Skip to content

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

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 \(O(1)\).

1
2
3
4
5
6
7
8
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]

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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};
            }
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
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};
            }
        }
    }
};

1
2
3
4
5
6
7
8
9
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}
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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];
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
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]
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * @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 \(O(1)\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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;
            }
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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;
            }
        }
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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--
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
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;
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
/**
 * @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;
        }
    }
};

Comments