Related Articles

# Two Pointers Technique

• Difficulty Level : Easy

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 ` `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

 `// C Program Illustrating Naive Approach to``// Find if There is a Pair in A[0..N-1] with Given Sum` `// Importing all libraries``#include ` `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);``  ` `    ``// 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` `// Importig 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[] = { ``3``, ``5``, ``9``, ``2``, ``8``, ``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 ``=` `[``3``, ``5``, ``9``, ``2``, ``8``, ``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 = { 3, 5, 9, 2, 8, 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` ``
Output
`1`

Time Complexity:  O(n2).

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 ` `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;``}`

## 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 ` `// 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[] = { 3, 5, 9, 2, 8, 10, 11 };``    ` `    ``// Custom value to be searched``    ``int` `val = 17;``    ` `    ``// size of the array``    ``int` `arrSize = ``sizeof``(arr) / ``sizeof``(arr);``  ` `    ``// 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[] = { ``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

 `# Java 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 ``=` `[``3``, ``5``, ``9``, ``2``, ``8``, ``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 = { 3, 5, 9, 2, 8, 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` ``
Output
`1`

Time Complexity:  O(n)

More problems based on two pointer technique.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up