LeetCode challenge – Two Sum – Java Solution
This is a solution for the LeetCode challenge – Two Sum written in Java ( Source )
Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Solution: O(n log n). Wrap index and element in a class and sort in increasing order. Do a two pointer sum and compare. An alternative solution is to use hashing which is a O(n) solution – For each element e check if element (target – e) is already found in hashset, if yes return their index, else add this to hash-set and continue.
import java.util.ArrayList;
import java.util.List;
public class TwoSum {
class NumIndex {
int i, e;
NumIndex(int i, int e) {
this.i = i;
this.e = e;
}
}
public static void main(String[] args) {
int[] nums = {3, 2, 4};
int[] ans = new TwoSum().twoSum(nums, 6);
for (int i : ans) System.out.println(i);
}
public int[] twoSum(int[] nums, int target) {
List<NumIndex> list = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
NumIndex n = new NumIndex(i, nums[i]);
list.add(n);
}
list.sort((o1, o2) -> Integer.compare(o1.e, o2.e));
int[] ans = new int[2];
for (int i = 0, j = nums.length - 1; i < j; ) {
NumIndex numi = list.get(i);
NumIndex numj = list.get(j);
int sum = numi.e + numj.e;
if (sum == target) {
ans[0] = numi.i;
ans[1] = numj.i;
return ans;
} else if (sum > target) {
j--;
} else i++;
}
return ans;
}
}
