Unique Pairs With Target Sum â Problem Statement & Solution Guide
Problem Description
You are provided with an integer array nums and an integer target. Your task is to identify all distinct pairs of elements within the array such that their sum equals target.
To ensure uniqueness, each pair must be represented in non-decreasing order (i.e., if a pair is (a, b), then a <= b). Furthermore, duplicate pairs with identical values must be excluded from the result. For instance, if the array contains multiple instances of the numbers 2 and 3, and target is 5, the pair [2, 3] should appear only once in the final output.
Return a list of lists, where each inner list contains two integers representing a valid unique pair. The order of the pairs in the returned list does not matter, but the elements within each pair must be sorted in non-decreasing order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Unique Pairs With Target Sum"
WHY DOES IT MATTER?
The twoâsum pattern is foundational because many realâworld problems reduce to finding complementary valuesâwhether matching transactions, pairing sensors, or reconciling logs. Mastering this pattern demonstrates a candidate's ability to convert a combinatorial explosion into a linear scan, a skill prized in performanceâcritical systems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that you do not need to compare every element with every other; you only need to know whether the complement of the current element has already been observed. This reduces the search space from O(n²) to O(n) by turning the problem into a constantâtime lookup.
REAL-WORLD CONNECTION
Imagine a distributed ledger where each node records a transaction amount. To detect fraudulent pairs that cancel each other out, you can hash each amount and instantly check for its opposite, mirroring the hashâbased twoâsum solution. This mirrors how highâfrequency trading platforms reconcile offsetting orders in microseconds.
During an interview, write the hashâset version first because it is concise and avoids sorting pitfalls. Immediately discuss how youâll enforce a ⤠b ordering and use a Set of strings (or tuples) to guarantee uniquenessâthis shows awareness of edge cases and clean code.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory â Why This Approach?
The problem of finding all unique pairs that sum to a target is a classic instance of the twoâsum family, which can be solved efficiently using hashâbased lookâups or the twoâpointer technique on a sorted array. A naive doubleâloop enumerates every possible pair, leading to O(n²) time, which quickly becomes prohibitive for large inputs (n can be up to 10âľ or more in interview settings). The optimal paradigm leverages the fact that we only need to know whether a complementary value (targetâcurrent) has been seen before; a hash set provides O(1) averageâcase lookâups, while sorting enables a deterministic linear scan with two pointers moving inward, both achieving O(n log n) or O(n) time respectively. By storing each discovered pair in a set (or by enforcing a ⤠b ordering), we guarantee uniqueness without extra postâprocessing.
When using a hash map, each element is processed once: we check if its complement exists, add the ordered pair to a result set, then insert the element itself for future matches. This yields O(n) time and O(n) auxiliary space. The twoâpointer method first sorts the array (O(n log n)), then advances pointers based on the current sum relative to the target, skipping duplicates to maintain distinctness. Both approaches illustrate the broader algorithmic principle of trading a modest amount of extra space for linear time, a pattern that recurs in many interview problems involving pairwise relationships.
Understanding why the naive O(n²) solution fails is crucial: with n = 10âľ, the double loop would perform ~10šⰠoperations, exceeding typical time limits and causing memory pressure when storing intermediate results. The optimal solutions avoid this explosion by reducing the search space dramaticallyâeither through constantâtime complement checks or by exploiting the order imposed by sorting. Mastery of these techniques not only solves this specific problem but also equips candidates to tackle a wide range of sumârelated challenges efficiently.
Interview Questions on This Problem
Q1How would you modify the solution to return the count of unique pairs instead of the pairs themselves, and what would be the impact on space complexity?
You can maintain a simple integer counter that increments each time you discover a new valid pair, eliminating the need to store the pair objects. With the hashâset approach, you still need a set to track seen numbers for complement checks, so space remains O(n). If you use the twoâpointer method on a sorted array, you can avoid any extra storage beyond the input array, achieving O(1) auxiliary space.
Q2Explain how you would adapt the algorithm to handle streaming data where the array cannot be fully loaded into memory.
In a streaming context, you can keep a hash set of numbers seen so far and, for each incoming element x, check if targetâx is already in the set. If it is, emit the ordered pair (min(x, targetâx), max(x, targetâx)). This yields O(1) amortized time per element and O(k) space where k is the number of distinct elements seen, which is the minimal state required for correctness.
Q3What changes are required if the problem asks for unique triplets that sum to the target, and how does the complexity evolve?
Finding unique triplets (the 3âsum problem) typically involves sorting the array and then fixing one element while applying the twoâpointer technique on the remaining subâarray. This results in O(n²) time and O(1) extra space after sorting. Unlike the pair version, you cannot achieve linear time because you must consider combinations of three elements, which inherently introduces a quadratic factor.
Examples
Input
nums = [1, 5, 3, 7, 2, 8, 4], target = 9
Output
[[1, 8], [2, 7], [3, 6], [4, 5]]
Explanation: 1. Sort the array: [1, 2, 3, 4, 5, 7, 8]. 2. Use two pointers: left=0 (1), right=6 (8). Sum=9. Add [1, 8]. Move left to 1, right to 5. 3. left=1 (2), right=5 (7). Sum=9. Add [2, 7]. Move left to 2, right to 4. 4. left=2 (3), right=4 (5). Sum=8. Move left to 3. 5. left=3 (4), right=4 (5). Sum=9. Add [4, 5]. Move left to 4, right to 3. Stop. 6. Note: 6 is not in the array, so [3, 6] is not a valid pair. Wait, let's re-evaluate the input. The input is [1, 5, 3, 7, 2, 8, 4]. The pairs summing to 9 are (1,8), (2,7), (5,4). [3,6] is invalid because 6 is not in the array. Let's correct the example output to match the input. Corrected Output: [[1, 8], [2, 7], [4, 5]] Corrected Explanation: 1. Sort: [1, 2, 3, 4, 5, 7, 8]. 2. (1,8) sum=9. 3. (2,7) sum=9. 4. (3,5) sum=8, move left. 5. (4,5) sum=9. 6. Stop. Result: [[1,8], [2,7], [4,5]].
Input
nums = [0, 0, 0, 0], target = 0
Output
[[0, 0]]
Explanation: 1. Sort the array: [0, 0, 0, 0]. 2. Two pointers: left=0, right=3. Sum=0. Add [0, 0]. 3. Skip duplicates: Move left past all 0s, move right past all 0s. 4. Pointers cross. Stop. 5. Only one unique pair [0, 0] is returned.
Input
nums = [-1, -2, -3, 1, 2, 3], target = 0
Output
[[-3, 3], [-2, 2], [-1, 1]]
Explanation: 1. Sort the array: [-3, -2, -1, 1, 2, 3]. 2. left=0 (-3), right=5 (3). Sum=0. Add [-3, 3]. 3. left=1 (-2), right=4 (2). Sum=0. Add [-2, 2]. 4. left=2 (-1), right=3 (1). Sum=0. Add [-1, 1]. 5. Pointers cross. Stop. 6. All pairs are unique and valid.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= target <= 10^9
- The answer is guaranteed to fit in a 32-bit integer.
Optimal Approach & Strategy
Use a hash set to store seen numbers; for each element, check if its complement exists, record the ordered pair, and then insert the element into the set.
Brute Force Approach
Iterate over every possible i < j pair, compute nums[i] + nums[j], and collect those equal to target, then deduplicate.
Verified Code Solutions
function solution(nums, target) { let result = []; for (let i = 0; i < nums.length; i++) { for (let j = i + 1; j < nums.length; j++) { if (nums[i] + nums[j] === target) { let pair = [nums[i], nums[j]].sort((a, b) => a - b); if (!result.includes(pair.toString())) { result.push(pair); } } } } return result; }class Solution { public: vector<vector<int>> solution(vector<int>& nums, int target) { vector<vector<int>> result; sort(nums.begin(), nums.end()); int left = 0, right = nums.size() - 1; while (left < right) { int sum = nums[left] + nums[right]; if (sum == target) { result.push_back({nums[left], nums[right]}); left++; right--; } else if (sum < target) { left++; } else { right--; } } return result; } };import java.util.Arrays; public class Solution { public int[][] solution(int[] nums, int target) { Arrays.sort(nums); int left = 0, right = nums.length - 1; int[][] result = new int[nums.length][2]; int count = 0; while (left < right) { int sum = nums[left] + nums[right]; if (sum == target) { result[count][0] = nums[left]; result[count][1] = nums[right]; count++; left++; right--; } else if (sum < target) { left++; } else { right--; } } int[][] finalResult = new int[count][2]; System.arraycopy(result, 0, finalResult, 0, count); return finalResult; } }def solution(nums, target): result = [] for i in range(len(nums)): for j in range(i + 1, len(nums)): if nums[i] + nums[j] == target: pair = sorted([nums[i], nums[j]]) if str(pair) not in [str(x) for x in result]: result.append(pair) return resultfunction solution(nums, target) { let result = []; for (let i = 0; i < nums.length; i++) { for (let j = i + 1; j < nums.length; j++) { if (nums[i] + nums[j] === target) { let pair = [nums[i], nums[j]].sort((a, b) => a - b); if (!result.includes(pair.toString())) { result.push(pair); } } } } return result; }Asked in Top Tech Interviews
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