N sum Question

N sum question from leetcode and lintcode

1 two sum

(1) Use HashMap to store (nums[i] , index); time O(n) space O(n)

one loop scan to find in map target - num[i] , if exist and index2 != index1( this is important for [3,2,4]  return [2,4] not [1,1] ) , return;

else not find

(2) two pointer, but need to store original index

use two pointer , we need to sort array firstly, but the problem is that we need return index not the value.

Solution: this has different way to implement, can write a pair class or copy original array to store index

My code here use array copy idea

// version one

 public int[] twoSum(int[] numbers, int target) {
       int res[] = new int[2];
       int sortarr[] = new int[numbers.length];
       sortarr = numbers.clone();
       
       Arrays.sort(sortarr);
       
       //two pointer
       int start = 0; 
       int end = numbers.length -1;
       while(start + 1 < end){
           if(sortarr[start] + sortarr[end] == target){
               break;
           }else if(sortarr[start] + sortarr[end] < target){
               start++;
           }else{
               end--;
           }
       }
       if(sortarr[start] + sortarr[end] == target || (start + 1 != end)){
          //find index , in case same value
          for(int i = 0; i< numbers.length; i++){
              if(numbers[i] == sortarr[start] && res[0] == 0){// res[0] not set yet
                  res[0] = i + 1;
              }else if(numbers[i] == sortarr[end]){
                  res[1] = i + 1;
              }
          }
       }
       Arrays.sort(res);
       return res;
    }

//code use extra class to implement 

3 Sum

find three numbers a + b + c = 0 in an array here may exists duplicate numbers need to filter duplicates

 [-2, 0, 0, 2,2], the expected output should be [-2,0,2]. 
 If no filter here, the output will be duplicate as [-2,0,2] and [-2,0,2]

Solution :

Two pointer

one loop get numbers[i] ( skip duplicate) , then use two pointer start = i+ 1, end  = n -1; 

find triplets num[i]  + num[start] + num[ end] == 0 

(if sum == 0) start++, end-- (Here: we also need skip duplicate numbers)

Code

3 Sum Closet

just return closet sum value to target

similar to 3 Sum Question, just keep a variable min to record closet sum

update if( |target - sum | < |target - min| ) min = sum;

One need to pay attention here, is we can assign res = Integer.MAX_VALUE / 2 to avoid overflow when | target - min| 

Code

4 Sum

find four number whose sum = 0,

similar to 3 sum question, but here we need fix two number: num[i] , num[ j ] (j start from i + 1), then use two pointer to locate another two number

K sum

extend the question to k  number

(i  )If will only need to return the number of possible solutions, this would be a DP Question.

f[ i ][ j ][ t ] 前 i 個數裏， 挑出 j 個數， 組合和爲t 有多少種方案

f[ i ][ j ][ t ] = f[ i-1 ][ j-1 ][ t- A[i]] + f[ i-1][ j ][ t ];
return f[ n ][ k ][ target ]

initial :
 f[0][0][0] = 1;
 f[0][>0][?] = 0;

(ii) if return all result list, extend from solution for 4 sum