Two Pointers Technique

Last Updated : 29 Mar, 2024

Two pointers is really an easy and effective technique that is typically used for searching pairs in a sorted array.
Given a sorted array A (sorted in ascending order), having N integers, find if there exists any pair of elements (A[i], A[j]) such that their sum is equal to X.

Illustration :

A[] = {10, 20, 35, 50, 75, 80}
X = =70
i = 0
j = 5

A[i] + A[j] = 10 + 80 = 90
Since A[i] + A[j] > X, j--
i = 0
j = 4

A[i] + A[j] = 10 + 75 = 85
Since A[i] + A[j] > X, j--
i = 0
j = 3

A[i] + A[j] = 10 + 50 = 60
Since A[i] + A[j] < X, i++
i = 1
j = 3
m
A[i] + A[j] = 20 + 50 = 70
Thus this signifies that Pair is Found.

Let us do discuss the working of two pointer algorithm in brief which is as follows. The algorithm basically uses the fact that the input array is sorted. We start the sum of extreme values (smallest and largest) and conditionally move both pointers. We move left pointer ‘i’ when the sum of A[i] and A[j] is less than X. We do not miss any pair because the sum is already smaller than X. Same logic applies for right pointer j.

Methods:

Here we will be proposing a two-pointer algorithm by starting off with the naïve approach only in order to showcase the execution of operations going on in both methods and secondary to justify how two-pointer algorithm optimizes code via time complexities across all dynamic programming languages such as C++, Java, Python, and even JavaScript

Naïve Approach using loops
Optimal approach using two pointer algorithm

Method 1: Naïve Approach

Below is the implementation:

C++

// C++ Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
  
// Importing all libraries 
#include <bits/stdc++.h> 
  
using namespace std; 
  
bool isPairSum(int A[], int N, int X) 
{ 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) { 
            // as equal i and j means same element 
            if (i == j) 
                continue; 
  
            // pair exists 
            if (A[i] + A[j] == X) 
                return true; 
  
            // as the array is sorted 
            if (A[i] + A[j] > X) 
                break; 
        } 
    } 
  
    // No pair found with given sum. 
    return false; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 2, 3, 5, 8, 9, 10, 11 }; 
    int val = 17; 
    int arrSize = *(&arr + 1) - arr; 
    sort(arr, arr + arrSize); // Sort the array 
    // Function call 
    cout << isPairSum(arr, arrSize, val); 
  
    return 0; 
}

C

// C Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
  
// Importing all libraries 
#include <stdio.h> 
  
int isPairSum(int A[], int N, int X) 
{ 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) { 
            // as equal i and j means same element 
            if (i == j) 
                continue; 
  
            // pair exists 
            if (A[i] + A[j] == X) 
                return 1; 
  
            // as the array is sorted 
            if (A[i] + A[j] > X) 
                break; 
        } 
    } 
  
    // No pair found with given sum. 
    return 0; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = { 2, 3, 5, 8, 9, 10, 11 }; 
    int val = 17; 
    int arrSize = sizeof(arr) / sizeof(arr[0]); 
  
    // Function call 
    printf("%d", isPairSum(arr, arrSize, val)); 
  
    return 0; 
}

Java

// Java Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
  
// Importing all input output classes 
import java.io.*; 
  
// Main class 
class GFG { 
  
    // Method 1 
    // Main driver method 
    public static void main(String[] args) 
    { 
        // Declaring and initializing array 
        int arr[] = { 2, 3, 5, 8, 9, 10, 11 }; 
  
        int val = 17; 
  
        System.out.println(isPairSum(arr, arr.length, val)); 
    } 
  
    // Method 2 
    //  To find Pairs in A[0..N-1] with given sum 
    private static int isPairSum(int A[], int N, int X) 
    { 
        // Nested for loops for iterations 
        for (int i = 0; i < N; i++) { 
            for (int j = i + 1; j < N; j++) { 
                // As equal i and j means same element 
                if (i == j) 
  
                    // continue keyword skips the execution 
                    // for following condition 
                    continue; 
  
                // Condition check if pair exists 
                if (A[i] + A[j] == X) 
                    return 1; 
  
                // By now the array is sorted 
                if (A[i] + A[j] > X) 
  
                    // Break keyword to hault the execution 
                    break; 
            } 
        } 
  
        // No pair found with given sum. 
        return 0; 
    } 
}

Python3

# Python Program Illustrating Naive Approach to 
# Find if There is a Pair in A[0..N-1] with Given Sum 
  
# Method 
  
  
def isPairSum(A, N, X): 
  
    for i in range(N): 
        for j in range(N): 
  
            # as equal i and j means same element 
            if(i == j): 
                continue
  
            # pair exists 
            if (A[i] + A[j] == X): 
                return True
  
            # as the array is sorted 
            if (A[i] + A[j] > X): 
                break
  
    # No pair found with given sum 
    return 0
  
  
# Driver code 
arr = [2, 3, 5, 8, 9, 10, 11] 
val = 17
  
print(isPairSum(arr, len(arr), val)) 
  
# This code is contributed by maheshwaripiyush9 

C#

// C# Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
using System; 
  
// Main class 
class GFG { 
  
    // Method 1 
    // Main driver method 
    public static void Main(String[] args) 
    { 
  
        // Declaring and initializing array 
        int[] arr = { 2, 3, 5, 8, 9, 10, 11 }; 
  
        int val = 17; 
  
        Console.Write(isPairSum(arr, arr.Length, val)); 
    } 
  
    // Method 2 
    //  To find Pairs in A[0..N-1] with given sum 
    private static int isPairSum(int[] A, int N, int X) 
    { 
  
        // Nested for loops for iterations 
        for (int i = 0; i < N; i++) { 
            for (int j = i + 1; j < N; j++) { 
  
                // As equal i and j means same element 
                if (i == j) 
  
                    // continue keyword skips the execution 
                    // for following condition 
                    continue; 
  
                // Condition check if pair exists 
                if (A[i] + A[j] == X) 
                    return 1; 
  
                // By now the array is sorted 
                if (A[i] + A[j] > X) 
  
                    // Break keyword to hault the execution 
                    break; 
            } 
        } 
  
        // No pair found with given sum. 
        return 0; 
    } 
} 
  
// This code is contributed by shivanisinghss2110

Javascript

// JavaScript Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
  
<script> 
  
 // Naive solution to find if there is a 
// pair in A[0..N-1] with given sum. 
  
function isPairSum(A, N, X) 
{ 
        for (var i = 0; i < N-1; i++) 
        { 
            for (var j = i+1; j < N; j++) 
            { 
                // as equal i and j means same element 
                if (i == j) 
                    continue; 
  
                // pair exists 
                if (A[i] + A[j] == X) 
                    return 1; 
  
                // as the array is sorted 
                if (A[i] + A[j] > X) 
                    break; 
            } 
        } 
  
        // No pair found with given sum. 
        return 0; 
} 
  
  
      
        var arr=[ 2, 3, 5, 8, 9, 10, 11 ]; 
          
        // value to search 
        var val = 17; 
      
        // size of the array 
        var arrSize = 7; 
      
        // Function call 
        document.write(isPairSum(arr, arrSize, val)); 
  
</script>

Output

Time Complexity: O(n²).
Auxiliary Space: O(1)

Method 2: Two Pointers Technique

Now let’s see how the two-pointer technique works. We take two pointers, one representing the first element and other representing the last element of the array, and then we add the values kept at both the pointers. If their sum is smaller than X then we shift the left pointer to right or if their sum is greater than X then we shift the right pointer to left, in order to get closer to the sum. We keep moving the pointers until we get the sum as X.

Below is the implementation:

C++

// C++ Program Illustrating Naive Approach to 
// Find if There is a Pair in A[0..N-1] with Given Sum 
// Using Two-pointers Technique 
  
// Importing required libraries 
#include <bits/stdc++.h> 
using namespace std; 
  
// Two pointer technique based solution to find 
// if there is a pair in A[0..N-1] with a given sum. 
int isPairSum(vector<int>& A, int N, int X) 
{ 
    // represents first pointer 
    int i = 0; 
  
    // represents second pointer 
    int j = N - 1; 
  
    while (i < j) { 
  
        // If we find a pair 
        if (A[i] + A[j] == X) 
            return 1; 
  
        // If sum of elements at current 
        // pointers is less, we move towards 
        // higher values by doing i++ 
        else if (A[i] + A[j] < X) 
            i++; 
  
        // If sum of elements at current 
        // pointers is more, we move towards 
        // lower values by doing j-- 
        else
            j--; 
    } 
    return 0; 
} 
  
// Driver code 
int main() 
{ 
    // array declaration 
    vector<int> arr = { 2, 3, 5, 8, 9, 10, 11 }; 
  
    // value to search 
    int val = 17; 
  
    // size of the array 
    int arrSize = arr.size(); 
  
    // array should be sorted before using two-pointer 
    // technique 
    sort(arr.begin(), arr.end()); 
  
    // Function call 
    cout << (isPairSum(arr, arrSize, val) ? "True"
                                          : "False"); 
  
    return 0; 
}

Java

import java.util.Arrays; 
import java.util.List; 
  
public class PairSum { 
    // Two pointer technique based solution to find 
    // if there is a pair in A[0..N-1] with a given sum. 
    public static int isPairSum(List<Integer> A, int N, 
                                int X) 
    { 
        // represents first pointer 
        int i = 0; 
  
        // represents second pointer 
        int j = N - 1; 
  
        while (i < j) { 
            // If we find a pair 
            if (A.get(i) + A.get(j) == X) 
                return 1; 
  
            // If sum of elements at current 
            // pointers is less, we move towards 
            // higher values by doing i++ 
            else if (A.get(i) + A.get(j) < X) 
                i++; 
  
            // If sum of elements at current 
            // pointers is more, we move towards 
            // lower values by doing j-- 
            else
                j--; 
        } 
        return 0; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        // array declaration 
        List<Integer> arr 
            = Arrays.asList(2, 3, 5, 8, 9, 10, 11); 
  
        // value to search 
        int val = 17; 
  
        // size of the array 
        int arrSize = arr.size(); 
  
        // array should be sorted before using the 
        // two-pointer technique 
        arr.sort(null); 
  
        // Function call 
        System.out.println(isPairSum(arr, arrSize, val) 
                           != 0); 
    } 
}

Python3

from typing import List
  
  
def isPairSum(A: List[int], N: int, X: int) -> bool: 
    # represents first pointer 
    i = 0
  
    # represents second pointer 
    j = N - 1
  
    while i < j: 
        # If we find a pair 
        if A[i] + A[j] == X: 
            return True
  
        # If sum of elements at current 
        # pointers is less, we move towards 
        # higher values by doing i++ 
        elif A[i] + A[j] < X: 
            i += 1
  
        # If sum of elements at current 
        # pointers is more, we move towards 
        # lower values by doing j-- 
        else: 
            j -= 1
  
    return False
  
  
# Driver code 
if __name__ == "__main__": 
    # array declaration 
    arr = [2, 3, 5, 8, 9, 10, 11] 
  
    # value to search 
    val = 17
  
    # size of the array 
    arrSize = len(arr) 
  
    # array should be sorted before using the two-pointer technique 
    arr.sort() 
  
    # Function call 
    print(isPairSum(arr, arrSize, val)) 

C#

using System; 
using System.Collections.Generic; 
  
class PairSum { 
    // Two pointer technique based solution to find 
    // if there is a pair in A[0..N-1] with a given sum. 
    static int IsPairSum(List<int> A, int N, int X) 
    { 
        // represents first pointer 
        int i = 0; 
  
        // represents second pointer 
        int j = N - 1; 
  
        while (i < j) { 
            // If we find a pair 
            if (A[i] + A[j] == X) 
                return 1; 
  
            // If sum of elements at current 
            // pointers is less, we move towards 
            // higher values by doing i++ 
            else if (A[i] + A[j] < X) 
                i++; 
  
            // If sum of elements at current 
            // pointers is more, we move towards 
            // lower values by doing j-- 
            else
                j--; 
        } 
        return 0; 
    } 
  
    // Driver code 
    static void Main(string[] args) 
    { 
        // array declaration 
        List<int> arr 
            = new List<int>{ 2, 3, 5, 8, 9, 10, 11 }; 
  
        // value to search 
        int val = 17; 
  
        // size of the array 
        int arrSize = arr.Count; 
  
        // array should be sorted before using the 
        // two-pointer technique 
        arr.Sort(); 
  
        // Function call 
        Console.WriteLine(IsPairSum(arr, arrSize, val) 
                          != 0); 
    } 
}

Javascript

function isPairSum(A, N, X) { 
    // represents first pointer 
    let i = 0; 
  
    // represents second pointer 
    let j = N - 1; 
  
    while (i < j) { 
        // If we find a pair 
        if (A[i] + A[j] === X) 
            return true; 
  
        // If sum of elements at current 
        // pointers is less, we move towards 
        // higher values by doing i++ 
        else if (A[i] + A[j] < X) 
            i++; 
  
        // If sum of elements at current 
        // pointers is more, we move towards 
        // lower values by doing j-- 
        else
            j--; 
    } 
    return false; 
} 
  
// Driver code 
const arr = [2, 3, 5, 8, 9, 10, 11]; 
const val = 17; 
const arrSize = arr.length; 
  
// array should be sorted before using the two-pointer technique 
arr.sort((a, b) => a - b); 
  
// Function call 
console.log(isPairSum(arr, arrSize, val)); 

Output

True

Time Complexity: O(n log n) (As sort function is used)
Auxiliary Space: O(1), since no extra space has been taken.

More problems based on two pointer technique.

Suggest improvement

Java Program for Two Pointers Technique

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