TWO SUM II -167

Two Sum II – Two Pointer Approach (Short Blog) Problem Given a sorted array numbers and a target, find two indices such that: numbers[i] + numbers[j] = target Return 1-based indices. Code class Solution : def twoSum ( self , numbers , target ): l = 0 r = len ( numbers ) - 1 while l < r : s = numbers [ l ] + numbers [ r ] if s > target : r -= 1 elif s < target : l += 1 else : return [ l + 1 , r + 1 ] Why This Approach? Array is sorted → we can avoid hash maps Use two pointers instead of nested loops More efficient → O(n) time, O(1) space Line-by-Line Idea l = 0 → start (smallest value) r = len(numbers)-1 → end (largest value) while l < r → ensure two different elements s = numbers[l] + numbers[r] → check current pair Conditions: If s > target → decrease sum → r -= 1 If s < target → increase sum → l += 1 If equal → return answer Key Logic Because the array is sorted: Move right pointer left → reduce sum Move left pointer right → increase sum Example numbers = [2, 7, 11, 15], target = 9 S