Question

167. Two Sum II - Input Array Is Sorted

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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 $O(1)$.

class Solution:
  def twoSum(self, numbers: List[int], target: int) -> List[int]:
    n = len(numbers)
    for i in range(n - 1):
      x = target - numbers[i]
      j = bisect_left(numbers, x, lo=i + 1)
      if j < n and numbers[j] == x:
        return [i + 1, j + 1]

class Solution {
  public int[] twoSum(int[] numbers, int target) {
    for (int i = 0, n = numbers.length;; ++i) {
      int x = target - numbers[i];
      int l = i + 1, r = n - 1;
      while (l < r) {
        int mid = (l + r) >> 1;
        if (numbers[mid] >= x) {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      if (numbers[l] == x) {
        return new int[] {i + 1, l + 1};
      }
    }
  }
}

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

func twoSum(numbers []int, target int) []int {
  for i, n := 0, len(numbers); ; i++ {
    x := target - numbers[i]
    j := sort.SearchInts(numbers[i+1:], x) + i + 1
    if j < n && numbers[j] == x {
      return []int{i + 1, j + 1}
    }
  }
}

function twoSum(numbers: number[], target: number): number[] {
  const n = numbers.length;
  for (let i = 0; ; ++i) {
    const x = target - numbers[i];
    let l = i + 1;
    let r = n - 1;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (numbers[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    if (numbers[l] === x) {
      return [i + 1, l + 1];
    }
  }
}

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]
  }
}

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (numbers, target) {
  const n = numbers.length;
  for (let i = 0; ; ++i) {
    const x = target - numbers[i];
    let l = i + 1;
    let r = n - 1;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (numbers[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    if (numbers[l] === x) {
      return [i + 1, l + 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:
  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

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;
      }
    }
  }
};

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;
    }
  }
};