Given an absolute sorted array and a number K, find the pair whose sum is K
Last Updated :
12 Jul, 2025
Given an absolute sorted array arr[] and a number target, the task is to find a pair of elements in the given array that sum to target. If no such pair exist in array the return an empty array.
An absolute sorted array is an array of numbers in which |arr[i]| <= |arr[j]|, for all i < j.
Examples:
Input: arr[] = [-8, 10, 12, -15, 15, 24], target = 0
Output: [-15, 15]
Explanation: The sum of elements at index 3 and 4 i.e., -15 and 15 is equal to target.
Input: arr[] = [-8, 10, 12, -15, 15, 24], target = -23
Output: [-8, -15]
Explanation: The sum of elements at index 0 and 3 i.e., -8 and -15 is equal to target.
Input: arr[] = [-8, 10, 12, -15, 15, 24], target = 20
Output: [ ]
Explanation: No pair exist whose sum is equal to target.
[Naive Approach] Using Nested Loop - O(n ^ 2) Time and O(1) Space
The simple approach is to check all possible pairs and identify the pair that adds up to the target. We can use nested loop, where the outer loop selects the first element and the inner loop selects the second element.
C++
// C++ program to find the pair which add up to target
// in absolute sorted array using nested loop
#include <iostream>
#include <vector>
using namespace std;
vector<int> findPair(vector<int> &arr, int target) {
vector<int> res;
// Outer loop will select first element and
// inner loop will select second element of pair
for (int i = 0; i < arr.size() - 1; i++) {
for (int j = i + 1; j < arr.size(); j++) {
// if found such pair
if (arr[i] + arr[j] == target) {
res.push_back(arr[i]);
res.push_back(arr[j]);
return res;
}
}
}
// if no pair exist
return res;
}
int main() {
vector<int> arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
vector<int> res = findPair(arr, target);
for (int ele : res)
cout << ele << " ";
return 0;
}
// C program to find the pair which add up to target
// in absolute sorted array using nested loop
#include <stdio.h>
int* findPair(int arr[], int n, int target, int *resSize) {
int* res = (int*) malloc(2 * sizeof(int));
// Outer loop will select first element and
// inner loop will select second element of pair
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// if found such pair return it
if (arr[i] + arr[j] == target) {
res[0] = arr[i];
res[1] = arr[j];
*resSize = 2;
return res;
}
}
}
return res;
}
int main() {
int arr[] = {-8, 10, -15, 15, 12, 24};
int target = 0;
int n = sizeof(arr) / sizeof(arr[0]);
int resSize = 0;
int* res = findPair(arr, n, target, &resSize);
for (int i = 0; i < resSize; ++i)
printf("%d ", res[i]);
return 0;
}
// Java program to find the pair which add up to target
// in absolute sorted array using nested loop
import java.util.*;
class GfG {
static ArrayList<Integer> findPair(int[] arr, int target) {
ArrayList<Integer> res = new ArrayList<>();
// Outer loop will select first element and
// inner loop will select second element of pair
for (int i = 0; i < arr.length - 1; i++) {
for (int j = i + 1; j < arr.length; j++) {
// if found such pair
if (arr[i] + arr[j] == target) {
res.add(arr[i]);
res.add(arr[j]);
return res;
}
}
}
// if no pair exist
return res;
}
public static void main(String[] args) {
int[] arr = {-8, 10, -15, 15, 12, 24};
int target = 0;
ArrayList<Integer> res = findPair(arr, target);
for (int ele : res)
System.out.print(ele + " ");
}
}
# Python program to find the pair which add up to target
# in absolute sorted array using nested loop
def findPair(arr, target):
res = []
# Outer loop will select first element and
# inner loop will select second element of pair
for i in range(len(arr) - 1):
for j in range(i + 1, len(arr)):
# if found such pair
if arr[i] + arr[j] == target:
res.append(arr[i])
res.append(arr[j])
return res
# if no pair exist
return res
if __name__ == "__main__":
arr = [-8, 10, -15, 15, 12, 24]
target = 0
res = findPair(arr, target)
for ele in res:
print(ele, end=" ")
// C# program to find the pair which add up to target
// in absolute sorted array using nested loop
using System;
using System.Collections.Generic;
class GfG {
static List<int> findPair(int[] arr, int target) {
List<int> res = new List<int>();
// Outer loop will select first element and
// inner loop will select second element of pair
for (int i = 0; i < arr.Length - 1; i++) {
for (int j = i + 1; j < arr.Length; j++) {
// if found such pair
if (arr[i] + arr[j] == target) {
res.Add(arr[i]);
res.Add(arr[j]);
return res;
}
}
}
// if no pair exist
return res;
}
static void Main() {
int[] arr = { -8, 10, -15, 15, 12, 24 };
int target = 0;
List<int> res = findPair(arr, target);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to find the pair which add up to target
// in absolute sorted array using nested loop
function findPair(arr, target) {
let res = [];
// Outer loop will select first element and
// inner loop will select second element of pair
for (let i = 0; i < arr.length - 1; i++) {
for (let j = i + 1; j < arr.length; j++) {
// if found such pair
if (arr[i] + arr[j] === target) {
res.push(arr[i], arr[j]);
return res;
}
}
}
// if no pair exist
return res;
}
// driver code
const arr = [-8, 10, -15, 15, 12, 24];
const target = 0;
const res = findPair(arr, target);
console.log(res.join(" "));
[Better Approach] Using Hash Set - O(n) Time and O(n) Space
The idea is that for every element x in the array arr[], we can check if its complement (target - x), exists in the array. To make this search faster, we can use a hash set.
C++
// C++ program to find the pair which add up to target
// in absolute sorted array using hash set
#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
vector<int> findPair(vector<int> &arr, int target) {
vector<int> res;
unordered_set<int> st;
for (int ele: arr) {
// calculate and search for complement of ele
// in already covered part of array
int complement = target - ele;
if (st.find(complement) != st.end()) {
res.push_back(complement);
res.push_back(ele);
return res;
}
// if complement is not found then store this element in hash set
st.insert(ele);
}
// if no pair exist
return res;
}
int main() {
vector<int> arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
vector<int> res = findPair(arr, target);
for (int ele : res)
cout << ele << " ";
return 0;
}
// Java program to find the pair which adds up to target
// in absolute sorted array using hash set
import java.util.ArrayList;
import java.util.HashSet;
class GfG {
static ArrayList<Integer> findPair(int[] arr, int target) {
ArrayList<Integer> res = new ArrayList<>();
HashSet<Integer> st = new HashSet<>();
for (int ele : arr) {
// Calculate and search for complement of ele
// in already covered part of the array
int complement = target - ele;
if (st.contains(complement)) {
res.add(complement);
res.add(ele);
return res;
}
// If complement is not found then store this element in hash set
st.add(ele);
}
// If no pair exists
return res;
}
public static void main(String[] args) {
int[] arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
ArrayList<Integer> res = findPair(arr, target);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
# Function to find the pair which adds up to target
# in absolute sorted array using hash set
def findPair(arr, target):
res = []
st = set()
for ele in arr:
# Calculate and search for complement of ele
# in already covered part of array
complement = target - ele
if complement in st:
res.append(complement)
res.append(ele)
return res
# If complement is not found then store this element in hash set
st.add(ele)
# If no pair exist
return res
if __name__ == "__main__":
arr = [-8, 10, 12, -15, 15, 24]
target = 0
res = findPair(arr, target)
for ele in res:
print(ele, end=" ")
// C# program to find the pair which adds up to target
// in absolute sorted array using hash set
using System;
using System.Collections.Generic;
class GfG {
static List<int> findPair(int[] arr, int target) {
List<int> res = new List<int>();
HashSet<int> st = new HashSet<int>();
foreach (int ele in arr) {
// Calculate and search for complement of ele
// in already covered part of array
int complement = target - ele;
if (st.Contains(complement)) {
res.Add(complement);
res.Add(ele);
return res;
}
// If complement is not found then store this element in hash set
st.Add(ele);
}
// If no pair exist
return res;
}
static void Main() {
int[] arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
List<int> res = findPair(arr, target);
foreach (int ele in res) {
Console.Write(ele + " ");
}
}
}
// Function to find the pair which adds up to target
// in absolute sorted array using hash set
function findPair(arr, target) {
const res = [];
const st = new Set();
for (let ele of arr) {
// Calculate and search for complement of ele
// in already covered part of array
const complement = target - ele;
if (st.has(complement)) {
res.push(complement);
res.push(ele);
return res;
}
// If complement is not found then store this element in hash set
st.add(ele);
}
// If no pair exists
return res;
}
// driver code
const arr = [-8, 10, 12, -15, 15, 24];
const target = 0;
const res = findPair(arr, target);
console.log(res.join(" "));
[Expected Approach] Using Two Pointers - O(n) Time and O(1) Space
For a sorted array, use the approach discussed in this article. In case of an absolute sorted array, there are generally three cases for pairs according to their property:
- Both the numbers in the pair are negative.
- Both the numbers in the pair are positive.
- One number is negative and the other is positive.
For cases (1) and (2), use the Two Pointer Approach separately by just limiting to consider either positive or negative numbers.
For case (3), use the same two pointer approach where we have one index for positive numbers and one index for negative numbers, and they both start from the leftmost possible index and then move towards right, skipping there respective opposite numbers.
Note: In absolute sorted array, the negative numbers were arranged in descending order.
Case 1: Both the numbers in the pair are positive, initialize i = 0, j = size - 1.
- Skip if either arr[i] or arr[j] is negative.
- If arr[i] + arr[j] < target, to increase the sum, increment i.
- If arr[i] + arr[j] > target, to decrease the sum, decrement j.
- If arr[i] + arr[j] == target, pair found.
Case 2: Both the numbers in the pair are negative, initialize i = 0, j = size - 1.
- Skip if either arr[i] or arr[j] is positive.
- If arr[i] + arr[j] < target, to increase the sum, decrement j.
- If arr[i] + arr[j] > target, to decrease the sum increment i.
- If arr[i] + arr[j] == target, pair found.
Case 3: One number is negative and the other is positive., initialize i = 0, j = 0.
- Skip if arr[i] is negative or arr[j] is positive.
- If arr[i] + arr[j] < target, to increase the sum, increment i.
- If arr[i] + arr[j] > target, to decrease the sum, increment j.
- If arr[i] + arr[j] == target, pair found.
C++
// C++ program to find the pair which add up to target
// in absolute sorted array using two pointers
#include <iostream>
#include <vector>
using namespace std;
vector<int> findPair(vector<int> &arr, int target) {
vector<int> res;
int i = 0, j = arr.size() - 1;
// Case 1: both elements in pair are positive
for (int i = 0, j = arr.size() - 1; i < j ; ) {
// skip negative elements
if (arr[i] < 0 ) {
i++;
} else if (arr[j] < 0) {
j--;
}
// if both elements are positive check further
else {
int sum = arr[i] + arr[j];
if(sum == target){
res.push_back(arr[i]);
res.push_back(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j--;
}
}
}
// Case 2: both elements in pair are negative
for (int i = 0, j = arr.size() - 1; i < j ; ) {
// skip positive elements
if (arr[i] > 0 ) {
i++;
} else if (arr[j] > 0) {
j--;
}
// if both elements are negative check further
else {
int sum = arr[i] + arr[j];
if(sum == target){
res.push_back(arr[i]);
res.push_back(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
j--;
}
// if sum is larger, move toward smaller value
else {
i++;
}
}
}
// Case 3: one element in pair is positive and other is negative
// i point to positive and j point to negative
for (int i = 0, j = 0; i < arr.size() && j < arr.size() ; ) {
// skip negative elements in arr[i] and skip positive in arr[j]
if (arr[i] < 0 ) {
i++;
} else if (arr[j] > 0) {
j++;
}
// if arr[i] is positive and arr[j] is negative, check further
else {
int sum = arr[i] + arr[j];
if(sum == target){
res.push_back(arr[min(i, j)]);
res.push_back(arr[max(i, j)]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j++;
}
}
}
return res;
}
int main() {
vector<int> arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
vector<int> res = findPair(arr, target);
for (int ele : res)
cout << ele << " ";
return 0;
}
// C program to find the pair which add up to target
// in absolute sorted array using two pointers
#include <stdio.h>
int* findPair(int arr[], int n, int target, int* resSize) {
int* res = (int*) malloc(2 * sizeof(int));
// Case 1: both elements in pair are positive
for (int i = 0, j = n - 1; i < j;) {
// skip negative elements
if (arr[i] < 0) {
i++;
} else if (arr[j] < 0) {
j--;
}
// if both elements are positive check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res[0] = arr[i];
res[1] = arr[j];
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j--;
}
}
}
// Case 2: both elements in pair are negative
for (int i = 0, j = n - 1; i < j;) {
// skip positive elements
if (arr[i] > 0) {
i++;
} else if (arr[j] > 0) {
j--;
}
// if both elements are negative check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res[0] = arr[i];
res[1] = arr[j];
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
j--;
}
// if sum is larger, move toward smaller value
else {
i++;
}
}
}
// Case 3: one element in pair is positive and other is negative
// i point to positive and j point to negative
for (int i = 0, j = 0; i < n && j < n;) {
// skip negative elements in arr[i] and skip positive in arr[j]
if (arr[i] < 0) {
i++;
} else if (arr[j] > 0) {
j++;
}
// if arr[i] is positive and arr[j] is negative, check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res[0] = arr[i < j ? i : j];
res[1] = arr[i > j ? i : j];
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j++;
}
}
}
// if no pair found
*resSize = 0;
return res;
}
int main() {
int arr[] = {-8, 10, 12, -15, 15, 24};
int n = sizeof(arr) / sizeof(arr[0]);
int target = 0;
int resSize = 2;
int* res = findPair(arr, n, target, &resSize);
for (int i=0; i < resSize; i++)
printf("%d ", res[i]);
return 0;
}
// Java program to find the pair which add up to target
// in absolute sorted array using two pointers
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findPair(int[] arr, int target) {
ArrayList<Integer> res = new ArrayList<>();
// Case 1: both elements in pair are positive
for (int i = 0, j = arr.length - 1; i < j;) {
// skip negative elements
if (arr[i] < 0) {
i++;
} else if (arr[j] < 0) {
j--;
}
// if both elements are positive check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.add(arr[i]);
res.add(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j--;
}
}
}
// Case 2: both elements in pair are negative
for (int i = 0, j = arr.length - 1; i < j;) {
// skip positive elements
if (arr[i] > 0) {
i++;
} else if (arr[j] > 0) {
j--;
}
// if both elements are negative check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.add(arr[i]);
res.add(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
j--;
}
// if sum is larger, move toward smaller value
else {
i++;
}
}
}
// Case 3: one element in pair is positive and other is negative
// i point to positive and j point to negative
for (int i = 0, j = 0; i < arr.length && j < arr.length;) {
// skip negative elements in arr[i] and skip positive in arr[j]
if (arr[i] < 0) {
i++;
} else if (arr[j] > 0) {
j++;
}
// if arr[i] is positive and arr[j] is negative, check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.add(arr[Math.min(i, j)]);
res.add(arr[Math.max(i, j)]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j++;
}
}
}
// no pair found
return res;
}
public static void main(String[] args) {
int[] arr = {-8, 10, 12, -15, 15, 24};
int target = 0;
ArrayList<Integer> res = findPair(arr, target);
for (int ele : res)
System.out.print(ele + " ");
}
}
# Python program to find the pair which add up to target
# in absolute sorted array using two pointers
def findPair(arr, target):
res = []
# Case 1: both elements in pair are positive
i, j = 0, len(arr) - 1
while i < j:
# skip negative elements
if arr[i] < 0:
i += 1
elif arr[j] < 0:
j -= 1
# if both elements are positive check further
else:
sum = arr[i] + arr[j]
if sum == target:
res.append(arr[i])
res.append(arr[j])
return res
# if sum is smaller, move to larger value
elif sum < target:
i += 1
# if sum is larger, move toward smaller value
else:
j -= 1
# Case 2: both elements in pair are negative
i, j = 0, len(arr) - 1
while i < j:
# skip positive elements
if arr[i] > 0:
i += 1
elif arr[j] > 0:
j -= 1
# if both elements are negative check further
else:
sum = arr[i] + arr[j]
if sum == target:
res.append(arr[i])
res.append(arr[j])
return res
# if sum is smaller, move to larger value
elif sum < target:
j -= 1
# if sum is larger, move toward smaller value
else:
i += 1
# Case 3: one element in pair is positive and other is negative
i, j = 0, 0
while i < len(arr) and j < len(arr):
# skip negative elements in arr[i] and skip positive in arr[j]
if arr[i] < 0:
i += 1
elif arr[j] > 0:
j += 1
# if arr[i] is positive and arr[j] is negative, check further
else:
sum = arr[i] + arr[j]
if sum == target:
res.append(arr[min(i, j)])
res.append(arr[max(i, j)])
return res
# if sum is smaller, move to larger value
elif sum < target:
i += 1
# if sum is larger, move toward smaller value
else:
j += 1
# no pair exist
return res
if __name__ == "__main__":
arr = [-8, 10, 12, -15, 15, 24]
target = 0
res = findPair(arr, target)
print(" ".join(map(str, res)))
// C# program to find the pair which add up to target
// in absolute sorted array using two pointers
using System;
using System.Collections.Generic;
class GfG {
static List<int> findPair(int[] arr, int target) {
List<int> res = new List<int>();
// Case 1: both elements in pair are positive
for (int i = 0, j = arr.Length - 1; i < j;) {
// skip negative elements
if (arr[i] < 0) {
i++;
} else if (arr[j] < 0) {
j--;
}
// if both elements are positive check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.Add(arr[i]);
res.Add(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j--;
}
}
}
// Case 2: both elements in pair are negative
for (int i = 0, j = arr.Length - 1; i < j;) {
// skip positive elements
if (arr[i] > 0) {
i++;
} else if (arr[j] > 0) {
j--;
}
// if both elements are negative check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.Add(arr[i]);
res.Add(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
j--;
}
// if sum is larger, move toward smaller value
else {
i++;
}
}
}
// Case 3: one element in pair is positive and other is negative
// i point to positive and j point to negative
for (int i = 0, j = 0; i < arr.Length && j < arr.Length;) {
// skip negative elements in arr[i] and skip positive in arr[j]
if (arr[i] < 0) {
i++;
} else if (arr[j] > 0) {
j++;
}
// if arr[i] is positive and arr[j] is negative, check further
else {
int sum = arr[i] + arr[j];
if (sum == target) {
res.Add(arr[Math.Min(i, j)]);
res.Add(arr[Math.Max(i, j)]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j++;
}
}
}
return res;
}
static void Main(string[] args) {
int[] arr = { -8, 10, 12, -15, 15, 24 };
int target = 0;
List<int> res = findPair(arr, target);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to find the pair which add up to target
// in absolute sorted array using two pointers
function findPair(arr, target) {
let res = [];
// Case 1: both elements in pair are positive
for (let i = 0, j = arr.length - 1; i < j;) {
// skip negative elements
if (arr[i] < 0) {
i++;
} else if (arr[j] < 0) {
j--;
}
// if both elements are positive check further
else {
let sum = arr[i] + arr[j];
if (sum === target) {
res.push(arr[i]);
res.push(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j--;
}
}
}
// Case 2: both elements in pair are negative
for (let i = 0, j = arr.length - 1; i < j;) {
// skip positive elements
if (arr[i] > 0) {
i++;
} else if (arr[j] > 0) {
j--;
}
// if both elements are negative check further
else {
let sum = arr[i] + arr[j];
if (sum === target) {
res.push(arr[i]);
res.push(arr[j]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
j--;
}
// if sum is larger, move toward smaller value
else {
i++;
}
}
}
// Case 3: one element in pair is positive and other is negative
// i point to positive and j point to negative
for (let i = 0, j = 0; i < arr.length && j < arr.length;) {
// skip negative elements in arr[i] and skip positive in arr[j]
if (arr[i] < 0) {
i++;
} else if (arr[j] > 0) {
j++;
}
// if arr[i] is positive and arr[j] is negative, check further
else {
let sum = arr[i] + arr[j];
if (sum === target) {
res.push(arr[Math.min(i, j)]);
res.push(arr[Math.max(i, j)]);
return res;
}
// if sum is smaller, move to larger value
else if (sum < target) {
i++;
}
// if sum is larger, move toward smaller value
else {
j++;
}
}
}
// no pair found
return res;
}
// driver code
let arr = [-8, 10, 12, -15, 15, 24];
let target = 0;
let res = findPair(arr, target);
console.log(res.join(" "));
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