167. Two Sum II  Input Array Is Sorted 🔗
Date: Jun 02, 2024
Tags: twopointers
Description
Given a 1indexed array of integers
numbers
that is already sorted in
nondecreasing order, find two numbers such that
they add up to a specific target
number. Let these
two numbers be numbers[index_{1}]
and
numbers[index_{2}]
where 1 <=
index_{1} < index_{2} <=
numbers.length
.
Return the indices of the two numbers,
index_{1}
and
index_{2}
, added by
one as an integer array [index_{1},
index_{2}]
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, index_{1} = 1, index_{2} = 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 index_{1} = 1, index_{2} = 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 index_{1} = 1, index_{2} = 2. We return [1, 2].
Constraints:

2 <= numbers.length <= 3 * 10^{4}

1000 <= numbers[i] <= 1000

numbers
is sorted in nondecreasing 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 };
BruteForce
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 };