Two Sum - Pair sum in sorted array

Given a 1-based indexed integer array arr[] that is sorted in non-decreasing order, along with an integer target. find two elements in the array such that their sum is equal to target. If such a pair exists, return the indices of the two elements in increasing order. If no such pair exists, return [-1, -1].

Examples:

Input: arr[] = [2, 7, 11, 15], target = 9
Output: 1 2
Explanation: Since the array is 1-indexed, arr[1] + arr[2] = 2 + 7 = 9

Input: arr[] = [1, 3, 4, 6, 8, 11] target = 10
Output: 3 4
Explanation: Since the array is 1-indexed, arr[3] + arr[5] = 4 + 6 = 10

[Naive Approach] Try all Pairs - O(n^2) Time and O(1) Space

The idea is to check every possible pair of elements using two nested loops. If any pair adds up to the target, we return their 1-based indices immediately.

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[Expected Approach] Using Two Pointers - O(n) Time and O(1) Space

The problem can be solved using two pointers technique. We can maintain two pointers, left = 0 and right = n - 1, and calculate their sum S = arr[left] + arr[right].

• If S = target, then return left and right.
• If S < target, then we need to increase sum S, so we will increment left = left + 1.
• If S > target, then we need to decrease sum S, so we will decrement right = right - 1.

If at any point left >= right, then no pair with sum = target is found.

Step-by-step algorithm:

• We need to initialize two pointers as left and right with position at the beginning and at the end of the array respectively.
• Need to calculate the sum of the elements in the array at these two positions of the pointers.
• If the sum equals to target value, return the 1-indexed positions of the two elements.
• If the sum comes out to be less than the target, move the left pointer to the right to the increase the sum.
• If the sum is greater than the target, move the right pointer to the left to decrease the sum.
• Repeat the following steps till both the pointers meet.

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