Two Pointers Technique
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 Python, JavaScript, PHP etc and all Statically typed languages such as C++, Java, C# etc.
- Naïve Approach using loops
- Optimal approach using two pointer algorithm
Method 1: Naïve Approach
Below is the implementation:
// 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 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 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;
}
}
# 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# 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 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++ 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;
}
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);
}
}
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))
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);
}
}
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.