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Sri Mahalakshmi
Sri Mahalakshmi

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Solving Two Sum II (Sorted Array) Using the Two Pointer Technique in Python

#python #beginners #coding

Introduction

In this problem, we need to find two numbers that add up to a given target.
Since the array is already sorted, we can use a more efficient approach instead of checking all possible pairs.

Problem Explanation

You are given:

  • A sorted array of integers (numbers)
  • A target value

Find two numbers such that their sum equals the target and return their indices using 1-based indexing.

Example 

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

Approach Used – Two Pointer Technique

  • Use two pointers:

    • left at the beginning
    • right at the end

Move the pointers based on the sum until the target is found.

Code (Python) 

class Solution:
    def twoSum(self, numbers: list[int], target: int) -> list[int]:
        left = 0
        right = len(numbers) - 1

        while left < right:
            current_sum = numbers[left] + numbers[right]

            if current_sum == target:
                return [left + 1, right + 1]

            elif current_sum < target:
                left += 1

            else:
                right -= 1
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Code Explanation (Step by Step)

Line 1 

class Solution:
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Defines the class required by LeetCode.

Line 2 

def twoSum(self, numbers: list[int], target: int) -> list[int]:
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Defines the function:

  • numbers is the input array
  • target is the required sum
  • Returns a list of indices

Line 3 

left = 0
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Initializes the left pointer at the beginning of the array.

Line 4 

right = len(numbers) - 1
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Initializes the right pointer at the end of the array.

Line 5 

while left < right:
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Runs the loop until both pointers meet.
Ensures we do not use the same element twice.

Line 6 

current_sum = numbers[left] + numbers[right]
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Calculates the sum of elements at the current pointers.

Line 7–8 

if current_sum == target:
    return [left + 1, right + 1]
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If the sum equals the target:

  • Return indices
  • Add 1 because the problem uses 1-based indexing

Line 9–10 

elif current_sum < target:
    left += 1
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If the sum is smaller than the target:

  • Move the left pointer forward
  • This increases the sum

Line 11–12 

else:
    right -= 1
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If the sum is greater than the target:

  • Move the right pointer backward
  • This decreases the sum

Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(1)

Conclusion

The two pointer technique efficiently solves this problem by using the sorted nature of the array, reducing time complexity and avoiding extra space.

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