Two Sum - Pair sum in sorted array

Last Updated : 26 Sep, 2024

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.

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: {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

Approach: To solve the problem, follow the below idea:

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.

Below is the implementation of the algorithm:

C++

#include <vector> using namespace std;  vector<int> twoSum(vector<int>& numbers, int target) {     int left = 0, right = numbers.size() - 1;      while (left < right) {         int current_sum = numbers[left] + numbers[right];         // If current sum = target, return left and right         if (current_sum == target) {             return { left + 1, right + 1 };         }         // If current sum < target, then increase the         // current sum by moving the left pointer by 1         else if (current_sum < target) {             left++;         }         else {             // If current sum > target, then decrease the             // current sum by moving the right pointer by 1             right--;         }     }      return {}; }  // Example usage #include <iostream> int main() {     vector<int> numbers = { 2, 7, 11, 15 };     int target = 9;     vector<int> result = twoSum(numbers, target);     for (int num : result) {         cout << num << " ";     }     cout << endl;     return 0; }  // This code is contributed by Shivam Gupta

Java

import java.util.*;  public class TwoSum {     public static List<Integer> twoSum(List<Integer> numbers, int target) {         int left = 0, right = numbers.size() - 1;          while (left < right) {             int current_sum = numbers.get(left) + numbers.get(right);             // If current sum = target, return left and right             if (current_sum == target) {                 return Arrays.asList(left + 1, right + 1);             }             // If current sum < target, then increase the             // current sum by moving the left pointer by 1             else if (current_sum < target) {                 left++;             } else {                 // If current sum > target, then decrease the                 // current sum by moving the right pointer by 1                 right--;             }         }          return Collections.emptyList();     }      public static void main(String[] args) {         List<Integer> numbers = Arrays.asList(2, 7, 11, 15);         int target = 9;         List<Integer> result = twoSum(numbers, target);         for (int num : result) {             System.out.print(num + " ");         }         System.out.println();     } } // This code is contributed by Ayush Mishra

Python

def twoSum(numbers, target):     left, right = 0, len(numbers) - 1      while left < right:         current_sum = numbers[left] + numbers[right]         # If current sum = target, return left and right         if current_sum == target:             return [left + 1, right + 1]         # If current sum < target, then increase the current sum by moving the left          # pointer by 1         elif current_sum < target:             left += 1         else:             # If current sum > target, then decrease the current sum by              # moving the right pointer by 1             right -= 1      return []   # Example usage numbers = [2, 7, 11, 15] target = 9 print(twoSum(numbers, target))

JavaScript

// Function to find two numbers such that they add up to a specific target function twoSum(numbers, target) {     let left = 0;     let right = numbers.length - 1;      // Continue searching while the left index is less than the right index     while (left < right) {         let currentSum = numbers[left] + numbers[right];                  // If the current sum equals the target, return the indices (1-indexed)         if (currentSum === target) {             return [left + 1, right + 1];         }         // If the current sum is less than the target, move the left index to the right         else if (currentSum < target) {             left++;         }         // If the current sum is greater than the target, move the right index to the left         else {             right--;         }     }      // Return an empty array if no two numbers sum up to the target     return []; }  const numbers = [2, 7, 11, 15]; const target = 9; const result = twoSum(numbers, target);  // Output the result result.forEach(num => {     console.log(num); });

Output

[1, 2]

Time Complexity: O(n), where n is declared to be as length of array because elements of array is traced only once.
Auxiliary Space: O(1)

Two Sum - Pair sum in sorted array

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