Leetcode 167 Two Sum II – Input Array Is Sorted

Leetcode 167 Two Sum II – Input Array Is Sorted

Two Sum II – Input Array Is Sorted – LeetCode

Two Sum II – Input Array Is Sorted

Problem Statement

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.

Your solution must use only constant extra space.

Examples

Example 1:

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].

Example 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].

Example 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.

Solution

Since the array is sorted, we can simply use 2 pointers to solve it.

1. Initialize Two Pointers: Set one pointer at the beginning (left) and one at the end (right) of the array.
2. Check the Sum: Calculate the sum of the numbers at these two pointers.
3. Adjust Pointers:
   • If the sum is equal to the target, return the indices (1-based index).
   • If the sum is less than the target, move the left pointer to the right to increase the sum.
   • If the sum is greater than the target, move the right pointer to the left to decrease the sum.
4. Repeat: Continue this process until you find the target sum or the pointers cross.

Python:

def twoSum(numbers, target):
    left = 0
    right = len(numbers) - 1
    
    while left < right:
        current_sum = numbers[left] + numbers[right]
        
        if current_sum == target:
            return [left + 1, right + 1]  # Return 1-based indices
        elif current_sum < target:
            left += 1
        else:
            right -= 1
    
    return []

Time Complexity:
The time complexity is O(n)since we are visiting all the elements in the array at most once

Space Complexity:
The space complexity of this brute force approach is O(1), as we are not using any additional data structures.

Java:

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int left = 0;
        int right = numbers.length - 1;
        
        while (left < right) {
            int currentSum = numbers[left] + numbers[right];
            
            if (currentSum == target) {
                return new int[]{left + 1, right + 1};  // Return 1-based indices
            } else if (currentSum < target) {
                left++;
            } else {
                right--;
            }
        }
        
        return new int[]{-1, -1};  // In case no solution is found
    }
}