Two Pointers

3-Sum

medium

DESCRIPTION (inspired by Leetcode.com)

Given an input integer array nums, write a function to find all unique triplets [nums[i], nums[j], nums[k]] such that i, j, and k are distinct indices, and the sum of nums[i], nums[j], and nums[k] equals zero. Ensure that the resulting list does not contain any duplicate triplets.

Input:

nums = [-1,0,1,2,-1,-1]

Output:

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

Explanation: Both nums[0], nums[1], nums[2] and nums[1], nums[2], nums[4] both include [-1, 0, 1] and sum to 0. nums[0], nums[3], nums[4] ([-1,-1,2]) also sum to 0.

Since we are looking for unique triplets, we can ignore the duplicate [-1, 0, 1] triplet and return [[-1, -1, 2], [-1, 0, 1]].

The order of the triplets and the order of the elements within the triplets do not matter.

Code Editor

Java

public class Solution {
    public List&amp;lt;List&amp;lt;Integer&amp;gt;&amp;gt; threeSum(int[] nums) {
        // Your code goes here
    }
}

Run your code to see results here

Have suggestions or found something wrong?

Solution

Visualization

Java

Try these examples:


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

3 sum

0 / 18

Explanation

We can leverage the two-pointer technique to solve this problem by first sorting the array. We can then iterate through each element in the array. The problem then reduces to finding two numbers in the rest of the array that sum to the negative of the current element, which follows the same logic as the Two Sum (Sorted Array) problem from the overview.

Result

Since our first triplet sums to 0, we can add it to our result set.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

3 sum

0 / 5

Avoiding Duplicates

As soon as we find a triplet that sums to 0, we can add it to our result set. We then have to move our left and right pointers to look for the next triplet while avoiding duplicate triplets. We can do this by moving the left and right pointers until they point to different numbers than the ones they were pointing to before.

Here we move the left pointer once until it reaches the last -1 in the array. Then, we can move both the left and right pointers so that they both point to new numbers.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

add triplet to output

0 / 2

Here we can do another iteration of the Two Sum problem using the new positions of the left and right pointers.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

move both pointers

0 / 3

At this point our left and right pointers have crossed, so we can move our iterator to the next number in the array.

Avoiding Duplicates II

In this case, since the next number in the array is the same as the previous number, we can skip it. We can do this by moving our iterator until it points to a new number.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

move both pointers

0 / 3

And we're ready to start the Two Sum algorithm again, so we reset our left and right pointers, and start the algorithm.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

initialize pointers

0 / 2

Termination

Our algorithm terminates when i reaches the 3rd to last element in the array (i.e., i < n - 2). This is because we need at least 2 more elements after i for left and right to form a triplet.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

move right pointer backward

0 / 2

What is the time complexity of this solution?

O(m * n)

O(n²)

O(n³)

O(n * logn)

Checking progress…

Two Pointers

3-Sum

medium

DESCRIPTION (inspired by Leetcode.com)

Input:

nums = [-1,0,1,2,-1,-1]

Output:

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

Explanation: Both nums[0], nums[1], nums[2] and nums[1], nums[2], nums[4] both include [-1, 0, 1] and sum to 0. nums[0], nums[3], nums[4] ([-1,-1,2]) also sum to 0.

Since we are looking for unique triplets, we can ignore the duplicate [-1, 0, 1] triplet and return [[-1, -1, 2], [-1, 0, 1]].

The order of the triplets and the order of the elements within the triplets do not matter.

Code Editor

Java

public class Solution {
    public List&amp;lt;List&amp;lt;Integer&amp;gt;&amp;gt; threeSum(int[] nums) {
        // Your code goes here
    }
}

Run your code to see results here

Have suggestions or found something wrong?

Solution

Visualization

Java

Try these examples:


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

3 sum

0 / 18

Explanation

Result

Since our first triplet sums to 0, we can add it to our result set.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

3 sum

0 / 5

Avoiding Duplicates

Here we move the left pointer once until it reaches the last -1 in the array. Then, we can move both the left and right pointers so that they both point to new numbers.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

add triplet to output

0 / 2

Here we can do another iteration of the Two Sum problem using the new positions of the left and right pointers.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

move both pointers

0 / 3

At this point our left and right pointers have crossed, so we can move our iterator to the next number in the array.

Avoiding Duplicates II

In this case, since the next number in the array is the same as the previous number, we can skip it. We can do this by moving our iterator until it points to a new number.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

move both pointers

0 / 3

And we're ready to start the Two Sum algorithm again, so we reset our left and right pointers, and start the algorithm.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

initialize pointers

0 / 2

Termination

Our algorithm terminates when i reaches the 3rd to last element in the array (i.e., i < n - 2). This is because we need at least 2 more elements after i for left and right to form a triplet.

Visualization

Java


public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        // Sort array to enable two-pointer technique and handle duplicates
        Arrays.sort(nums);
        
        // Fix the first element and use two pointers for the remaining two
        for (int i = 0; i < nums.length - 2; i++) {
            // Skip duplicate values for the first element to avoid duplicate triplets
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            // Initialize two pointers for the remaining subarray
            int left = i + 1;
            int right = nums.length - 1;
            
            // Use two-pointer technique to find pairs that sum to -nums[i]
            while (left < right) {
                int total = nums[i] + nums[left] + nums[right];
                
                if (total < 0) {
                    // Sum too small, move left pointer right to increase sum
                    left++;
                } else if (total > 0) {
                    // Sum too large, move right pointer left to decrease sum
                    right--;
                } else {
                    // Found a valid triplet
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    
                    // Skip all duplicate values to avoid duplicate triplets
                    while (left < right && nums[left] == nums[left + 1]) {
                        left++;
                    }
                    while (left < right && nums[right] == nums[right - 1]) {
                        right--;
                    }
                    // Move both pointers to continue searching
                    left++;
                    right--;
                }
            }
        }
        return result;
    }
}

Result

[-1, -1, 2]

[-1, 0, 1]

move right pointer backward

0 / 2

What is the time complexity of this solution?

O(m * n)

O(n²)

O(n³)

O(n * logn)

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3-Sum

DESCRIPTION (inspired by Leetcode.com)

Solution

Result

Explanation

Result

Result

Avoiding Duplicates

Result

Result

Avoiding Duplicates II

Result

Result

Termination

Result

What is the time complexity of this solution?

Top pages

Top content

3-Sum

DESCRIPTION (inspired by Leetcode.com)

Solution

Result

Explanation

Result

Result

Avoiding Duplicates

Result

Result

Avoiding Duplicates II

Result

Result

Termination

Result

What is the time complexity of this solution?