Array Reverse - Complete Tutorial
Last Updated :
23 Jul, 2025
Given an array arr[], the task is to reverse the array. Reversing an array means rearranging the elements such that the first element becomes the last, the second element becomes second last and so on.
Examples:
Input: arr[] = {1, 4, 3, 2, 6, 5}
Output: {5, 6, 2, 3, 4, 1}
Explanation: The first element 1 moves to last position, the second element 4 moves to second-last and so on.
Input: arr[] = {4, 5, 1, 2}
Output: {2, 1, 5, 4}
Explanation: The first element 4 moves to last position, the second element 5 moves to second last and so on.
[Naive Approach] Using a temporary array - O(n) Time and O(n) Space
The idea is to use a temporary array to store the reverse of the array.
- Create a temporary array of same size as the original array.
- Now, copy all elements from original array to the temporary array in reverse order.
- Finally, copy all the elements from temporary array back to the original array.
Working:
Below is the implementation of the algorithm:
C++
// C++ Program to reverse an array using temporary array
#include <iostream>
#include <vector>
using namespace std;
// function to reverse an array
void reverseArray(vector<int> &arr) {
int n = arr.size();
// Temporary array to store elements in reversed order
vector<int> temp(n);
// Copy elements from original array to temp in reverse order
for(int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for(int i = 0; i < n; i++)
arr[i] = temp[i];
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C Program to reverse an array using temporary array
#include <stdio.h>
#include <stdlib.h>
// function to reverse an array
void reverseArray(int arr[], int n) {
// Temporary array to store elements in reversed order
int temp[n];
// Copy elements from original array to temp in reverse order
for(int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for(int i = 0; i < n; i++)
arr[i] = temp[i];
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for(int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java Program to reverse an array using temporary array
import java.util.Arrays;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.length;
// Temporary array to store elements in reversed order
int[] temp = new int[n];
// Copy elements from original array to temp in reverse order
for (int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (int i = 0; i < n; i++)
arr[i] = temp[i];
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
# Python Program to reverse an array using temporary array
# function to reverse an array
def reverseArray(arr):
n = len(arr)
# Temporary array to store elements in reversed order
temp = [0] * n
# Copy elements from original array to temp in reverse order
for i in range(n):
temp[i] = arr[n - i - 1]
# Copy elements back to original array
for i in range(n):
arr[i] = temp[i]
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
// C# Program to reverse an array using temporary array
using System;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.Length;
// Temporary array to store elements in reversed order
int[] temp = new int[n];
// Copy elements from original array to temp in reverse order
for (int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (int i = 0; i < n; i++)
arr[i] = temp[i];
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to reverse an array using temporary array
// function to reverse an array
function reverseArray(arr) {
let n = arr.length;
// Temporary array to store elements in reversed order
let temp = new Array(n);
// Copy elements from original array to temp in reverse order
for (let i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (let i = 0; i < n; i++)
arr[i] = temp[i];
}
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), Copying elements to a new array is a linear operation.
Auxiliary Space: O(n), as we are using an extra array to store the reversed array.
[Expected Approach - 1] Using Two Pointers - O(n) Time and O(1) Space
The idea is to maintain two pointers: left and right, such that left points at the beginning of the array and right points to the end of the array.
While left pointer is less than the right pointer, swap the elements at these two positions. After each swap, increment the left pointer and decrement the right pointer to move towards the center of array. This will swap all the elements in the first half with their corresponding element in the second half.
Working:
Below is the implementation of the algorithm:
C++
// C++ Program to reverse an array using Two Pointers
#include <iostream>
#include <vector>
using namespace std;
// function to reverse an array
void reverseArray(vector<int> &arr) {
// Initialize left to the beginning and right to the end
int left = 0, right = arr.size() - 1;
// Iterate till left is less than right
while(left < right) {
// Swap the elements at left and right position
swap(arr[left], arr[right]);
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C Program to reverse an array using Two Pointers
#include <stdio.h>
// Function to swap two numbers
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
// function to reverse an array
void reverseArray(int arr[], int n) {
// Initialize left to the beginning and right to the end
int left = 0, right = n - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left and right position
swap(&arr[left], &arr[right]);
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java Program to reverse an array using Two Pointers
import java.util.Arrays;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
// Initialize left to the beginning and right to the end
int left = 0, right = arr.length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left and right position
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
# Python Program to reverse an array using Two Pointers
# function to reverse an array
def reverseArray(arr):
# Initialize left to the beginning and right to the end
left = 0
right = len(arr) - 1
# Iterate till left is less than right
while left < right:
# Swap the elements at left and right position
arr[left], arr[right] = arr[right], arr[left]
# Increment the left pointer
left += 1
# Decrement the right pointer
right -= 1
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
// C# Program to reverse an array using Two Pointers
using System;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
// Initialize left to the beginning and right to the end
int left = 0, right = arr.Length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left and right position
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to reverse an array using Two Pointers
// function to reverse an array
function reverseArray(arr) {
// Initialize left to the beginning and right to the end
let left = 0, right = arr.length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left and right position
[arr[left], arr[right]] = [arr[right], arr[left]];
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), as we are visiting each element exactly once.
Auxiliary Space: O(1)
[Expected Approach - 2] By Swapping Elements - O(n) Time and O(1) Space
The idea is to iterate over the first half of the array and swap each element with its corresponding element from the end. So, while iterating over the first half, any element at index i is swapped with the element at index (n - i - 1).
Working:
Below is the implementation of the algorithm:
C++
// C++ Program to reverse an array by swapping elements
#include <iostream>
#include <vector>
using namespace std;
// function to reverse an array
void reverseArray(vector<int> &arr) {
int n = arr.size();
// Iterate over the first half and for every index i,
// swap arr[i] with arr[n - i - 1]
for(int i = 0; i < n/2; i++) {
swap(arr[i], arr[n - i - 1]);
}
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C Program to reverse an array by swapping elements
#include <stdio.h>
// function to reverse an array
void reverseArray(int arr[], int n) {
// Iterate over the first half and for every index i,
// swap arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java Program to reverse an array by swapping elements
import java.util.Arrays;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.length;
// Iterate over the first half and for every index i,
// swap arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
# Python Program to reverse an array by swapping elements
def reverseArray(arr):
n = len(arr)
# Iterate over the first half and for every index i,
# swap arr[i] with arr[n - i - 1]
for i in range(n // 2):
temp = arr[i]
arr[i] = arr[n - i - 1]
arr[n - i - 1] = temp
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
// C# Program to reverse an array by swapping elements
using System;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.Length;
// Iterate over the first half and for every index i,
// swap arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to reverse an array by swapping elements
// function to reverse an array
function reverseArray(arr) {
let n = arr.length;
// Iterate over the first half and for every index i,
// swap arr[i] with arr[n - i - 1]
for (let i = 0; i < n / 2; i++) {
let temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), the loop runs through half of the array, so it's linear with respect to the array size.
Auxiliary Space: O(1), no extra space is required, therefore we are reversing the array in-place.
[Alternate Approach] Using Recursion - O(n) Time and O(n) Space
The idea is to use recursion and define a recursive function that takes a range of array elements as input and reverses it. Inside the recursive function,
- Swap the first and last element.
- Recursively call the function with the remaining subarray.
C++
// C++ Program to reverse an array using Recursion
#include <iostream>
#include <vector>
using namespace std;
// recursive function to reverse an array from l to r
void reverseArrayRec(vector<int> &arr, int l, int r) {
if(l >= r)
return;
// Swap the elements at the ends
swap(arr[l], arr[r]);
// Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1);
}
// function to reverse an array
void reverseArray(vector<int> &arr) {
int n = arr.size();
reverseArrayRec(arr, 0, n - 1);
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// C Program to reverse an array using Recursion
#include <stdio.h>
// recursive function to reverse an array from l to r
void reverseArrayRec(int arr[], int l, int r) {
if(l >= r)
return;
// Swap the elements at the ends
int temp = arr[l];
arr[l] = arr[r];
arr[r] = temp;
// Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1);
}
// function to reverse an array
void reverseArray(int arr[], int n) {
reverseArrayRec(arr, 0, n - 1);
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for(int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
// Java Program to reverse an array using Recursion
import java.util.Arrays;
class GfG {
// recursive function to reverse an array from l to r
static void reverseArrayRec(int[] arr, int l, int r) {
if (l >= r)
return;
// Swap the elements at the ends
int temp = arr[l];
arr[l] = arr[r];
arr[r] = temp;
// Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1);
}
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.length;
reverseArrayRec(arr, 0, n - 1);
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
# Python Program to reverse an array using Recursion
# recursive function to reverse an array from l to r
def reverseArrayRec(arr, l, r):
if l >= r:
return
# Swap the elements at the ends
arr[l], arr[r] = arr[r], arr[l]
# Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1)
# function to reverse an array
def reverseArray(arr):
n = len(arr)
reverseArrayRec(arr, 0, n - 1)
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
// C# Program to reverse an array using Recursion
using System;
class GfG {
// recursive function to reverse an array from l to r
static void reverseArrayRec(int[] arr, int l, int r) {
if (l >= r)
return;
// Swap the elements at the ends
int temp = arr[l];
arr[l] = arr[r];
arr[r] = temp;
// Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1);
}
// function to reverse an array
static void reverseArray(int[] arr) {
int n = arr.Length;
reverseArrayRec(arr, 0, n - 1);
}
static void Main(string[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to reverse an array using Recursion
// recursive function to reverse an array from l to r
function reverseArrayRec(arr, l, r) {
if (l >= r)
return;
// Swap the elements at the ends
[arr[l], arr[r]] = [arr[r], arr[l]];
// Recur for the remaining array
reverseArrayRec(arr, l + 1, r - 1);
}
// function to reverse an array
function reverseArray(arr) {
let n = arr.length;
reverseArrayRec(arr, 0, n - 1);
}
let arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), the recurrence relation will be T(n) = T(n - 2) + O(1), which can be simplified to O(n).
Auxiliary Space: O(n), as we are using recursion stack.
Using Inbuilt Methods - O(n) Time and O(1) Space
The idea is to use inbuilt reverse methods available across different languages.
C++
// C++ Program to reverse an array using inbuilt methods
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// function to reverse an array
void reverseArray(vector<int> &arr) {
reverse(arr.begin(), arr.end());
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
// Java Program to reverse an array using inbuilt methods
import java.util.*;
class GfG {
// function to reverse an array
static void reverseArray(List<Integer> arr) {
Collections.reverse(arr);
}
public static void main(String[] args) {
List<Integer> arr =
new ArrayList<>(Arrays.asList(1, 4, 3, 2, 6, 5));
reverseArray(arr);
for (int i = 0; i < arr.size(); i++)
System.out.print(arr.get(i) + " ");
}
}
# Python Program to reverse an array using inbuilt methods
# function to reverse an array
def reverse_array(arr):
arr.reverse()
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverse_array(arr)
print(" ".join(map(str, arr)))
// C# Program to reverse an array using inbuilt methods
using System;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
Array.Reverse(arr);
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// JavaScript Program to reverse an array using inbuilt methods
// function to reverse an array
function reverseArray(arr) {
arr.reverse();
}
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), the reverse method has linear time complexity.
Auxiliary Space: O(1) Additional space is not used to store the reversed array, as the in-built array method swaps the values in-place.
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Second Largest Element in an ArrayGiven an array of positive integers arr[] of size n, the task is to find second largest distinct element in the array.Note: If the second largest element does not exist, return -1. Examples:Input: arr[] = [12, 35, 1, 10, 34, 1]Output: 34Explanation: The largest element of the array is 35 and the sec
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Move all zeros to end of arrayGiven an array of integers arr[], the task is to move all the zeros to the end of the array while maintaining the relative order of all non-zero elements.Examples: Input: arr[] = [1, 2, 0, 4, 3, 0, 5, 0]Output: arr[] = [1, 2, 4, 3, 5, 0, 0, 0]Explanation: There are three 0s that are moved to the end
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Rearrange array such that even positioned are greater than oddGiven an array arr[], sort the array according to the following relations: arr[i] >= arr[i - 1], if i is even, ∀ 1 <= i < narr[i] <= arr[i - 1], if i is odd, ∀ 1 <= i < nFind the resultant array.[consider 1-based indexing]Examples: Input: arr[] = [1, 2, 2, 1]Output: [1 2 1 2] Expla
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Rearrange an array in maximum minimum form using Two Pointer TechniqueGiven a sorted array of positive integers, rearrange the array alternately i.e first element should be a maximum value, at second position minimum value, at third position second max, at fourth position second min, and so on. Examples: Input: arr[] = {1, 2, 3, 4, 5, 6, 7} Output: arr[] = {7, 1, 6, 2
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Segregate even and odd numbers using Lomuto’s Partition SchemeGiven an array arr[] of integers, segregate even and odd numbers in the array such that all the even numbers should be present first, and then the odd numbers.Examples: Input: arr[] = {7, 2, 9, 4, 6, 1, 3, 8, 5}Output: 2 4 6 8 7 9 1 3 5Input: arr[] = {1, 3, 2, 4, 7, 6, 9, 10}Output: 2 4 6 10 7 1 9 3
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Reversal algorithm for Array rotationGiven an array arr[] of size N, the task is to rotate the array by d position to the left. Examples: Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, d = 2Output: 3, 4, 5, 6, 7, 1, 2Explanation: If the array is rotated by 1 position to the left, it becomes {2, 3, 4, 5, 6, 7, 1}.When it is rotated further by 1
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Print left rotation of array in O(n) time and O(1) spaceGiven an array of size n and multiple values around which we need to left rotate the array. How to quickly print multiple left rotations?Examples : Input : arr[] = {1, 3, 5, 7, 9}k1 = 1k2 = 3k3 = 4k4 = 6Output : 3 5 7 9 17 9 1 3 59 1 3 5 73 5 7 9 1Input : arr[] = {1, 3, 5, 7, 9}k1 = 14 Output : 9 1
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Sort an array which contain 1 to n valuesWe are given an array that contains 1 to n elements, our task is to sort this array in an efficient way. We are not allowed to simply copy the numbers from 1 to n.Examples : Input : arr[] = {2, 1, 3};Output : {1, 2, 3}Input : arr[] = {2, 1, 4, 3};Output : {1, 2, 3, 4} Native approach - O(n Log n) Ti
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Count Possible TrianglesGiven an unsorted array of positive integers, the task is to find the number of triangles that can be formed with three different array elements as three sides of triangles. For a triangle to be possible from 3 values as sides, the sum of the two values (or sides) must always be greater than the thi
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Find all distinct elements in a given arrayGiven an integer array arr[], find all the distinct elements of the given array. The given array may contain duplicates and the output should contain every element only once.Examples: Input: arr[] = [2, 2, 3, 3, 7, 5]Output: [2, 3, 7, 5]Explanation : After removing the duplicates 2 and 3 we get 2 3
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Unique Number IGiven an array of integers, every element in the array appears twice except for one element which appears only once. The task is to identify and return the element that occurs only once.Examples: Input: arr[] = [2, 3, 5, 4, 5, 3, 4]Output: 2 Explanation: Since 2 occurs once, while other numbers occu
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Leaders in an arrayGiven an array arr[] of size n, the task is to find all the Leaders in the array. An element is a Leader if it is greater than or equal to all the elements to its right side. Note: The rightmost element is always a leader. Examples: Input: arr[] = [16, 17, 4, 3, 5, 2]Output: [17 5 2]Explanation: 17
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Subarray with Given SumGiven a 1-based indexing array arr[] of non-negative integers and an integer sum. You mainly need to return the left and right indexes(1-based indexing) of that subarray. In case of multiple subarrays, return the subarray indexes which come first on moving from left to right. If no such subarray exi
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Intermediate problems on Array
Rearrange an array such that arr[i] = iGiven an array of elements of length n, ranging from 0 to n - 1. All elements may not be present in the array. If the element is not present then there will be -1 present in the array. Rearrange the array such that arr[i] = i and if i is not present, display -1 at that place.Examples: Input: arr[] =
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Alternate Rearrangement of Positives and NegativesAn array contains both positive and negative numbers in random order. Rearrange the array elements so that positive and negative numbers are placed alternatively. A number of positive and negative numbers need not be equal. If there are more positive numbers they appear at the end of the array. If t
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Reorder an array according to given indexesGiven two integer arrays of the same length, arr[] and index[], the task is to reorder the elements in arr[] such that after reordering, each element from arr[i] moves to the position index[i]. The new arrangement reflects the values being placed at their target indices, as described by index[] arra
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Find the smallest missing numberGiven a sorted array of n distinct integers where each integer is in the range from 0 to m-1 and m > n. Find the smallest number that is missing from the array. Examples: Input: {0, 1, 2, 6, 9}, n = 5, m = 10 Output: 3 Input: {4, 5, 10, 11}, n = 4, m = 12 Output: 0 Input: {0, 1, 2, 3}, n = 4, m =
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1D Difference ArrayIn many problems, we’re asked to perform multiple range updates like adding a value to all elements from index l to r. A direct approach updates each element in the range, leading to a time complexity of O(k × n) for k updates, which becomes inefficient for large inputs.The 1D Difference Array optim
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Stock Buy and Sell – Max 2 Transactions AllowedIn the stock market, a person buys a stock and sells it on some future date. Given the stock prices of n days in an array prices[ ]. Find out the maximum profit a person can make in at most 2 transactions. A transaction is equivalent to (buying + selling) of a stock and a new transaction can start o
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Smallest subarray with sum greater than a given valueGiven an array arr[] of integers and a number x, the task is to find the smallest subarray with a sum strictly greater than x.Examples:Input: x = 51, arr[] = [1, 4, 45, 6, 0, 19]Output: 3Explanation: Minimum length subarray is [4, 45, 6]Input: x = 100, arr[] = [1, 10, 5, 2, 7]Output: 0Explanation: N
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Count Inversions of an ArrayGiven an integer array arr[] of size n, find the inversion count in the array. Two array elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j.Note: Inversion Count for an array indicates that how far (or close) the array is from being sorted. If the array is already sorted
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Merge Two Sorted Arrays Without Extra SpaceGiven two sorted arrays a[] and b[] of size n and m respectively, the task is to merge both the arrays and rearrange the elements such that the smallest n elements are in a[] and the remaining m elements are in b[]. All elements in a[] and b[] should be in sorted order.Examples: Input: a[] = [2, 4,
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Majority ElementGiven an array arr[] of size n, find the element that appears more than ⌊n/2⌋ times. If no such element exists, return -1.Examples:Input: arr[] = [1, 1, 2, 1, 3, 5, 1]Output: 1Explanation: Element 1 appears 4 times. Since ⌊7/2⌋ = 3, and 4 > 3, it is the majority element.Input: arr[] = [7]Output:
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Two Pointers TechniqueTwo pointers is really an easy and effective technique that is typically used for Two Sum in Sorted Arrays, Closest Two Sum, Three Sum, Four Sum, Trapping Rain Water and many other popular interview questions. Given a sorted array arr (sorted in ascending order) and a target, find if there exists an
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3 Sum - Triplet Sum in ArrayGiven an array arr[] of size n and an integer sum, the task is to check if there is a triplet in the array which sums up to the given target sum.Examples: Input: arr[] = [1, 4, 45, 6, 10, 8], target = 13Output: true Explanation: The triplet [1, 4, 8] sums up to 13Input: arr[] = [1, 2, 4, 3, 6, 7], t
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Equilibrium IndexGiven an array arr[] of size n, the task is to return an equilibrium index (if any) or -1 if no equilibrium index exists. The equilibrium index of an array is an index such that the sum of all elements at lower indexes equals the sum of all elements at higher indexes. Note: When the index is at the
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Hard problems on Array
MO's Algorithm (Query Square Root Decomposition) | Set 1 (Introduction)Let us consider the following problem to understand MO's Algorithm. We are given an array and a set of query ranges, we are required to find the sum of every query range.Example: Input: arr[] = {1, 1, 2, 1, 3, 4, 5, 2, 8}; query[] = [0, 4], [1, 3] [2, 4]Output: Sum of arr[] elements in range [0, 4]
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Square Root (Sqrt) Decomposition AlgorithmSquare Root Decomposition Technique is one of the most common query optimization techniques used by competitive programmers. This technique helps us to reduce Time Complexity by a factor of sqrt(N) The key concept of this technique is to decompose a given array into small chunks specifically of size
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Sparse TableSparse table concept is used for fast queries on a set of static data (elements do not change). It does preprocessing so that the queries can be answered efficiently.Range Minimum Query Using Sparse TableYou are given an integer array arr of length n and an integer q denoting the number of queries.
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Range sum query using Sparse TableWe have an array arr[]. We need to find the sum of all the elements in the range L and R where 0 <= L <= R <= n-1. Consider a situation when there are many range queries. Examples: Input : 3 7 2 5 8 9 query(0, 5) query(3, 5) query(2, 4) Output : 34 22 15Note : array is 0 based indexed and q
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Range LCM QueriesGiven an array arr[] of integers of size N and an array of Q queries, query[], where each query is of type [L, R] denoting the range from index L to index R, the task is to find the LCM of all the numbers of the range for all the queries.Examples: Input: arr[] = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 4
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Jump Game - Minimum Jumps to Reach EndGiven an array arr[] of non-negative numbers. Each number tells you the maximum number of steps you can jump forward from that position.For example:If arr[i] = 3, you can jump to index i + 1, i + 2, or i + 3 from position i.If arr[i] = 0, you cannot jump forward from that position.Your task is to fi
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Space optimization using bit manipulationsThere are many situations where we use integer values as index in array to see presence or absence, we can use bit manipulations to optimize space in such problems.Let us consider below problem as an example.Given two numbers say a and b, mark the multiples of 2 and 5 between a and b using less than
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Maximum value of Sum(i*arr[i]) with array rotations allowedGiven an array arr[], the task is to determine the maximum possible value of the expression i*arr[i] after rotating the array any number of times (including zero).Note: In each rotation, every element of the array shifts one position to the right, and the last element moves to the front.Examples : I
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Construct an array from its pair-sum arrayGiven a pair-sum array arr[], where each element represents the sum of a unique pair of elements from an unknown original array res[].The pairs are considered in a specific order:First: res[0] + res[1]Then: res[0] + res[2], res[0] + res[3], ..., res[0] + res[n-1]Then: res[1] + res[2], res[1] + res[3
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Maximum equilibrium sum in an arrayGiven an array arr[]. Find the maximum value of prefix sum which is also suffix sum for index i in arr[].Examples : Input : arr[] = {-1, 2, 3, 0, 3, 2, -1}Output : 4Explanation : Prefix sum of arr[0..3] = Suffix sum of arr[3..6]Input : arr[] = {-3, 5, 3, 1, 2, 6, -4, 2}Output : 7Explanation : Prefix
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Smallest Difference Triplet from Three arraysThree arrays of same size are given. Find a triplet such that maximum - minimum in that triplet is minimum of all the triplets. A triplet should be selected in a way such that it should have one number from each of the three given arrays. If there are 2 or more smallest difference triplets, then the
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Top 50 Array Coding Problems for Interviews Array is one of the most widely used data structure and is frequently asked in coding interviews to the problem solving skills. The following list of 50 array coding problems covers a range of difficulty levels, from easy to hard, to help candidates prepare for interviews.Easy ProblemsSecond Largest
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