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Two Pointers Technique

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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

  1. Naïve Approach using loops
  2. Optimal approach using two pointer algorithm

Implementation:

Method 1: Naïve Approach 

Examples 

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 true;
 
            // 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
1
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. 
Examples
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 <iostream>
#include <algorithm>
 
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(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
    int arr[] = { 2, 3, 5, 8, 9, 10, 11 };
     
    // value to search
    int val = 17;
     
    // size of the array
    int arrSize = *(&arr + 1) - arr;
     
      // array should be sorted before using two-pointer technique
      sort(arr, arr+7);
   
    // Function call
    cout << (bool)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
// Using Two-pointers Technique
 
// Importing I/O libraries
#include <stdio.h>
 
// Two pointer technique based solution to find
// if there is a pair in A[0..N-1] with a given sum.
int isPairSum(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;
}
 
// Main method
int main()
{
    // Array declaration
    int arr[] = { 2, 3, 5, 8, 9, 10, 11 };
     
    // Custom value to be searched
    int val = 17;
     
    // size of the array
    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
// Using Two-pointers Technique
 
// Importing all utility classes
import java.io.*;
 
// Main class
class GFG
{
     // 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(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
    public static void main(String[] args)
    {
        // array declaration
        int arr[] = { 2, 3, 5, 8, 9, 10, 11 };
         
        // value to search
        int val = 17;
       
        // size of the array
        int arrSize = arr.length;
       
        // Function call
        System.out.println(isPairSum(arr, arrSize, val));
    }
}
Python3



# Python Program Illustrating Naive Approach to
# Find if There is a Pair in A[0..N-1] with Given Sum
# Using Two-pointers Technique
 
# Method
def isPairSum(A, N, X):
 
    # 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 += 1
        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 -= 1
        else:
            j -= 1
    return 0
 
# array declaration
arr = [2, 3, 5, 8, 9, 10, 11]
 
# value to search
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 Two-pointers Technique
 
// Importing all utility classes
using System;
 
// Main class
class GFG
{
     // 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(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
    public static void Main(String[] args)
    {
        // array declaration
        int []arr = { 2, 3, 5, 8, 9, 10, 11 };
         
        // value to search
        int val = 17;
       
        // size of the array
        int arrSize = arr.Length;
       
        // Function call
        Console.Write(isPairSum(arr, arrSize, val));
    }
}
 
// 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
// Using Two-pointers Technique
 
<script>
// Two pointer technique based solution to find
// if there is a pair in A[0..N-1] with a given sum.
function isPairSum(A, N, X)
{
 
    // represents first pointer
    var i = 0;
 
    // represents second pointer
    var 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
 
    // array declaration
    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));
     
    // This Code is Contributed by Harshit Srivastava
 
   </script>
Output
1
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. 

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