0167 - Two Sum II - Input Array Is Sorted

0167 - Two Sum II - Input Array Is Sorted

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Examples

Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints

2 <= numbers.length <= 3 * 104 -1000 <= numbers[i] <= 1000 numbers is sorted in non-decreasing order. -1000 <= target <= 1000 The tests are generated such that there is exactly one solution.

Java Solution (Two Pointer)

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int p1 = 0;
        int p2 = numbers.length - 1;
        
        while (p1 < p2) {
            if(numbers[p1] + numbers[p2] == target) return new int[]{++p1, ++p2};
            else if(numbers[p1] + numbers[p2] > target) p2--;
            else p1++;
        }
        return new int[] {};
    }
}

Java Solution (Hashmap)

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        Map<Integer, Integer> hm = new HashMap<>();
        
        for(int i = 0; i < numbers.length; i++) {
            if(hm.containsKey(numbers[i])) return new int[] {hm.get(numbers[i])+1, i+1};
            else hm.put(target - numbers[i], i);
        }
        return new int[]{};
    }
}