Leetcode -- Day13

Question 1

Add Binary

Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".

Firslty I wirte a very direct and kind of brute forch method. Just consider each case of sum 1, 0, 2, 3 for aNum + bNum + carry by switch case.

public class Solution {
    
    String result = "";
    int carry = 0;
    
    public String addBinary(String a, String b) {
        if (a.length() == 0 )
            return b;
        else if (b.length() == 0)
            return a;
        
        
        int alen = a.length()-1;
        int blen = b.length()-1;
        
        
        while (alen >= 0 && blen >=0){
            int aNum = a.charAt(alen) - '0';
            int bNum = b.charAt(blen) - '0';
            calculator(aNum + bNum + carry);
            alen --;
            blen --;
        }
        
        while (alen >= 0){
            int aNum = a.charAt(alen) - '0';
            calculator(aNum + carry);
            alen --;
        }
        
        while (blen >= 0){
            int bNum = b.charAt(blen) - '0';
            calculator(bNum + carry);
            blen --;
        }
        
        if (carry == 1)
            result = "1" + result;
        
        return result;
    }
    
    public void calculator(int sum){
        switch(sum){
            case 0:
                result = "0" + result;
                break;
            case 1:
                result = "1" + result;
                carry = 0;
                break;
            case 2:
                result = "0" + result;
                carry = 1;
                break;
            case 3:
                result = "1" + result;
                carry = 1;
                break;
        }
    }
}

The other way does not use switch case but use %2 or /2 to defind carry and sum, which is much clear and shorter. 

public String addBinary(String a, String b) {
        if (a.length() == 0 )
            return b;
        else if (b.length() == 0)
            return a;
            
        String result = "";
        int carry = 0;
        
        int len = Math.max(a.length(), b.length());
        
        for (int i = 0; i < len; i ++){
            int aNum = 0;
            int bNum = 0;
            if (i < a.length())
                aNum = a.charAt(a.length() - 1 - i) - '0';
            if (i < b.length())
                bNum = b.charAt(b.length() - 1 - i) - '0';
            
            int sum = aNum + bNum + carry;
            result = sum%2 + result;
            carry = sum / 2;
        }
        
        if (carry == 1)
            result = "1" + result;
        
        return result;
        
    }

n-sum problem 

Question 2

Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

solution 1: Brute force. O(n*n)

    public int[] twoSum(int[] nums, int target) {
        if(nums.length == 0)
            return null;
       
        for (int i = 0; i < nums.length; i ++){
            for (int j = i+1; j < nums.length; j ++){
                if (nums[i] + nums[j] == target){
                    int[] result = new int[]{i+1, j+1};
                    return result;
                }
            }
        }
        return null;
    }

solution 2: sort and binary search. O(n+nlogn+n+n) = O(nlogn)

public int[] twoSum(int[] nums, int target) {
        if(nums.length == 0)
            return null;
            
        int[] result = new int[2];
       
        int[] copyNums = new int[nums.length];
        System.arraycopy(nums, 0, copyNums, 0, nums.length);  
        Arrays.sort(copyNums);
        
        int low = 0; 
        int high = copyNums.length-1;
        
        while (low < high){
            int sum = copyNums[low] + copyNums[high];
            if (sum == target){
                result[0] = copyNums[low];
                result[1] = copyNums[high];
                break;
            }
            else if (sum < target)
                low ++;
            else
                high --;
        }
        
        if (result.length == 0)
            return null;
            
        else{
            int index1 = -1, index2 = -1;
            for (int i = 0; i < nums.length; i ++){
                if (result[0] == nums[i] && index1 == -1)
                    index1 = i + 1;
                else if (result[1] == nums[i] && index2 == -1)
                    index2 = i + 1;
            }
            
            result[0] = index1;
            result[1] = index2;
            Arrays.sort(result);
            return result;
        }
    }

solution 2: hashmap. O(n)

    public int[] twoSum(int[] nums, int target) {
        if(nums.length < 2 || nums == null)
            return null;
            
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        int[] res = new int[2];
        
        for (int i = 0; i < nums.length; i ++){
            if (!map.containsKey(target-nums[i]))
                map.put(nums[i], i + 1);
            else{
                res[0] = map.get(target-nums[i]);
                res[1] = i + 1;
                break;
            }
        }
        return res;
    }

 Question 3

3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},
    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

 I feel this is much more difficult than the before 2 sum one. We need three pointers here. And I found a trick, with which you did not write tree times list.add(element).

Arrays.asList(a, b, c) 

    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> triples = new ArrayList();
        if (nums.length < 3)
            return triples;
        
        Arrays.sort(nums);
        int i = 0, last = nums.length - 1;
        while (i < last) {
            int a = nums[i], j = i+1, k = last;
            while (j < k) {
                int b = nums[j], c = nums[k], sum = a+b+c;
                if (sum == 0) triples.add(Arrays.asList(a, b, c));
                if (sum <= 0) while (nums[j] == b && j < k) j++;
                if (sum >= 0) while (nums[k] == c && j < k) k--;
            }
            while (nums[i] == a && i < last) i++;
        }
        return triples;
    }

Question 4

3Sum Closest

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

public int threeSumClosest(int[] nums, int target) {
        if (nums.length < 3)
            return Integer.MAX_VALUE;
        
        Arrays.sort(nums);
        int result = nums[0] + nums[1] + nums[2];
            
        int i = 0;
        int last = nums.length - 1;
        while (i < last){
            int a = nums[i];
            int j = i + 1;
            int k = last;
            while (j < k){
                int b = nums[j];
                int c = nums[k];
                int sum = a + b + c;
                int diff = Math.abs(target - sum);
                
                if (diff < Math.abs(target - result)){
                    result = sum;
                }
                
                if (diff == 0)
                    return sum;
                else if (sum > target){
                    while (nums[k] == c && j < k)
                        k --;
                }
                else if (sum < target){
                    while (nums[j] == b && j < k)
                        j ++;
                }
            }
            while (nums[i] == a && i < k)
                i ++;
        }
        return result;
        
    }