# Minimize Steps required to obtain Sorted Order of an Array

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:

Input: arr[ ] = {3, 6, 5, 4, 1, 2}
Output:
Explanation:
Increasing Permutation: {1, 2, 3, 4, 5, 6}
Step 1 : arr[] = {3, 6, 5, 4, 1, 2} (Given array)
Step 2 : arr[] = {5, 2, 1, 4, 3, 6}
Step 3 : arr[] = {1, 6, 3, 4, 5, 2}
Step 4 : arr[] = {3, 2, 5, 4, 1, 6}
Step 5 : arr[] = {5, 6, 1, 4, 3, 2}
Step 6 : arr[] = {1, 2, 3, 4, 5, 6} (Increasing Permutation)
Therefore, the total number of steps required are 6.

Input: arr[ ] = [5, 1, 4, 3, 2]
Output:

Approach:
This problem can be solved simply by using the concept of Direct Addressing. Follow the steps given below to solve the problem:

• Initialize an array dat[] for direct addressing.
• Iterate over [1, N] and calculate the difference of current index every element from its index in 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:

 `// C++ Program to implement  ` `// the above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to find  ` `// GCD of two numbers  ` `int` `gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == 0)  ` `        ``return` `a;  ` ` `  `    ``return` `gcd(b, a % b);  ` `}  ` ` `  `// Function to calculate the  ` `// LCM of array elements  ` `int` `findlcm(``int` `arr[], ``int` `n)  ` `{  ` `    ``// Initialize result  ` `    ``int` `ans = 1;  ` ` `  `    ``for` `(``int` `i = 1; i <= n; i++)  ` `        ``ans = (((arr[i] * ans))  ` `            ``/ (gcd(arr[i], ans)));  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Function to find minimum steps  ` `// required to obtain sorted sequence  ` `void` `minimumSteps(``int` `arr[], ``int` `n)  ` `{  ` ` `  `    ``// Inititalize dat[] array for  ` `    ``// Direct Address Table.  ` `    ``int` `i, dat[n + 1];  ` ` `  `    ``for` `(i = 1; i <= n; i++)  ` ` `  `        ``dat[arr[i - 1]] = i;  ` ` `  `    ``int` `b[n + 1], 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  ` `    ``cout << findlcm(b, n);  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` ` `  `    ``int` `arr[] = { 5, 1, 4, 3, 2, 7, 6 };  ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);  ` ` `  `    ``minimumSteps(arr, N);  ` ` `  `    ``return` `0;  ` `}  `

 `// Java program to implement ` `// the above approach ` `class` `GFG{ ` `     `  `// Function to find ` `// GCD of two numbers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``) ` `        ``return` `a; ` ` `  `    ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to calculate the ` `// LCM of array elements ` `static` `int` `findlcm(``int` `arr[], ``int` `n) ` `{ ` `     `  `    ``// Initialize result ` `    ``int` `ans = ``1``; ` ` `  `    ``for``(``int` `i = ``1``; i <= n; i++) ` `        ``ans = (((arr[i] * ans)) /  ` `            ``(gcd(arr[i], ans))); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Function to find minimum steps ` `// required to obtain sorted sequence ` `static` `void` `minimumSteps(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Inititalize dat[] array for ` `    ``// Direct Address Table. ` `    ``int` `i; ` `    ``int` `dat[] = ``new` `int``[n + ``1``]; ` ` `  `    ``for``(i = ``1``; i <= n; i++) ` `        ``dat[arr[i - ``1``]] = i; ` ` `  `    ``int` `b[] = ``new` `int``[n + ``1``]; ` `    ``int` `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 ` `    ``System.out.println(findlcm(b, n)); ` `} ` ` `  `// Driver code     ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``5``, ``1``, ``4``, ``3``, ``2``, ``7``, ``6` `}; ` ` `  `    ``int` `N = arr.length; ` ` `  `    ``minimumSteps(arr, N); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

 `# Python3 program to implement  ` `# the above approach  ` ` `  `# Function to find  ` `# GCD of two numbers  ` `def` `gcd(a, b):  ` ` `  `    ``if``(b ``=``=` `0``):  ` `        ``return` `a  ` ` `  `    ``return` `gcd(b, a ``%` `b)  ` ` `  `# Function to calculate the  ` `# LCM of array elements  ` `def` `findlcm(arr, n):  ` ` `  `    ``# Initialize result  ` `    ``ans ``=` `1` ` `  `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``ans ``=` `((arr[i] ``*` `ans) ``/``/` `            ``(gcd(arr[i], ans)))  ` ` `  `    ``return` `ans  ` ` `  `# Function to find minimum steps  ` `# required to obtain sorted sequence  ` `def` `minimumSteps(arr, n):  ` ` `  `    ``# Inititalize dat[] array for  ` `    ``# Direct Address Table.  ` `    ``dat ``=` `[``0``] ``*` `(n ``+` `1``)  ` ` `  `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``dat[arr[i ``-` `1``]] ``=` `i  ` ` `  `    ``b ``=` `[``0``] ``*` `(n ``+` `1``)  ` `    ``j ``=` `0` ` `  `    ``# 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]  ` ` `  `        ``b[i] ``=` `c  ` ` `  `    ``# Calculate LCM of the array  ` `    ``print``(findlcm(b, n))  ` ` `  `# Driver Code  ` `arr ``=` `[ ``5``, ``1``, ``4``, ``3``, ``2``, ``7``, ``6` `]  ` ` `  `N ``=` `len``(arr)  ` ` `  `minimumSteps(arr, N)  ` ` `  `# This code is contributed by Shivam Singh  `

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to find ` `// GCD of two numbers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` ` `  `    ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to calculate the ` `// LCM of array elements ` `static` `int` `findlcm(``int` `[]arr, ``int` `n) ` `{ ` `     `  `    ``// Initialize result ` `    ``int` `ans = 1; ` ` `  `    ``for``(``int` `i = 1; i <= n; i++) ` `        ``ans = (((arr[i] * ans)) /  ` `            ``(gcd(arr[i], ans))); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Function to find minimum steps ` `// required to obtain sorted sequence ` `static` `void` `minimumSteps(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// Inititalize dat[] array for ` `    ``// Direct Address Table. ` `    ``int` `i; ` `    ``int` `[]dat = ``new` `int``[n + 1]; ` ` `  `    ``for``(i = 1; i <= n; i++) ` `        ``dat[arr[i - 1]] = i; ` ` `  `    ``int` `[]b = ``new` `int``[n + 1]; ` `    ``int` `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 ` `    ``Console.WriteLine(findlcm(b, n)); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 5, 1, 4, 3, 2, 7, 6 }; ` ` `  `    ``int` `N = arr.Length; ` ` `  `    ``minimumSteps(arr, N); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1  `

Output:
```6
```

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

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