167. Two Sum II - Input Array Is Sorted 🔗
Date: Jun 02, 2024
Tags: two-pointers
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 -
numbersis sorted in non-decreasing order. -
-1000 <= target <= 1000 - The tests are generated such that there is exactly one solution.
Optimal
function twoSum(numbers: number[], target: number): number[] {
let l = 0;
let r = numbers.length - 1;
while (l < r) {
const sum = numbers[l] + numbers[r];
if (sum === target) return [l + 1, r + 1];
if (sum > target) r--;
if (sum < target) l++;
}
throw new Error('No Two Sum solution');
}
export { twoSum };
Brute-Force
function twoSum(numbers: number[], target: number): number[] {
let result: number[] = [];
for (let x = 0; x < numbers.length; x++) {
for (let y = x + 1; y < numbers.length; y++) {
if (numbers[x] + numbers[y] === target) {
result = [x + 1, y + 1];
break;
}
}
}
return result;
}
export { twoSum };
