# Count minimum number of moves to front or end to sort an array

• Difficulty Level : Medium
• Last Updated : 19 Jun, 2022

Given an array arr[] of size N, the task is to find the minimum moves to the beginning or end of the array required to make the array sorted in non-decreasing order.

Examples:

Input: arr[] = {4, 7, 2, 3, 9}
Output:
Explanation:
Perform the following operations:
Step 1: Move the element 3 to the start of the array. Now, arr[] modifies to {3, 4, 7, 2, 9}.
Step 2: Move the element 2 to the start of the array. Now, arr[] modifies to {2, 3, 4, 7, 9}.
Now, the resultant array is sorted.
Therefore, the minimum moves required is 2.

Input: arr[] = {1, 4, 5, 7, 12}
Output:
Explanation:
The array is already sorted. Therefore, no moves required.

Naive Approach: The simplest approach is to check for every array element, how many moves are required to sort the given array arr[]. For each array element, if it is not at its sorted position, the following possibilities arise:

• Either move the current element to the front.
• Otherwise, move the current element to the end.

After performing the above operations, print the minimum number of operations required to make the array sorted. Below is the recurrence relation of the same:

• If the array arr[] is equal to the array brr[], then return 0.
• If arr[i] < brr[j], then count of operation will be:

1 + recursive_function(arr, brr, i + 1, j + 1)

• Otherwise, the count of operation can be calculated by taking the maximum of the following states:
1. recursive_function(arr, brr, i + 1, j)
2. recursive_function(arr, brr, i, j + 1)

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``int` `minOperations(``int` `arr1[], ``int` `arr2[],``                  ``int` `i, ``int` `j,``                  ``int` `n)``{``  ``// Base Case``  ``int` `f = 0;``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``if` `(arr1[i] != arr2[i])``      ``f = 1;``    ``break``;``  ``}``  ``if` `(f == 0)``    ``return` `0;` `  ``if` `(i >= n || j >= n)``    ``return` `0;` `  ``// If arr[i] < arr[j]``  ``if` `(arr1[i] < arr2[j])` `    ``// Include the current element``    ``return` `1 + minOperations(arr1, arr2,``                             ``i + 1, j + 1, n);` `  ``// Otherwise, excluding the current element``  ``return` `max(minOperations(arr1, arr2,``                           ``i, j + 1, n),``             ``minOperations(arr1, arr2,``                           ``i + 1, j, n));``}` `// Function that counts the minimum``// moves required to sort the array``void` `minOperationsUtil(``int` `arr[], ``int` `n)``{``  ``int` `brr[n];` `  ``for` `(``int` `i = 0; i < n; i++)``    ``brr[i] = arr[i];` `  ``sort(brr, brr + n);``  ``int` `f = 0;` `  ``// If both the arrays are equal``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``if` `(arr[i] != brr[i])` `      ``// No moves required``      ``f = 1;``    ``break``;``  ``}``  ` `  ``// Otherwise``  ``if` `(f == 1)``    ` `    ``// Print minimum``    ``// operations required``    ``cout << (minOperations(arr, brr,``                           ``0, 0, n));``  ``else``    ``cout << ``"0"``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = {4, 7, 2, 3, 9};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``minOperationsUtil(arr, n);``}` `// This code is contributed by Chitranayal`

## Java

 `// Java program for the above approach``import` `java.util.*;``import` `java.io.*;``import` `java.lang.Math;` `class` `GFG{` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``static` `int` `minOperations(``int` `arr1[], ``int` `arr2[],``                         ``int` `i, ``int` `j)``{``    ` `    ``// Base Case``    ``if` `(arr1.equals(arr2))``        ``return` `0``;``          ` `    ``if` `(i >= arr1.length || j >= arr2.length)``        ``return` `0``;``      ` `    ``// If arr[i] < arr[j]``    ``if` `(arr1[i] < arr2[j])``    ` `        ``// Include the current element``        ``return` `1` `+ minOperations(arr1, arr2,``                                 ``i + ``1``, j + ``1``);``         ` `    ``// Otherwise, excluding the current element``    ``return` `Math.max(minOperations(arr1, arr2,``                                  ``i, j + ``1``),``                    ``minOperations(arr1, arr2,``                                  ``i + ``1``, j));``}` `// Function that counts the minimum``// moves required to sort the array``static` `void` `minOperationsUtil(``int``[] arr)``{``    ``int` `brr[] = ``new` `int``[arr.length];``    ` `    ``for``(``int` `i = ``0``; i < arr.length; i++)``        ``brr[i] = arr[i];``        ` `    ``Arrays.sort(brr);``      ` `    ``// If both the arrays are equal``    ``if` `(arr.equals(brr))``    ` `        ``// No moves required``        ``System.out.print(``"0"``);``        ` `    ``// Otherwise``    ``else``    ` `        ``// Print minimum operations required``        ``System.out.println(minOperations(arr, brr,``                                         ``0``, ``0``));``}` `// Driver code``public` `static` `void` `main(``final` `String[] args)``{``    ``int` `arr[] = { ``4``, ``7``, ``2``, ``3``, ``9` `};``    ` `    ``minOperationsUtil(arr);``}``}` `// This code is contributed by bikram2001jha`

## Python3

 `# Python3 program for the above approach` `# Function that counts the minimum``# moves required to convert arr[] to brr[]``def` `minOperations(arr1, arr2, i, j):``    ` `    ``# Base Case``    ``if` `arr1 ``=``=` `arr2:``        ``return` `0``        ` `    ``if` `i >``=` `len``(arr1) ``or` `j >``=` `len``(arr2):``        ``return` `0``    ` `    ``# If arr[i] < arr[j]``    ``if` `arr1[i] < arr2[j]:``        ` `        ``# Include the current element``        ``return` `1` `\``        ``+` `minOperations(arr1, arr2, i ``+` `1``, j ``+` `1``)``        ` `    ``# Otherwise, excluding the current element``    ``return` `max``(minOperations(arr1, arr2, i, j ``+` `1``),``               ``minOperations(arr1, arr2, i ``+` `1``, j))``    ` `# Function that counts the minimum``# moves required to sort the array``def` `minOperationsUtil(arr):``    ` `    ``brr ``=` `sorted``(arr);``    ` `    ``# If both the arrays are equal``    ``if``(arr ``=``=` `brr):``        ` `        ``# No moves required``        ``print``(``"0"``)` `    ``# Otherwise``    ``else``:``        ` `        ``# Print minimum operations required``        ``print``(minOperations(arr, brr, ``0``, ``0``))` `# Driver Code` `arr ``=` `[``4``, ``7``, ``2``, ``3``, ``9``]` `minOperationsUtil(arr)`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``static` `int` `minOperations(``int``[] arr1, ``int``[] arr2,``                         ``int` `i, ``int` `j)``{``    ` `    ``// Base Case``    ``if` `(arr1.Equals(arr2))``        ``return` `0;``           ` `    ``if` `(i >= arr1.Length ||``        ``j >= arr2.Length)``        ``return` `0;``       ` `    ``// If arr[i] < arr[j]``    ``if` `(arr1[i] < arr2[j])``     ` `        ``// Include the current element``        ``return` `1 + minOperations(arr1, arr2,``                                 ``i + 1, j + 1);``          ` `    ``// Otherwise, excluding the current element``    ``return` `Math.Max(minOperations(arr1, arr2,``                                  ``i, j + 1),``                    ``minOperations(arr1, arr2,``                                  ``i + 1, j));``}` `// Function that counts the minimum``// moves required to sort the array``static` `void` `minOperationsUtil(``int``[] arr)``{``    ``int``[] brr = ``new` `int``[arr.Length];``     ` `    ``for``(``int` `i = 0; i < arr.Length; i++)``        ``brr[i] = arr[i];``         ` `    ``Array.Sort(brr);``       ` `    ``// If both the arrays are equal``    ``if` `(arr.Equals(brr))``     ` `        ``// No moves required``        ``Console.Write(``"0"``);``         ` `    ``// Otherwise``    ``else``     ` `        ``// Print minimum operations required``        ``Console.WriteLine(minOperations(arr, brr,``                                         ``0, 0));``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 4, 7, 2, 3, 9 };``     ` `    ``minOperationsUtil(arr);``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

`2`

Time Complexity: O(2N
Auxiliary Space: O(N)

Efficient Approach: The above approach has many overlapping subproblems. Therefore, the above approach can be optimized using Dynamic programming. Follow the steps below to solve the problem:

• Maintain a 2D array table[][] to store the computed results.
• Apply recursion to solve the problem using the results of smaller subproblems.
• If arr1[i] < arr2[j], then return 1 + minOperations(arr1, arr2, i + 1, j – 1, table)
• Otherwise, either move the i-th element of the array to the end or the j-th element of the array to the front. Therefore, the recurrence relation is:

table[i][j] = max(minOperations(arr1, arr2, i, j + 1, table), minOperations(arr1, arr2, i + 1, j, table))

• Finally, print the value stored in table[0][N – 1].

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``int` `minOperations(``int` `arr1[], ``int` `arr2[],``                  ``int` `i, ``int` `j,``                  ``int` `n)``{``  ``// Base Case``  ``int` `f = 0;``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``if` `(arr1[i] != arr2[i])``      ``f = 1;``    ``break``;``  ``}``  ``if` `(f == 0)``    ``return` `0;` `  ``if` `(i >= n || j >= n)``    ``return` `0;` `  ``// If arr[i] < arr[j]``  ``if` `(arr1[i] < arr2[j])` `    ``// Include the current element``    ``return` `1 + minOperations(arr1, arr2,``                             ``i + 1, j + 1, n);` `  ``// Otherwise, excluding the current element``  ``return` `max(minOperations(arr1, arr2,``                           ``i, j + 1, n),``             ``minOperations(arr1, arr2,``                           ``i + 1, j, n));``}` `// Function that counts the minimum``// moves required to sort the array``void` `minOperationsUtil(``int` `arr[], ``int` `n)``{``  ``int` `brr[n];` `  ``for` `(``int` `i = 0; i < n; i++)``    ``brr[i] = arr[i];` `  ``sort(brr, brr + n);``  ``int` `f = 0;` `  ``// If both the arrays are equal``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``if` `(arr[i] != brr[i])` `      ``// No moves required``      ``f = 1;``    ``break``;``  ``}``  ` `  ``// Otherwise``  ``if` `(f == 1)``    ` `    ``// Print minimum``    ``// operations required``    ``cout << (minOperations(arr, brr,``                           ``0, 0, n));``  ``else``    ``cout << ``"0"``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = {4, 7, 2, 3, 9};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``minOperationsUtil(arr, n);``}` `// This code is contributed by Chitranayal`

## Java

 `// Java program for the above approach``import` `java.util.*;``import` `java.io.*;``import` `java.lang.Math;` `class` `GFG{` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``static` `int` `minOperations(``int` `arr1[], ``int` `arr2[],``                         ``int` `i, ``int` `j)``{``    ` `    ``// Base Case``    ``if` `(arr1.equals(arr2))``        ``return` `0``;``          ` `    ``if` `(i >= arr1.length || j >= arr2.length)``        ``return` `0``;``      ` `    ``// If arr[i] < arr[j]``    ``if` `(arr1[i] < arr2[j])``    ` `        ``// Include the current element``        ``return` `1` `+ minOperations(arr1, arr2,``                                 ``i + ``1``, j + ``1``);``         ` `    ``// Otherwise, excluding the current element``    ``return` `Math.max(minOperations(arr1, arr2,``                                  ``i, j + ``1``),``                    ``minOperations(arr1, arr2,``                                  ``i + ``1``, j));``}` `// Function that counts the minimum``// moves required to sort the array``static` `void` `minOperationsUtil(``int``[] arr)``{``    ``int` `brr[] = ``new` `int``[arr.length];``    ` `    ``for``(``int` `i = ``0``; i < arr.length; i++)``        ``brr[i] = arr[i];``        ` `    ``Arrays.sort(brr);``      ` `    ``// If both the arrays are equal``    ``if` `(arr.equals(brr))``    ` `        ``// No moves required``        ``System.out.print(``"0"``);``        ` `    ``// Otherwise``    ``else``    ` `        ``// Print minimum operations required``        ``System.out.println(minOperations(arr, brr,``                                         ``0``, ``0``));``}` `// Driver code``public` `static` `void` `main(``final` `String[] args)``{``    ``int` `arr[] = { ``4``, ``7``, ``2``, ``3``, ``9` `};``    ` `    ``minOperationsUtil(arr);``}``}` `// This code is contributed by bikram2001jha`

## Python3

 `# Python3 program for the above approach` `# Function that counts the minimum``# moves required to convert arr[] to brr[]``def` `minOperations(arr1, arr2, i, j):``    ` `    ``# Base Case``    ``if` `arr1 ``=``=` `arr2:``        ``return` `0``        ` `    ``if` `i >``=` `len``(arr1) ``or` `j >``=` `len``(arr2):``        ``return` `0``    ` `    ``# If arr[i] < arr[j]``    ``if` `arr1[i] < arr2[j]:``        ` `        ``# Include the current element``        ``return` `1` `\``        ``+` `minOperations(arr1, arr2, i ``+` `1``, j ``+` `1``)``        ` `    ``# Otherwise, excluding the current element``    ``return` `max``(minOperations(arr1, arr2, i, j ``+` `1``),``               ``minOperations(arr1, arr2, i ``+` `1``, j))``    ` `# Function that counts the minimum``# moves required to sort the array``def` `minOperationsUtil(arr):``    ` `    ``brr ``=` `sorted``(arr);``    ` `    ``# If both the arrays are equal``    ``if``(arr ``=``=` `brr):``        ` `        ``# No moves required``        ``print``(``"0"``)` `    ``# Otherwise``    ``else``:``        ` `        ``# Print minimum operations required``        ``print``(minOperations(arr, brr, ``0``, ``0``))` `# Driver Code` `arr ``=` `[``4``, ``7``, ``2``, ``3``, ``9``]` `minOperationsUtil(arr)`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function that counts the minimum``// moves required to convert arr[] to brr[]``static` `int` `minOperations(``int``[] arr1, ``int``[] arr2,``                         ``int` `i, ``int` `j)``{``    ` `    ``// Base Case``    ``if` `(arr1.Equals(arr2))``        ``return` `0;``           ` `    ``if` `(i >= arr1.Length ||``        ``j >= arr2.Length)``        ``return` `0;``       ` `    ``// If arr[i] < arr[j]``    ``if` `(arr1[i] < arr2[j])``     ` `        ``// Include the current element``        ``return` `1 + minOperations(arr1, arr2,``                                 ``i + 1, j + 1);``          ` `    ``// Otherwise, excluding the current element``    ``return` `Math.Max(minOperations(arr1, arr2,``                                  ``i, j + 1),``                    ``minOperations(arr1, arr2,``                                  ``i + 1, j));``}` `// Function that counts the minimum``// moves required to sort the array``static` `void` `minOperationsUtil(``int``[] arr)``{``    ``int``[] brr = ``new` `int``[arr.Length];``     ` `    ``for``(``int` `i = 0; i < arr.Length; i++)``        ``brr[i] = arr[i];``         ` `    ``Array.Sort(brr);``       ` `    ``// If both the arrays are equal``    ``if` `(arr.Equals(brr))``     ` `        ``// No moves required``        ``Console.Write(``"0"``);``         ` `    ``// Otherwise``    ``else``     ` `        ``// Print minimum operations required``        ``Console.WriteLine(minOperations(arr, brr,``                                         ``0, 0));``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 4, 7, 2, 3, 9 };``     ` `    ``minOperationsUtil(arr);``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

`2`

Time Complexity: O(N2)

Auxiliary Space: O(N2)

We are using two variables namely i and j to determine a unique state of the DP(Dynamic Programming) stage, and each one of i and j can attain N values from 0 to N-1. Thus the recursion will have N*N number of transitions each of O(1) cost. Hence the time complexity is O(N*N).

More Efficient Approach: Sort the given array keeping index of elements aside, now find the longest streak of increasing values of index. This longest streak reflects that these elements are already sorted and do the above operation on rest of the elements. Let’s take above array as an example, arr = [8, 2, 1, 5, 4] after sorting their index values will be – [2, 1, 4, 3, 0], here longest streak is 2(1, 4) which means except these 2 numbers we have to follow the above operation and sort the array therefore, 5(arr.length) – 2 = 3 will be the answer.

Implementation of above approach:

## C++

 `// C++ algorithm of above approach` `#include ``#include ``using` `namespace` `std;` `// Function to find minimum number of operation required``// so that array becomes meaningful``int` `minOperations(``int` `arr[], ``int` `n)``{``    ``// Initializing vector of pair type which contains value``    ``// and index of arr``    ``vector> vect;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``vect.push_back(make_pair(arr[i], i));``    ``}` `    ``// Sorting array num on the basis of value``    ``sort(vect.begin(), vect.end());` `    ``// Initializing variables used to find maximum``    ``// length of increasing streak in index``    ``int` `res = 1;``    ``int` `streak = 1;``    ``int` `prev = vect[0].second;``    ``for` `(``int` `i = 1; i < n; i++) {``        ``if` `(prev < vect[i].second) {``            ``res++;` `            ``// Updating streak``            ``streak = max(streak, res);``        ``}``        ``else``            ``res = 1;``        ``prev = vect[i].second;``    ``}` `    ``// Returning number of elements left except streak``    ``return` `n - streak;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 4, 7, 2, 3, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `count = minOperations(arr, n);``    ``cout << count;``}`

## Java

 `// Java algorithm for above approach` `import` `java.util.*;` `class` `GFG {` `    ``// Pair class which will store element of array with its``    ``// index``    ``public` `static` `class` `Pair {``        ``int` `val;``        ``int` `idx;``        ``Pair(``int` `val, ``int` `idx)``        ``{``            ``this``.val = val;``            ``this``.idx = idx;``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``5``;``        ``int``[] arr = { ``4``, ``7``, ``2``, ``3``, ``9` `};``        ``System.out.println(minOperations(arr, n));``    ``}` `    ``// Function to find minimum number of operation required``    ``// so that array becomes meaningful``    ``public` `static` `int` `minOperations(``int``[] arr, ``int` `n)``    ``{``        ``// Initializing array of Pair type which can be used``        ``// to sort arr with respect to its values``        ``Pair[] num = ``new` `Pair[n];``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``num[i] = ``new` `Pair(arr[i], i);``        ``}` `        ``// Sorting array num on the basis of value``        ``Arrays.sort(num, (Pair a, Pair b) -> a.val - b.val);` `        ``// Initializing variables used to find maximum``        ``// length of increasing streak in index``        ``int` `res = ``1``;``        ``int` `streak = ``1``;``        ``int` `prev = num[``0``].idx;``        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``if` `(prev < num[i].idx) {``                ``res++;` `                ``// Updating streak``                ``streak = Math.max(res, streak);``            ``}``            ``else``                ``res = ``1``;``            ``prev = num[i].idx;``        ``}` `        ``// Returning number of elements left except streak``        ``return` `n - streak;``    ``}``}`

## Python3

 `# Python algorithm of above approach` `# Function to find minimum number of operation required``# so that array becomes meaningful``def` `minOperations(arr, n):` `    ``# Initializing vector of pair type which contains value``    ``# and index of arr``    ``vect ``=` `[]``    ``for` `i ``in` `range``(n):``        ``vect.append([arr[i], i])` `    ``# Sorting array num on the basis of value``    ``vect.sort()` `    ``# Initializing variables used to find maximum``    ``# length of increasing streak in index``    ``res ``=` `1``    ``streak ``=` `1``    ``prev ``=` `vect[``0``][``1``]``    ``for` `i ``in` `range``(``1``,n):``        ``if` `(prev < vect[i][``1``]):``            ``res ``+``=` `1` `            ``# Updating streak``            ``streak ``=` `max``(streak, res)``        ``else``:``            ``res ``=` `1``        ``prev ``=` `vect[i][``1``]` `    ``# Returning number of elements left except streak``    ``return` `n ``-` `streak` `# Driver code``arr ``=` `[ ``4``, ``7``, ``2``, ``3``, ``9` `]``n ``=` `len``(arr)``count ``=` `minOperations(arr, n)``print``(count)` `# This code is contributed by shinjanpatra.`

## C#

 `// C# program to implement above approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `class` `GFG``{``  ``// Pair class which will store element of array with its``  ``// index``  ``class` `Pair {``    ``public` `int` `val;``    ``public` `int` `idx;``    ``public` `Pair(``int` `val, ``int` `idx)``    ``{``      ``this``.val = val;``      ``this``.idx = idx;``    ``}``  ``}` `  ``// Comparator Function``  ``class` `Comp : IComparer{``    ``public` `int` `Compare(Pair a, Pair b)``    ``{``      ``return` `a.val - b.val;``    ``}``  ``}` `  ``// Function to find minimum number of operation required``  ``// so that array becomes meaningful``  ``public` `static` `int` `minOperations(``int``[] arr, ``int` `n)``  ``{``    ``// Initializing array of Pair type which can be used``    ``// to sort arr with respect to its values``    ``Pair[] num = ``new` `Pair[n];``    ``for` `(``int` `i = 0 ; i < n ; i++) {``      ``num[i] = ``new` `Pair(arr[i], i);``    ``}` `    ``// Sorting array num on the basis of value``    ``Array.Sort(num, ``new` `Comp());` `    ``// Initializing variables used to find maximum``    ``// length of increasing streak in index``    ``int` `res = 1;``    ``int` `streak = 1;``    ``int` `prev = num[0].idx;``    ``for` `(``int` `i = 1 ; i < n ; i++) {``      ``if` `(prev < num[i].idx) {``        ``res++;` `        ``// Updating streak``        ``streak = Math.Max(res, streak);``      ``}``      ``else``{``        ``res = 1;``      ``}``      ``prev = num[i].idx;``    ``}` `    ``// Returning number of elements left except streak``    ``return` `n - streak;``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(``string``[] args){` `    ``int` `n = 5;``    ``int``[] arr = ``new` `int``[]{ 4, 7, 2, 3, 9 };``    ``Console.WriteLine(minOperations(arr, n));` `  ``}``}` `// This code is contributed by subhamgoyal2014.`

## Javascript

 ``

Output

`2`

Time complexity: O(nlogn)

Auxiliary Space: O(n)

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