ACAlgoGeeks

1left = 0, right = n - 1   → left = 0, right = 3
2while left < right:
3  sum = arr[left] + arr[right]   2 + 15 = 17
4  if sum == target: return [left, right]   
5  if sum < target: left++
6  else: right--
7return [] (no pair)

1function twoSum(arr: number[], target: number): [number, number] | [] {
2  let left = 0, right = arr.length - 1;
3  while (left < right) {
4    const sum = arr[left] + arr[right];
5    if (sum === target) return [left, right];
6    if (sum < target) left++;
7    else right--;
8  }
9  return [];
10}

About

Two Pointers finds two numbers in a sorted array that sum to a target by placing one pointer at the start and one at the end, then moving them toward each other based on whether the current sum is too small or too large.

Step-by-step explanation

Step 1 — Sorted input: The array must be sorted (or use indices from a sorted copy). Place left = 0 and right = n - 1. The idea is that if arr[left] + arr[right] is less than the target, we need a larger sum, so we move left forward (larger value). If the sum is greater than the target, we move right backward (smaller value).

Step 2 — Loop: While left < right, compute sum = arr[left] + arr[right]. If sum equals the target, we have found a pair; return [left, right]. If sum < target, increment left. If sum > target, decrement right.

Step 3 — Why it works: Because the array is sorted, moving left increases the sum and moving right decreases it. So we never skip a valid pair. Each step either finds the answer or eliminates one position from consideration.

Step 4 — Termination: If the loop exits without finding a pair, no two elements sum to the target; return empty or -1. Time O(n), space O(1). Works for "Two Sum II" style problems.

What's happening

Sum = 17 > 9 → move right backward.

Time Complexity

O(n)

Space Complexity

O(1)

Output

Sum = 17 > 9 → move right backward.