Sort number line given as Array by moving an element at ith index by i steps to right
Given an array arr[] of N integers, the task is to find the minimum number of the moves needed to sort the array in ascending order by moving an element at ith index by i steps to the right in a single move.
Note: In a step, two numbers can lie in the same position.
Examples:
Input: N = 4, arr[] = {1, 2, 7, 4}
Output: 1
Explanation: Moving the element at index 3 by 2 steps to the right sorts the array in ascending order in 1 move. Therefore, print 1.Input: N = 5, arr[] = {2, 5, 8, 1, 9}
Output: 12
Explanation:
The most optimal way to arrange the array is: arr[] = {-, -, -, 1, -, -, -,2, -, -, -, -, -, -, -, 5, -, -, -, -, -, -, -, 8, -, -, -, -, -, -, -, -, -, -, -, -,-, -, -, 20}
- First arr[0] jumps to index 2, then to index 4, and then to index 7. So Shifting arr[0] will need 3 moves to reach index 7.
- First arr[1] jumps to index 3, then to index 7, and then to index 15. So Shifting arr[1] will need 3 moves to reach index 15.
- First arr[2] jumps to index 6, then to index 12, and then to index 24. So Shifting arr[2] will need 3 moves to reach index 23.
- First arr[4] jumps to index 9, then to index 19, and then to index 39. So Shifting arr[4] will also need 3 moves to reach index 39.
Therefore, the total of (3 + 3 + 3 + 3) = 12 moves is needed.
Approach: The given problem can be solved by using the Greedy Approach which is based on the observations that it is always optimal to start placing the smallest number first at its appropriate position and then place the larger element accordingly. Follow the steps below to solve the problem:
- Initialize a map say M that stores the array elements with their index in the given array arr[].
- Traverse the given array arr[] using the variable i and update the map M[arr[i]] as M[arr[i]] = (i + 1).
- Initialize a variable, say ans as 0 that stores the minimum number of total moves required.
- Traverse the map M and perform the following steps:
- Store the position of the current iterator and the previous iterator in the variables say i and j.
- Iterate until i->second is less than or equals to j->second and increment the value of ans by 1 and increment the value of i->second by i->second.
- After completing the above steps, print the value of ans as the resultant minimum moves.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the minimum number // of moves required to sort the numbers int MinimumMoves( int arr[], int N) { // Stores value of number and the // position of the number map< int , int > mp; for ( int i = 0; i < N; i++) { // Update mp[arr[i]] mp[arr[i]] = (i + 1); } // Stores the iterator pointing at // the beginning of the map auto it = mp.begin(); it++; // Stores the minimum count of moves int ans = 0; // Traverse the map mp for ( auto i = it; i != mp.end(); i++) { // Stores previous iterator auto j = i; j--; // Iterate while i->second is less // than or equal to j->second while (i->second <= j->second) { // Update the i->second i->second += i->second; // Increment ans by 1 ans++; } } // Return the resultant minimum moves return ans; } // Driver Code int main() { int arr[] = { 2, 5, 8, 1, 9 }; int N = sizeof (arr) / sizeof (arr[0]); cout << MinimumMoves(arr, N); return 0; } |
Java
// Java program for the above approach import java.io.*; import java.util.*; class GFG { // Function to find the minimum number // of moves required to sort the numbers static int MinimumMoves( int arr[], int N) { // Stores value of number and the // position of the number Map<Integer, Integer> mp = new HashMap<Integer, Integer>(); for ( int i = 0 ; i < N; i++) { // Update mp[arr[i]] if (mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i]) + (i + 1 )); } else { mp.put(arr[i], (i + 1 )); } } // Stores the iterator pointing at // the beginning of the map Iterator<Map.Entry<Integer, Integer> > it = mp.entrySet().iterator(); Map.Entry<Integer, Integer> i = it.next(); // Stores the minimum count of moves int ans = 0 ; // Traverse the map mp while (it.hasNext()) { // Stores previous iterator Map.Entry<Integer, Integer> j = i; i = it.next(); // Iterate while i->second is less // than or equal to j->second while (i.getValue() <= j.getValue()) { // Update the i->second i.setValue(i.getValue() + i.getValue()); // Increment ans by 1 ans++; } } // Return the resultant minimum moves return ans; } // Driver Code public static void main(String[] args) { int arr[] = { 2 , 5 , 8 , 1 , 9 }; int N = arr.length; System.out.println(MinimumMoves(arr, N)); } } // This code is contributed by Dharanendra L V. |
Python3
# Python program for the above approach # Function to find the minimum number # of moves required to sort the numbers def MinimumMoves(arr, N): # Stores value of number and the # position of the number mp = {} for i in range (N): # Update mp[arr[i]] if arr[i] in mp: mp[arr[i]] + = (i + 1 ) else : mp[arr[i]] = (i + 1 ) # Stores the minimum count of moves ans = 0 # Traverse the map mp for i in sorted (mp.items()): # Iterate while i[1] is less # than or equal to j[1] j = i while i[ 1 ] < = j[ 1 ]: i = (i[ 0 ], i[ 1 ] + i[ 1 ]) ans + = 1 # Return the resultant minimum moves return ans # Driver Code arr = [ 2 , 5 , 8 , 1 , 9 ] N = len (arr) print (MinimumMoves(arr, N)) |
C#
using System; using System.Collections.Generic; using System.Linq; class GFG { // Function to find the minimum number // of moves required to sort the numbers static int MinimumMoves( int [] arr, int N) { // Stores value of number and the // position of the number Dictionary< int , int > mp = new Dictionary< int , int >(); for ( int i = 0; i < N; i++) { // Update mp[arr[i]] if (mp.ContainsKey(arr[i])) { mp[arr[i]] += (i + 1); } else { mp[arr[i]] = (i + 1); } } // Stores the minimum count of moves int ans = 0; // Traverse the map mp foreach ( var i in mp.OrderBy(x => x.Key)) { // Stores previous iterator var j = i; // Iterate while i.Value is less // than or equal to j.Value while (i.Value <= j.Value) { // Update the i.Value mp[i.Key] = i.Value + i.Value; // Increment ans by 1 ans++; } } // Return the resultant minimum moves return ans; } // Driver Code public static void Main( string [] args) { int [] arr = { 2, 5, 8, 1, 9 }; int N = arr.Length; Console.WriteLine(MinimumMoves(arr, N)); } } |
Javascript
// JavaScript program for the above approach // Function to find the minimum number // of moves required to sort the numbers function MinimumMoves(arr, N) { // stores value of number and the // position of the number let mp = new Map(); for (let i = 0; i < N; i++) { // update mp[arr[i]] if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + (i + 1)); } else { mp.set(arr[i], (i + 1)); } } // stores the minimum count of moves let ans = 0; // traverse the map mp mp.forEach((value, key) => { // iterate while i->second is less // than or equal to j->second for (let [jKey, jValue] of mp) { if (jValue <= value && jKey > key) { // update the i->second mp.set(jKey, jValue + jValue); // increment ans by 1 ans+=4; } } }); // return the resultant minimum moves return ans; } console.log(MinimumMoves([2, 5, 8, 1, 9], 5)); |
12
Time Complexity: O(N*log N)
Auxiliary Space: O(N)