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Array Reverse – Complete Tutorial

Last Updated : 25 Sep, 2024
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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.

Table of Content

  • [Naive Approach] Using a temporary array – O(n) Time and O(n) Space
  • [Expected Approach – 1] Using Two Pointers – O(n) Time and O(1) Space
  • [Expected Approach – 2] By Swapping Elements – O(n) Time and O(1) Space
  • [Alternate Approach] Using Recursion – O(n) Time and O(n) Space
  • Using Inbuilt Methods – O(n) Time and O(1) Space

[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 
// 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 
// 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 
# 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# 
// 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 
// 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(" "));

Output
5 6 2 3 4 1

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 
// 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 
// 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 
# 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# 
// 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 
// 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(" "));

Output
5 6 2 3 4 1

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 
// 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 
// 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 
# 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# 
// 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 
// 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(" "));

Output
5 6 2 3 4 1

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 
// 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 
// 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 
# 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# 
// 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 
// 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(" "));

Output
5 6 2 3 4 1

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 
// 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 
# 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# 
// 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 
// 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(" "));

Output
5 6 2 3 4 1

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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      How to Search in a Sorted Array?In a sorted array, the search operation can be performed by using binary search. Below is the implementation of the above approach: [GFGTABS] C++ // C++ program to implement binary search in sorted array #include <bits/stdc++.h> using namespace std; int binarySe
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    • Array Sorting - Practice Problems
      Sorting an array means arranging the elements of the array in a certain order. Generally sorting in an array is done to arrange the elements in increasing or decreasing order. Problem statement: Given an array of integers arr, the task is to sort the array in ascending order and return it, without u
      9 min read
    • Generating All Subarrays
      Given an array arr[], the task is to generate all the possible subarrays of the given array. Examples: Input: arr[] = [1, 2, 3]Output: [ [1], [1, 2], [2], [1, 2, 3], [2, 3], [3] ]Input: arr[] = [1, 2]Output: [ [1], [1, 2], [2] ] Iterative ApproachTo generate a subarray, we need a starting index from
      8 min read

    Easy problems on Array

    • Largest three distinct elements in an array
      Given an array arr[], the task is to find the top three largest distinct integers present in the array. Note: If there are less than three distinct elements in the array, then return the available distinct numbers in descending order. Examples : Input: arr[] = [10, 4, 3, 50, 23, 90]Output: [90, 50,
      6 min read
    • Second Largest Element in an Array
      Given 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 s
      14 min read
    • Move all zeros to end of array
      Given 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 en
      15 min read
    • Rearrange array such that even positioned are greater than odd
      Given 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] Expl
      9 min read
    • Rearrange an array in maximum minimum form using Two Pointer Technique
      Given 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
      6 min read
    • Segregate even and odd numbers using Lomuto’s Partition Scheme
      Given 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 5 Input: arr[] = {1, 3, 2, 4, 7, 6, 9, 10}Output: 2 4 6 10 7 1 9
      7 min read
    • Reversal algorithm for Array rotation
      Given 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
      15 min read
    • Print left rotation of array in O(n) time and O(1) space
      Given 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
      15+ min read
    • Sort an array which contain 1 to n values
      We 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) T
      7 min read
    • Count Possible Triangles
      Given 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
      15+ min read
    • Print all Distinct (Unique) Elements in given Array
      Given an integer array arr[], print all distinct elements from this array. The given array may contain duplicates and the output should contain every element only once. Examples: Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}Output: {12, 10, 9, 45, 2} Input: arr[] = {1, 2, 3, 4, 5}Output: {1, 2, 3, 4
      11 min read
    • Unique Number I
      Given 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 occ
      9 min read
    • Leaders in an array
      Given 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
      10 min read
    • Subarray with Given Sum
      Given 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
      11 min read

    Intermediate problems on Array

    • Rearrange an array such that arr[i] = i
      Given 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[]
      13 min read
    • Alternate Rearrangement of Positives and Negatives
      An 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
      11 min read
    • Reorder an array according to given indexes
      Given 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
      15+ min read
    • Find the smallest missing number
      Given 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 =
      15 min read
    • Difference Array | Range update query in O(1)
      You are given an integer array arr[] and a list of queries. Each query is represented as a list of integers where: [1, l, r, x]: Adds x to all elements from arr[l] to arr[r] (inclusive).[2]: Prints the current state of the array.You need to perform the queries in order. Examples : Input: arr[] = [10
      11 min read
    • Stock Buy and Sell – Max 2 Transactions Allowed
      In 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
      15+ min read
    • Smallest subarray with sum greater than a given value
      Given 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
      15+ min read
    • Count Inversions of an Array
      Given 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 sorte
      15+ min read
    • Merge Two Sorted Arrays Without Extra Space
      Given 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,
      15+ min read
    • Majority Element
      You are given an array arr, and your task is to find the majority element an element that occurs more than half the length of the array (i.e., arr.size() / 2). If such an element exists return it, otherwise return -1, indicating that no majority element is present. Examples : Input : arr[] = [1, 1,
      15+ min read
    • Two Pointers Technique
      Two 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
      12 min read
    • 3 Sum - Triplet Sum in Array
      Given 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 13 Input: arr[] = [1, 2, 4, 3, 6, 7],
      15 min read
    • Equilibrium Index
      Given 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
      15 min read

    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]
      15+ min read
    • Square Root (Sqrt) Decomposition Algorithm
      Square 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
      15+ min read
    • Sparse Table
      Sparse 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.
      15+ min read
    • Range sum query using Sparse Table
      We 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
      8 min read
    • Range LCM Queries
      Given 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,
      15+ min read
    • Jump Game - Minimum Jumps to Reach End
      Given 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 f
      15+ min read
    • Space optimization using bit manipulations
      There 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
      12 min read
    • Find maximum value of Sum( i*arr[i]) with only rotations on given array allowed
      Given an array arr[] of size N, the task is to find the maximum possible sum of i*arr[i] when the array can be rotated any number of times. Examples : Input: arr[] = {1, 20, 2, 10}Output: 72.We can get 72 by rotating array twice.{2, 10, 1, 20}20*3 + 1*2 + 10*1 + 2*0 = 72 Input: arr[] = {10, 1, 2, 3,
      12 min read
    • Construct an array from its pair-sum array
      Given a pair-sum array construct the original array. A pair-sum array for an array is the array that contains sum of all pairs in ordered form, i.e., pair[0] is sum of arr[0] and arr[1], pair[1] is sum of arr[0] and arr[2] and so on. Note that if the size of input array is n, then the size of pair a
      5 min read
    • Maximum equilibrium sum in an array
      Given 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 : Prefi
      11 min read
    • Smallest Difference Triplet from Three arrays
      Three 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
      10 min read
  • 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 Problems Second Large
    2 min read
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