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: 2
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: 0
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:
- recursive_function(arr, brr, i + 1, j)
- 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 arrayvoid 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 codeint 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 codepublic 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 arraydef 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 Codearr = [4, 7, 2, 3, 9]minOperationsUtil(arr) |
C#
// C# program for the above approachusing 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 codestatic 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 arrayfunction 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 codelet 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 arrayvoid 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 codeint 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 codepublic 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 arraydef 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 Codearr = [4, 7, 2, 3, 9]minOperationsUtil(arr) |
C#
// C# program for the above approachusing 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 codestatic 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 arrayfunction 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 codelet 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 meaningfulint 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 Codeint 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 approachimport 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 meaningfuldef 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 codearr = [ 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 approachusing 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 meaningfulfunction 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 ocdelet 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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