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LeetCode

Approach 1: Brute Force

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class Solution {
    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[j] == target - nums[i]) {
                    return new int[] { i, j };
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

Approach2: One-pass Hash Table, Time complexity : O(n)

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class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer,Integer> map=new HashMap<>();
        for(int i=0;i<nums.length;i++){
            int complement=target-nums[i];
            if(map.containsKey(complement)) return new int[]{map.get(complement),i};
            map.put(nums[i],i);
        }
        return null;
    }
}

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