Related Articles

# Move all values equal to K to the end of the Array

• Difficulty Level : Hard
• Last Updated : 26 Mar, 2021

Given an array arr[] of size N and an integer K, the task is to print the array after moving all value equal to K at the end of the array.
Examples:

Input: arr = [2, 1, 2, 2, 2, 3, 4, 2], K = 2
Output: [4, 1, 3, 2, 2, 2, 2, 2]
Explanation:
2 is the number which has to be moved to the end of the array arr[]. Therefore, after making the change the array is [4, 1, 3, 2, 2, 2, 2, 2]. The numbers 4, 1, and 3 could be ordered differently.
Input: arr = [1, 1, 3, 5, 6], K = 1
Output: [6, 5, 3, 1, 1 ]
Explanation:
1 is the number which has to be moved to the end of the array arr[]. Therefore, after making the change the array is [6, 5, 3, 1, 1 ].

Approach: To solve the problem mentioned above we use Two Pointer Technique.

1. Initialize two pointers where the left pointer marks the start of the array and the other one that is right one marks the end of the array, respectively.
2. Decrement the count of right pointer long as it points to K, and increment the left pointer as long as it doesn’t point to the integer m.
3. When both pointers aren’t moving, swap their values in place.
4. Repeat this process until the pointers pass each other.

Below is the implementation of the above approach:

## C++

 `// C++ program to move all values``// equal to K to the end of the Array` `#include ``using` `namespace` `std;` `// Function to move the element to the end``vector<``int``> moveElementToEnd(``    ``vector<``int``> array, ``int` `toMove)``{``    ``// Mark left pointer``    ``int` `i = 0;` `    ``// Mark the right pointer``    ``int` `j = array.size() - 1;` `    ``// Iterate untill left pointer``    ``// crosses the right pointer``    ``while` `(i < j) {` `        ``while` `(i < j && array[j] == toMove)` `            ``// decrement right pointer``            ``j--;` `        ``if` `(array[i] == toMove)` `            ``// swap the two elements``            ``// in the array``            ``swap(array[i], array[j]);` `        ``// increment left pointer``        ``i++;``    ``}` `    ``// return the result``    ``return` `array;``}` `// Driver code``int` `main(``int` `argc, ``char``* argv[])``{` `    ``vector<``int``> arr = { 1, 1, 3, 5, 6 };``    ``int` `K = 1;` `    ``vector<``int``> ans``        ``= moveElementToEnd(arr, K);` `    ``for` `(``int` `i = 0; i < arr.size(); i++)``        ``cout << ans[i] << ``" "``;` `    ``return` `0;``}`

## Java

 `// Java program to move all values``// equal to K to the end of the Array``class` `GFG{` `// Function to move the element to the end``static` `int``[] moveElementToEnd(``int` `[]array,``                              ``int` `toMove)``{``    ``// Mark left pointer``    ``int` `i = ``0``;` `    ``// Mark the right pointer``    ``int` `j = array.length - ``1``;` `    ``// Iterate untill left pointer``    ``// crosses the right pointer``    ``while` `(i < j)``    ``{``        ``while` `(i < j && array[j] == toMove)` `            ``// Decrement right pointer``            ``j--;` `        ``if` `(array[i] == toMove)` `            ``// Swap the two elements``            ``// in the array``            ``swap(array, i, j);` `        ``// Increment left pointer``        ``i++;``    ``}` `    ``// Return the result``    ``return` `array;``}` `static` `int``[] swap(``int` `[]arr, ``int` `i, ``int` `j)``{``    ``int` `temp = arr[i];``    ``arr[i] = arr[j];``    ``arr[j] = temp;``    ``return` `arr;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[]arr = { ``1``, ``1``, ``3``, ``5``, ``6` `};``    ``int` `K = ``1``;``    ``int` `[]ans = moveElementToEnd(arr, K);` `    ``for``(``int` `i = ``0``; i < arr.length; i++)``       ``System.out.print(ans[i] + ``" "``);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program to move all values``# equal to K to the end of the Array`` ` `# Function to move the element to the end``def` `moveElementToEnd(array, toMove):``    ` `    ``# Mark left pointer``    ``i ``=` `0`` ` `    ``# Mark the right pointer``    ``j ``=` `len``(array) ``-` `1`` ` `    ``# Iterate untill left pointer``    ``# crosses the right pointer``    ``while` `(i < j):`` ` `        ``while` `(i < j ``and` `array[j] ``=``=` `toMove):`` ` `            ``# decrement right pointer``            ``j``-``=``1`` ` `        ``if` `(array[i] ``=``=` `toMove):`` ` `            ``# swap the two elements``            ``# in the array``            ``array[i], array[j] ``=` `array[j] , array[i]`` ` `        ``# increment left pointer``        ``i ``+``=` `1`` ` `    ``# return the result``    ``return` `array`` ` `# Driver code``if` `__name__ ``=``=``"__main__"``:`` ` `    ``arr ``=` `[ ``1``, ``1``, ``3``, ``5``, ``6` `]``    ``K ``=` `1``    ``ans ``=` `moveElementToEnd(arr, K)``    ``for` `i ``in` `range``(``len``(arr)):``        ``print``(ans[i] ,end``=` `" "``)`` ` `# This code is contributed by chitranayal`

## C#

 `// C# program to move all values``// equal to K to the end of the Array``using` `System;``class` `GFG{` `// Function to move the element to the end``static` `int``[] moveElementToEnd(``int` `[]array,``                              ``int` `toMove)``{``    ``// Mark left pointer``    ``int` `i = 0;` `    ``// Mark the right pointer``    ``int` `j = array.Length - 1;` `    ``// Iterate untill left pointer``    ``// crosses the right pointer``    ``while` `(i < j)``    ``{``        ``while` `(i < j && array[j] == toMove)` `            ``// Decrement right pointer``            ``j--;` `        ``if` `(array[i] == toMove)` `            ``// Swap the two elements``            ``// in the array``            ``swap(array, i, j);` `        ``// Increment left pointer``        ``i++;``    ``}` `    ``// Return the result``    ``return` `array;``}` `static` `int``[] swap(``int` `[]arr, ``int` `i, ``int` `j)``{``    ``int` `temp = arr[i];``    ``arr[i] = arr[j];``    ``arr[j] = temp;``    ``return` `arr;``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `[]arr = { 1, 1, 3, 5, 6 };``    ``int` `K = 1;``    ``int` `[]ans = moveElementToEnd(arr, K);` `    ``for``(``int` `i = 0; i < arr.Length; i++)``    ``Console.Write(ans[i] + ``" "``);``}``}` `// This code is contributed by rock_cool`

## Javascript

 ``
Output:
`6 5 3 1 1`

Time complexity: O(N), where N is the length of the array.
Space complexity: O(1)

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

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

My Personal Notes arrow_drop_up