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Two Pointers Technique
  • Difficulty Level : Easy

Two pointers is really an easy and effective technique which 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.

Let’s see the naive solution.  

C++




// Naive solution to find if there is a
// pair in A[0..N-1] with given sum.
#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[] = { 3, 5, 9, 2, 8, 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




// Naive solution to find if there is a
// pair in A[0..N-1] with given sum.
#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[]={3,5,9,2,8,10,11};
    int val=17;
    int arrSize = sizeof(arr)/sizeof(arr[0]);
   
    // Function call
    printf("%d",isPairSum(arr,arrSize,val));
 
    return 0;
}

Java




// Naive solution to find if there is a
// pair in A[0..N-1] with given sum.
import java.io.*;
 
class GFG {
    public static void main(String[] args)
    {
        int arr[] = { 3, 5, 9, 2, 8, 10, 11 };
        int val = 17;
 
        System.out.println(isPairSum(arr, arr.length, val));
    }
     
    private static 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;
    }
}

Python3




# Naive solution to find if there is a
# pair in A[0..N-1] with given sum.
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 = [3, 5, 9, 2, 8, 10, 11]
val = 17
 
print(isPairSum(arr, len(arr), val))
 
# This code is contributed by maheshwaripiyush9
Output
1

Time Complexity:  O(n2).

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. 



C




#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;
}
 
// Driver code
int main()
{
    // array declaration
    int arr[] = { 3, 5, 9, 2, 8, 10, 11 };
     
    // value to search
    int val = 17;
     
    // size of the array
    int arrSize = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    printf("%d", isPairSum(arr, arrSize, val));
 
    return 0;
}

C++




#include <iostream>
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[] = { 3, 5, 9, 2, 8, 10, 11 };
     
    // value to search
    int val = 17;
     
    // size of the array
    int arrSize = *(&arr + 1) - arr;
     
    // Function call
    cout << (bool)isPairSum(arr, arrSize, val);
 
    return 0;
}

Java




import java.io.*;
 
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[] = { 3, 5, 9, 2, 8, 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




# Two pointer technique based solution to find
# if there is a pair in A[0..N-1] with a given sum.
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 = [3, 5, 9, 2, 8, 10, 11]
 
# value to search
val = 17
 
print(isPairSum(arr, len(arr), val))
 
# This code is contributed by maheshwaripiyush9.
Output
1

Illustration : 
 

Time Complexity:  O(n)

How does this work? 
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.

More problems based on two pointer technique. 

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