# My accepted java code with O(NlgN) complexity without using java default sort or hash map

• This piece of code is overly complex (I always do that) but somehow accepted by the system. I think the complexity is O(NLgN). There are more elegant code available in the forum. I posted here for your thought and discussion.

Basically the concept is

1. Copy the original array to a temporary array
2. Sort the copied array using MergeSort constructed (tried QuickSort, if the array is pre-sorted, the time limit will be reached)
3. Find location of the value which is smaller than target (assume it is m) and find the location of the index of half of target (ms)
4. start from ms (most likely will be in the middle in normal case) add values from both side to match target.

I know, it is overly complex, just for your thought.

``````public class Solution {
public int[] twoSum(int[] numbers, int target) {

int[] result = new int[2];

//option 1: brutal force (O(N^2))
/*
for(int i=0; i<numbers.length; i++){
if(numbers[i]>=target)
continue;

for(int j=i+1; j<numbers.length; j++){
if(numbers[j]>=target-numbers[i])
continue;

if(numbers[i]+numbers[j]==target){
result[0]=i+1;
result[1]=j+1;
return result;
}
}
}*/

//option 2:
int size = numbers.length;
int[] numCopies = new int[size];
for(int i=0; i<size; i++)           //Cost N
numCopies[i] = numbers[i];

//QuickSort(numCopies, 0, size-1);    //Cost O(NlgN) //Worst Case O(N^2)
int[] aux = new int[size];
MergeSort(numCopies, aux, 0, size-1); //Worst Case

int m = FindIndex(numCopies, target, 0, size-1);      //O(lgN)
int ms = FindIndex(numCopies, (int)target/2, 0, m);   //O(lgN)

int i=ms;
int j=ms+1;
while(true){                            //O(N)
if(i<0||j>m) break;
int tmp = numCopies[i]+numCopies[j];
if(tmp ==target) break;
else if(tmp>target) i--;
else j++;
}

boolean indexOneSet = false;
boolean indexTwoSet = false;
for(int index=0;index <size; index++){  //O(N)
if(!indexOneSet && (numbers[index]==numCopies[i]||numbers[index]==numCopies[j]))
{
result[0] = index+1;
indexOneSet = true;
}

else if(!indexTwoSet && (numbers[index]==numCopies[i]||numbers[index]==numCopies[j]))
{
result[1] = index+1;
indexTwoSet = true;
break;
}
}
return result;
}

//Binary Search
public int FindIndex(int[] numbers, int n, int start, int end){
if(start==end) return start;

int mid = (int)((start+end)/2);

if(numbers[mid]==n) return mid;
else if (numbers[mid] < n && numbers[mid + 1] > n) return mid;
else if(numbers[mid]>n) return FindIndex(numbers, n, start, mid);
else return FindIndex(numbers, n, mid+1, end);
}

//MergeSort
public void MergeSort(int[] numbers, int[] aux, int start, int end){
if(start==end) return;

int mid = (int) (start+end)/2 ;
MergeSort(numbers, aux, start, mid);
MergeSort(numbers, aux, mid+1, end);

for(int i=start; i<=end; i++)
aux[i] = numbers[i];

int i= start;
int j= mid+1;

for(int n=start; n<=end; n++){
if(i>mid) numbers[n] = aux[j++];
else if(j>end) numbers[n] = aux[i++];
else if(aux[i]<aux[j]) numbers[n] = aux[i++];
else numbers[n] = aux[j++];
}
}

//QuickSort
public void QuickSort(int[] numbers,int left, int right)
{
if(left>=right) return;
if(left+1==right){
if(numbers[left]>numbers[right])
Swap(numbers,left, right);
return;
}

int i=left+1;
int j=right;
while(i<j){
while(numbers[i]<=numbers[left] && i<j) i++;
while(numbers[j]>numbers[left] && j>i) j--;
Swap(numbers,i, j);
}
Swap(numbers,left, i);
QuickSort(numbers, left, i-1);
QuickSort(numbers, i+1, right);
}

public void Swap(int[] numbers, int i, int j){
int tmp = numbers[i];
numbers[i]=numbers[j];
numbers[j]=tmp;
}

}``````

• IMHO, there's no point in implementing merge sort by your self, as the Arrays.sort() already does that for you. Not using STL, however, brings little benefit but more chance for error.

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