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 codeint 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 Codeint 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 classesimport java.io.*;// Main classclass 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# Methoddef 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 codearr = [2, 3, 5, 8, 9, 10, 11]val = 17print(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 Sumusing System;// Main classclass 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.
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 codeint 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 Listdef 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 codeif __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 codeconst arr = [2, 3, 5, 8, 9, 10, 11];const val = 17;const arrSize = arr.length;// array should be sorted before using the two-pointer techniquearr.sort((a, b) => a - b);// Function callconsole.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.
- Find the closest pair from two sorted arrays
- Find the pair in array whose sum is closest to x
- Find all triplets with zero sum
- Find a triplet that sum to a given value
- Find a triplet such that sum of two equals to third element
- Find four elements that sum to a given value
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