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167. Two Sum II - Input Array Is Sorted

Description

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.

 

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.

Solutions

Solution 1: Binary Search

Note that the array is sorted in non-decreasing order, so for each numbers[i], we can find the position of target - numbers[i] by binary search, and return $[i + 1, j + 1]$ if it exists.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array numbers. The space complexity is $O(1)$.

Solution 2: Two Pointers

We define two pointers $i$ and $j$, which point to the first element and the last element of the array respectively. Each time we calculate $numbers[i] + numbers[j]$. If the sum is equal to the target value, return $[i + 1, j + 1]$ directly. If the sum is less than the target value, move $i$ to the right by one position, and if the sum is greater than the target value, move $j$ to the left by one position.

The time complexity is $O(n)$, where $n$ is the length of the array numbers. The space complexity is $O(1)$.

  • class Solution {
    public int[] twoSum(int[] numbers, int target) {
        for (int i = 0, j = numbers.length - 1;;) {
            int x = numbers[i] + numbers[j];
            if (x == target) {
                return new int[] {i + 1, j + 1};
            }
            if (x < target) {
                ++i;
            } else {
                --j;
            }
        }
    }
}

  • class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        for (int i = 0, j = numbers.size() - 1;;) {
            int x = numbers[i] + numbers[j];
            if (x == target) {
                return {i + 1, j + 1};
            }
            if (x < target) {
                ++i;
            } else {
                --j;
            }
        }
    }
};

  • class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        i, j = 0, len(numbers) - 1
        while i < j:
            x = numbers[i] + numbers[j]
            if x == target:
                return [i + 1, j + 1]
            if x < target:
                i += 1
            else:
                j -= 1


  • func twoSum(numbers []int, target int) []int {
	for i, j := 0, len(numbers)-1; ; {
		x := numbers[i] + numbers[j]
		if x == target {
			return []int{i + 1, j + 1}
		}
		if x < target {
			i++
		} else {
			j--
		}
	}
}

  • function twoSum(numbers: number[], target: number): number[] {
    for (let i = 0, j = numbers.length - 1; ; ) {
        const x = numbers[i] + numbers[j];
        if (x === target) {
            return [i + 1, j + 1];
        }
        if (x < target) {
            ++i;
        } else {
            --j;
        }
    }
}


  • /**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (numbers, target) {
    for (let i = 0, j = numbers.length - 1; ; ) {
        const x = numbers[i] + numbers[j];
        if (x === target) {
            return [i + 1, j + 1];
        }
        if (x < target) {
            ++i;
        } else {
            --j;
        }
    }
};


  • use std::cmp::Ordering;

impl Solution {
    pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
        let n = numbers.len();
        let mut l = 0;
        let mut r = n - 1;
        loop {
            match (numbers[l] + numbers[r]).cmp(&target) {
                Ordering::Less => {
                    l += 1;
                }
                Ordering::Greater => {
                    r -= 1;
                }
                Ordering::Equal => {
                    break;
                }
            }
        }
        vec![(l as i32) + 1, (r as i32) + 1]
    }
}

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