Given an array arr[] consisting of a permutation of integers [1, N], derived by rearranging the sorted order [1, N], the task is to find the minimum number of steps after which the sorted order [1, N] is repeated, by repeating the same process by which arr[] is obtained from the sorted sequence at each step.
Examples:Â
Input: arr[ ] = {3, 6, 5, 4, 1, 2}Â
Output: 6Â
Explanation:Â
Increasing Permutation: {1, 2, 3, 4, 5, 6}Â
Step 1 : arr[] = {3, 6, 5, 4, 1, 2} (Given array)Â
Step 2 : arr[] = {5, 2, 1, 4, 3, 6}Â
Step 3 : arr[] = {1, 6, 3, 4, 5, 2}Â
Step 4 : arr[] = {3, 2, 5, 4, 1, 6}Â
Step 5 : arr[] = {5, 6, 1, 4, 3, 2}Â
Step 6 : arr[] = {1, 2, 3, 4, 5, 6} (Increasing Permutation)Â
Therefore, the total number of steps required are 6.
Input: arr[ ] = [5, 1, 4, 3, 2]Â
Output: 6Â
Approach:Â
This problem can be solved simply by using the concept of Direct Addressing. Follow the steps given below to solve the problem:Â Â
- Initialize an array dat[] for direct addressing.
- Iterate over [1, N] and calculate the difference of the current index of every element from its index in the sorted sequence.
- Calculate the LCM of the array dat[].
- Now, print the obtained LCM as the minimum steps required to obtain the sorted order.
Below is the implementation of the above approach:Â
C++14
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; Â
// Function to find // GCD of two numbers int gcd(int a, int b) { Â Â Â Â if (b == 0) Â Â Â Â Â Â Â Â return a; Â
    return gcd(b, a % b); } Â
// Function to calculate the // LCM of array elements int findlcm(int arr[], int n) {     // Initialize result     int ans = 1; Â
    for (int i = 1; i <= n; i++)         ans = (((arr[i] * ans))             / (gcd(arr[i], ans))); Â
    return ans; } Â
// Function to find minimum steps // required to obtain sorted sequence void minimumSteps(int arr[], int n) { Â
    // Initialize dat[] array for     // Direct Address Table.     int i, dat[n + 1]; Â
    for (i = 1; i <= n; i++) Â
        dat[arr[i - 1]] = i; Â
    int b[n + 1], j = 0, c; Â
    // Calculating steps required     // for each element to reach     // its sorted position     for (i = 1; i <= n; i++) {         c = 1;         j = dat[i];         while (j != i) {             c++;             j = dat[j];         }         b[i] = c;     } Â
    // Calculate LCM of the array     cout << findlcm(b, n); } Â
// Driver Code int main() { Â
    int arr[] = { 5, 1, 4, 3, 2, 7, 6 }; Â
    int N = sizeof(arr) / sizeof(arr[0]); Â
    minimumSteps(arr, N); Â
    return 0; } |
Java
// Java program to implement// the above approachclass GFG{Â Â Â Â Â // Function to find// GCD of two numbersstatic int gcd(int a, int b){Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â return a;Â
    return gcd(b, a % b);}Â
// Function to calculate the// LCM of array elementsstatic int findlcm(int arr[], int n){         // Initialize result    int ans = 1;Â
    for(int i = 1; i <= n; i++)        ans = (((arr[i] * ans)) /             (gcd(arr[i], ans)));Â
    return ans;}Â
// Function to find minimum steps// required to obtain sorted sequencestatic void minimumSteps(int arr[], int n){Â
    // Initialize dat[] array for    // Direct Address Table.    int i;    int dat[] = new int[n + 1];Â
    for(i = 1; i <= n; i++)        dat[arr[i - 1]] = i;Â
    int b[] = new int[n + 1];    int j = 0, c;Â
    // Calculating steps required    // for each element to reach    // its sorted position    for(i = 1; i <= n; i++)    {        c = 1;        j = dat[i];                 while (j != i)         {            c++;            j = dat[j];        }        b[i] = c;    }Â
    // Calculate LCM of the array    System.out.println(findlcm(b, n));}Â
// Driver code   public static void main(String[] args){    int arr[] = { 5, 1, 4, 3, 2, 7, 6 };Â
    int N = arr.length;Â
    minimumSteps(arr, N);}}Â
// This code is contributed by rutvik_56 |
Python3
# Python3 program to implement # the above approach Â
# Function to find # GCD of two numbers def gcd(a, b): Â
    if(b == 0):         return a Â
    return gcd(b, a % b) Â
# Function to calculate the # LCM of array elements def findlcm(arr, n): Â
    # Initialize result     ans = 1Â
    for i in range(1, n + 1):         ans = ((arr[i] * ans) //            (gcd(arr[i], ans))) Â
    return ans Â
# Function to find minimum steps # required to obtain sorted sequence def minimumSteps(arr, n): Â
    # Initialize dat[] array for     # Direct Address Table.     dat = [0] * (n + 1) Â
    for i in range(1, n + 1):         dat[arr[i - 1]] = i Â
    b = [0] * (n + 1)     j = 0Â
    # Calculating steps required     # for each element to reach     # its sorted position     for i in range(1, n + 1):         c = 1        j = dat[i]         while(j != i):             c += 1            j = dat[j] Â
        b[i] = c Â
    # Calculate LCM of the array     print(findlcm(b, n)) Â
# Driver Code arr = [ 5, 1, 4, 3, 2, 7, 6 ] Â
N = len(arr) Â
minimumSteps(arr, N) Â
# This code is contributed by Shivam Singh |
C#
// C# program to implement// the above approachusing System;Â
class GFG{Â Â Â Â Â // Function to find// GCD of two numbersstatic int gcd(int a, int b){Â Â Â Â if (b == 0)Â Â Â Â Â Â Â Â return a;Â
    return gcd(b, a % b);}Â
// Function to calculate the// LCM of array elementsstatic int findlcm(int []arr, int n){         // Initialize result    int ans = 1;Â
    for(int i = 1; i <= n; i++)        ans = (((arr[i] * ans)) /             (gcd(arr[i], ans)));Â
    return ans;}Â
// Function to find minimum steps// required to obtain sorted sequencestatic void minimumSteps(int []arr, int n){Â
    // Initialize dat[] array for    // Direct Address Table.    int i;    int []dat = new int[n + 1];Â
    for(i = 1; i <= n; i++)        dat[arr[i - 1]] = i;Â
    int []b = new int[n + 1];    int j = 0, c;Â
    // Calculating steps required    // for each element to reach    // its sorted position    for(i = 1; i <= n; i++)    {        c = 1;        j = dat[i];                 while (j != i)         {            c++;            j = dat[j];        }        b[i] = c;    }Â
    // Calculate LCM of the array    Console.WriteLine(findlcm(b, n));}Â
// Driver code public static void Main(String[] args){Â Â Â Â int []arr = { 5, 1, 4, 3, 2, 7, 6 };Â
    int N = arr.Length;Â
    minimumSteps(arr, N);}}Â
// This code is contributed by gauravrajput1 |
Javascript
<script>// JavaScript program for the above approachÂ
// Function to find// GCD of two numbersfunction gcd(a, b){    if (b == 0)        return a;       return gcd(b, a % b);}   // Function to calculate the// LCM of array elementsfunction findlcm(arr, n){           // Initialize result    let ans = 1;       for(let i = 1; i <= n; i++)        ans = (((arr[i] * ans)) /             (gcd(arr[i], ans)));       return ans;}   // Function to find minimum steps// required to obtain sorted sequencefunction minimumSteps(arr, n){       // Initialize dat[] array for    // Direct Address Table.    let i;    let dat = Array.from({length: n+1}, (_, i) => 0);       for(i = 1; i <= n; i++)        dat[arr[i - 1]] = i;       let b = Array.from({length: n+1}, (_, i) => 0);    let j = 0, c;       // Calculating steps required    // for each element to reach    // its sorted position    for(i = 1; i <= n; i++)    {        c = 1;        j = dat[i];                   while (j != i)         {            c++;            j = dat[j];        }        b[i] = c;    }       // Calculate LCM of the array    document.write(findlcm(b, n));}     // Driver Code            let arr = [ 5, 1, 4, 3, 2, 7, 6 ];       let N = arr.length;       minimumSteps(arr, N);                   </script> |
Output:Â
6
Time Complexity: O(NlogN)Â
Auxiliary Space: O(N)
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