class Solution {

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

/**

1. understanding

- find the indices where the numbers of that index pair sum upto the target number.

- exactly one solution

- use each element only once

- return in any order

2. strategy

- brute force:

- validate for all pair sum

- hashtable:

- assing hashtable variable to save the nums and index

- while iterate over the nums,

- calculate the diff, and check if the diff in the hashtable.

- or save new entry.

3. complexity

- time: O(N)

- space: O(N)

*/

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

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

int diff = target - nums[i];

if (map.containsKey(diff)) {

return new int[] {map.get(diff), i};

}

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

}

return new int[] {};

}

}