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

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  • Difficulty Level : Medium
  • Last Updated : 19 Jun, 2022
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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 <bits/stdc++.h>
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




<script>
 
// Javascript program for the above approach
 
// Function that counts the minimum
// moves required to convert arr[] to brr[]
function minOperations(arr1, arr2,
                       i, j, n)
{
     
    // Base Case
    let f = 0;
    for(let 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 Math.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
function minOperationsUtil(arr, n)
{
    let brr = new Array(n);
     
    for(let i = 0; i < n; i++)
        brr[i] = arr[i];
     
    brr.sort();
    let f = 0;
     
    // If both the arrays are equal
    for(let i = 0; i < n; i++)
    {
        if (arr[i] != brr[i])
     
        // No moves required
        f = 1;
        break;
    }
     
    // Otherwise
    if (f == 1)
         
        // Print minimum
        // operations required
        document.write(minOperations(arr, brr,
                                     0, 0, n));
    else
        cout << "0";
}
 
// Driver code
let arr = [ 4, 7, 2, 3, 9 ];
let n = arr.length;
 
minOperationsUtil(arr, n);
     
// This code is contributed by Mayank Tyagi
     
</script>

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 <bits/stdc++.h>
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




<script>
 
// Javascript program for the above approach
 
// Function that counts the minimum
// moves required to convert arr[] to brr[]
function minOperations(arr1, arr2,
                       i, j, n)
{
     
    // Base Case
    let f = 0;
    for(let 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 Math.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
function minOperationsUtil(arr, n)
{
    let brr = new Array(n);
     
    for(let i = 0; i < n; i++)
        brr[i] = arr[i];
     
    brr.sort();
    let f = 0;
     
    // If both the arrays are equal
    for(let i = 0; i < n; i++)
    {
        if (arr[i] != brr[i])
     
        // No moves required
        f = 1;
        break;
    }
     
    // Otherwise
    if (f == 1)
         
        // Print minimum
        // operations required
        document.write(minOperations(arr, brr,
                                     0, 0, n));
    else
        cout << "0";
}
 
// Driver code
let arr = [ 4, 7, 2, 3, 9 ];
let n = arr.length;
 
minOperationsUtil(arr, n);
     
// This code is contributed by Mayank Tyagi
     
</script>

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 <bits/stdc++.h>
#include <vector>
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<pair<int, int>> 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<Pair>{
    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




<script>
 
// JavaScript algorithm of above approach
 
 
// Function to find minimum number of operation required
// so that array becomes meaningful
function minOperations(arr, n)
{
    // Initializing vector of pair type which contains value
    // and index of arr
    let vect = [];
    for (let i = 0; i < n; i++) {
        vect.push([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
    let res = 1;
    let streak = 1;
    let prev = vect[0][1];
    for (let i = 1; i < n; i++) {
        if (prev < vect[i][1]) {
            res++;
 
            // Updating streak
            streak = Math.max(streak, res);
        }
        else
            res = 1;
        prev = vect[i][1];
    }
 
    // Returning number of elements left except streak
    return n - streak;
}
 
// driver ocde
let arr = [ 4, 7, 2, 3, 9 ];
let n = arr.length;
let count = minOperations(arr, n);
document.write(count,"</br>");
 
// This code is contributed by shinjanpatra.
</script>

Output

2

Time complexity: O(nlogn)

Auxiliary Space: O(n)


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