[/dsa/leetcode]

167 - Two Sum II - Input Array Is Sorted

Wednesday, 16 September 2026

<medium> [problem]

Two pointers from both ends of the sorted array, dropping whichever end can no longer be part of the pair.

{array}{two pointers}

Problem

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.

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]

Input: numbers = [2,3,4], target = 6
Output: [1,3]

Input: numbers = [-1,0], target = -1
Output: [1,2]

Constraints:

  • 2 <= numbers.length <= 3 * 10^4
  • -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.

Approach

Start with l at the first element and r at the last, and look at sum = numbers[l] + numbers[r].

  • sum == target: found it, return {l + 1, r + 1}.
  • sum < target: numbers[r] is the largest value still in range, and even that isn’t enough for numbers[l]. So numbers[l] can’t be in the answer with any element. Drop it: l++.
  • sum > target: numbers[l] is the smallest value still in range, and even that is too much for numbers[r]. So numbers[r] can’t be in the answer. Drop it: r--.
numbers = [2, 7, 11, 15], target = 9
l = 0, r = 3  ->  2 + 15 = 17 > 9  ->  r--
l = 0, r = 2  ->  2 + 11 = 13 > 9  ->  r--
l = 0, r = 1  ->  2 + 7  = 9       ->  return [1, 2]
class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int l = 0, r = numbers.size() - 1;
        while (l < r) {
            int sum = numbers[l] + numbers[r];
            if (sum == target) return {l + 1, r + 1};
            if (sum < target) l++;
            else r--;
        }
        return {};
    }
};
  • time: O(n) — every step moves one pointer inward, so at most n - 1 steps
  • space: O(1) — just the two pointers
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