import java.util.*;

public class twoSum {

public static void main(String[] args){

twoSum test = new twoSum();

System.out.println(Arrays.toString(test.twoSum(new int[]{1,2,7,9},9)));

System.out.println(Arrays.toString(test.twoSumv2(new int[]{1,2,7,9},9)));

}

// Approach 1: one-pass HashMap

/* While we iterate elements in the array, we look back whether current element's complement exists.

Use a hashmap to help.

Time complexity: O(n)

Space complexity: O(n)

* */

private int[] twoSum(int[] nums, int target){

Map <Integer, Integer> map = new HashMap<>();

for(int i = 0; i < nums.length; i++){

if(map.containsKey(target - nums[i])){

return new int[] {map.get(target - nums[i]), i};

}else{

map.put(nums[i],i);

}

}

throw new IllegalArgumentException("No valid solutions");

}

// Approach 2: two pointers

/* Sort array first. Then use two pointers.

Time complexity: O(nlogn)

Space complesity: O(1)

* */

private int[] twoSumv2(int[] nums, int target){

Arrays.sort(nums);

int lo = 0, hi = nums.length-1;

while(lo < hi){

if(nums[lo] + nums[hi] == target){

return new int[] {lo, hi};

}

if(nums[lo] + nums[hi] < target){

lo ++;

}else{

hi --;

}

}

throw new IllegalArgumentException("No valid solutions");

}

}