

The Problem

Problem Statement

Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary positive integer k and created two new 0-indexed integer arrays lower and higher in the following manner:

lower[i] = arr[i] - k, for every index i where 0 <= i < n
higher[i] = arr[i] + k, for every index i where 0 <= i < n

Unfortunately, Alice lost all three arrays. However, she remembers the integers that were present in the arrays lower and higher, but not the array each integer belonged to. Help Alice and recover the original array.

Given an array nums consisting of 2n integers, where exactly n of the integers were present in lower and the remaining in higher, return the original array arr. In case the answer is not unique, return any valid array.

Note: The test cases are generated such that there exists at least one valid array arr.

Example 1:

Input: nums = [2,10,6,4,8,12]
Output: [3,7,11]
Explanation:
If arr = [3,7,11] and k = 1, we get lower = [2,6,10] and higher = [4,8,12].
Combining lower and higher gives us [2,6,10,4,8,12], which is a permutation of nums.
Another valid possibility is that arr = [5,7,9] and k = 3. In that case, lower = [2,4,6] and higher = [8,10,12].

Example 2:

Input: nums = [1,1,3,3]
Output: [2,2]
Explanation:
If arr = [2,2] and k = 1, we get lower = [1,1] and higher = [3,3].
Combining lower and higher gives us [1,1,3,3], which is equal to nums.
Note that arr cannot be [1,3] because in that case, the only possible way to obtain [1,1,3,3] is with k = 0.
This is invalid since k must be positive.

Example 3:

Input: nums = [5,435]
Output: [220]
Explanation:
The only possible combination is arr = [220] and k = 215. Using them, we get lower = [5] and higher = [435].

Constraints:

2 * n == nums.length
1 <= n <= 1000
1 <= nums[i] <= 10⁹
The test cases are generated such that there exists at least one valid array arr.

Step 01

Brute Force Baseline

Problem summary: Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary positive integer k and created two new 0-indexed integer arrays lower and higher in the following manner: lower[i] = arr[i] - k, for every index i where 0 <= i < n higher[i] = arr[i] + k, for every index i where 0 <= i < n Unfortunately, Alice lost all three arrays. However, she remembers the integers that were present in the arrays lower and higher, but not the array each integer belonged to. Help Alice and recover the original array. Given an array nums consisting of 2n integers, where exactly n of the integers were present in lower and the remaining in higher, return the original array arr. In case the answer is not unique, return any valid array. Note: The test cases are generated such that there exists at least one valid array arr.

Baseline thinking

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

Pattern signal: Array · Hash Map · Two Pointers

Example 1

[2,10,6,4,8,12]

Example 2

[1,1,3,3]

Example 3

[5,435]

Core Insight

What unlocks the optimal approach

If we fix the value of k, how can we check if an original array exists for the fixed k?
The smallest value of nums is obtained by subtracting k from the smallest value of the original array. How can we use this to reduce the search space for finding a valid k?
You can compute every possible k by using the smallest value of nums (as lower[i]) against every other value in nums (as the corresponding higher[i]).
For every computed k, greedily pair up the values in nums. This can be done sorting nums, then using a map to store previous values and searching that map for a corresponding lower[i] for the current nums[j] (as higher[i]).

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 #2122: Recover the Original Array
class Solution {
    public int[] recoverArray(int[] nums) {
        Arrays.sort(nums);
        for (int i = 1, n = nums.length; i < n; ++i) {
            int d = nums[i] - nums[0];
            if (d == 0 || d % 2 == 1) {
                continue;
            }
            boolean[] vis = new boolean[n];
            vis[i] = true;
            List<Integer> t = new ArrayList<>();
            t.add((nums[0] + nums[i]) >> 1);
            for (int l = 1, r = i + 1; r < n; ++l, ++r) {
                while (l < n && vis[l]) {
                    ++l;
                }
                while (r < n && nums[r] - nums[l] < d) {
                    ++r;
                }
                if (r == n || nums[r] - nums[l] > d) {
                    break;
                }
                vis[r] = true;
                t.add((nums[l] + nums[r]) >> 1);
            }
            if (t.size() == (n >> 1)) {
                int[] ans = new int[t.size()];
                int idx = 0;
                for (int e : t) {
                    ans[idx++] = e;
                }
                return ans;
            }
        }
        return null;
    }
}

// Accepted solution for LeetCode #2122: Recover the Original Array
func recoverArray(nums []int) []int {
	sort.Ints(nums)
	for i, n := 1, len(nums); i < n; i++ {
		d := nums[i] - nums[0]
		if d == 0 || d%2 == 1 {
			continue
		}
		vis := make([]bool, n)
		vis[i] = true
		ans := []int{(nums[0] + nums[i]) >> 1}
		for l, r := 1, i+1; r < n; l, r = l+1, r+1 {
			for l < n && vis[l] {
				l++
			}
			for r < n && nums[r]-nums[l] < d {
				r++
			}
			if r == n || nums[r]-nums[l] > d {
				break
			}
			vis[r] = true
			ans = append(ans, (nums[l]+nums[r])>>1)
		}
		if len(ans) == (n >> 1) {
			return ans
		}
	}
	return []int{}
}

# Accepted solution for LeetCode #2122: Recover the Original Array
class Solution:
    def recoverArray(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        for i in range(1, n):
            d = nums[i] - nums[0]
            if d == 0 or d % 2 == 1:
                continue
            vis = [False] * n
            vis[i] = True
            ans = [(nums[0] + nums[i]) >> 1]
            l, r = 1, i + 1
            while r < n:
                while l < n and vis[l]:
                    l += 1
                while r < n and nums[r] - nums[l] < d:
                    r += 1
                if r == n or nums[r] - nums[l] > d:
                    break
                vis[r] = True
                ans.append((nums[l] + nums[r]) >> 1)
                l, r = l + 1, r + 1
            if len(ans) == (n >> 1):
                return ans
        return []

// Accepted solution for LeetCode #2122: Recover the Original Array
/**
 * [2122] Recover the Original Array
 *
 * Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary positive integer k and created two new 0-indexed integer arrays lower and higher in the following manner:
 * <ol>
 * 	lower[i] = arr[i] - k, for every index i where 0 <= i < n
 * 	higher[i] = arr[i] + k, for every index i where 0 <= i < n
 * </ol>
 * Unfortunately, Alice lost all three arrays. However, she remembers the integers that were present in the arrays lower and higher, but not the array each integer belonged to. Help Alice and recover the original array.
 * Given an array nums consisting of 2n integers, where exactly n of the integers were present in lower and the remaining in higher, return the original array arr. In case the answer is not unique, return any valid array.
 * Note: The test cases are generated such that there exists at least one valid array arr.
 *  
 * Example 1:
 *
 * Input: nums = [2,10,6,4,8,12]
 * Output: [3,7,11]
 * Explanation:
 * If arr = [3,7,11] and k = 1, we get lower = [2,6,10] and higher = [4,8,12].
 * Combining lower and higher gives us [2,6,10,4,8,12], which is a permutation of nums.
 * Another valid possibility is that arr = [5,7,9] and k = 3. In that case, lower = [2,4,6] and higher = [8,10,12].
 *
 * Example 2:
 *
 * Input: nums = [1,1,3,3]
 * Output: [2,2]
 * Explanation:
 * If arr = [2,2] and k = 1, we get lower = [1,1] and higher = [3,3].
 * Combining lower and higher gives us [1,1,3,3], which is equal to nums.
 * Note that arr cannot be [1,3] because in that case, the only possible way to obtain [1,1,3,3] is with k = 0.
 * This is invalid since k must be positive.
 *
 * Example 3:
 *
 * Input: nums = [5,435]
 * Output: [220]
 * Explanation:
 * The only possible combination is arr = [220] and k = 215. Using them, we get lower = [5] and higher = [435].
 *
 *  
 * Constraints:
 *
 * 	2 * n == nums.length
 * 	1 <= n <= 1000
 * 	1 <= nums[i] <= 10^9
 * 	The test cases are generated such that there exists at least one valid array arr.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/recover-the-original-array/
// discuss: https://leetcode.com/problems/recover-the-original-array/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

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

// submission codes end

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

    #[test]
    #[ignore]
    fn test_2122_example_1() {
        let nums = vec![2, 10, 6, 4, 8, 12];

        let result = vec![3, 7, 11];

        assert_eq!(Solution::recover_array(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2122_example_2() {
        let nums = vec![1, 1, 3, 3];

        let result = vec![2, 2];

        assert_eq!(Solution::recover_array(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2122_example_3() {
        let nums = vec![5, 435];

        let result = vec![220];

        assert_eq!(Solution::recover_array(nums), result);
    }
}

// Accepted solution for LeetCode #2122: Recover the Original Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2122: Recover the Original Array
// class Solution {
//     public int[] recoverArray(int[] nums) {
//         Arrays.sort(nums);
//         for (int i = 1, n = nums.length; i < n; ++i) {
//             int d = nums[i] - nums[0];
//             if (d == 0 || d % 2 == 1) {
//                 continue;
//             }
//             boolean[] vis = new boolean[n];
//             vis[i] = true;
//             List<Integer> t = new ArrayList<>();
//             t.add((nums[0] + nums[i]) >> 1);
//             for (int l = 1, r = i + 1; r < n; ++l, ++r) {
//                 while (l < n && vis[l]) {
//                     ++l;
//                 }
//                 while (r < n && nums[r] - nums[l] < d) {
//                     ++r;
//                 }
//                 if (r == n || nums[r] - nums[l] > d) {
//                     break;
//                 }
//                 vis[r] = true;
//                 t.add((nums[l] + nums[r]) >> 1);
//             }
//             if (t.size() == (n >> 1)) {
//                 int[] ans = new int[t.size()];
//                 int idx = 0;
//                 for (int e : t) {
//                     ans[idx++] = e;
//                 }
//                 return ans;
//             }
//         }
//         return null;
//     }
// }

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.