Two Sum, 3Sum, and 4Sum

Two Sum (LeetCode 1)

Using a Hash Map allows for finding the complement in O(1), leading to a total time complexity of O(n):

int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{};
}

Variant: Two Sum II - Input Array Is Sorted (LeetCode 167): Use the two-pointer approach since the array is sorted.

int[] twoSumSorted(int[] numbers, int target) {
    int l = 0, r = numbers.length - 1;
    while (l < r) {
        int sum = numbers[l] + numbers[r];
        if (sum == target) return new int[]{l + 1, r + 1};
        if (sum < target) l++; 
        else r--;
    }
    return new int[]{};
}

3Sum (LeetCode 15)

Sort the array first, then iterate with one fixed number and utilize the two-pointer technique for the remaining two numbers. Focus on skipping duplicates to ensure unique triplets.

List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
    for (int i = 0; i < nums.length - 2; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;   // Deduplicate pivot
        if (nums[i] > 0) break;                            // Optimization: sum can't be 0 if smallest is > 0
        
        int l = i + 1, r = nums.length - 1;
        while (l < r) {
            int sum = nums[i] + nums[l] + nums[r];
            if (sum == 0) {
                res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                while (l < r && nums[l] == nums[l + 1]) l++;  // Deduplicate left
                while (l < r && nums[r] == nums[r - 1]) r--;  // Deduplicate right
                l++; r--;
            } else if (sum < 0) l++;
            else r--;
        }
    }
    return res;
}

3Sum Closest (LeetCode 16)

int threeSumClosest(int[] nums, int target) {
    Arrays.sort(nums);
    int closest = nums[0] + nums[1] + nums[2];
    for (int i = 0; i < nums.length - 2; i++) {
        int l = i + 1, r = nums.length - 1;
        while (l < r) {
            int sum = nums[i] + nums[l] + nums[r];
            if (Math.abs(sum - target) < Math.abs(closest - target)) closest = sum;
            if (sum < target) l++;
            else if (sum > target) r--;
            else return sum; // Found exact match
        }
    }
    return closest;
}

4Sum (LeetCode 18)

Append another layer of iteration; the two-pointer logic remains the same.

List<List<Integer>> fourSum(int[] nums, int target) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
    int n = nums.length;
    for (int i = 0; i < n - 3; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;
        for (int j = i + 1; j < n - 2; j++) {
            if (j > i + 1 && nums[j] == nums[j - 1]) continue;
            int l = j + 1, r = n - 1;
            while (l < r) {
                // Use long to avoid overflow when summing four numbers
                long sum = (long) nums[i] + nums[j] + nums[l] + nums[r];
                if (sum == target) {
                    res.add(Arrays.asList(nums[i], nums[j], nums[l], nums[r]));
                    while (l < r && nums[l] == nums[l + 1]) l++;
                    while (l < r && nums[r] == nums[r - 1]) r--;
                    l++; r--;
                } else if (sum < target) l++;
                else r--;
            }
        }
    }
    return res;
}

Note: Be careful with potential integer overflows in 4Sum; use long for the sum calculation.

Summary