# Minimum inverting factor in an array

• Difficulty Level : Easy
• Last Updated : 31 May, 2021

Given an array of n positive integers, the task is to find the minimum inverting factor in the given array.
Inverting factor is defined as the absolute difference between the reverse of any two numbers arri and arrj where i != j.
Note : Trailing zeroes should be ignored while reversing the digits i.e. 1200 becomes 21 when reversed.
Examples:

Input : arr[] = { 56, 20, 47, 93, 45 }
Output : 9
The minimum inverting factor is 9, of the pair (56, 47).
Input : arr[] = { 26, 15, 45, 150 }
Output : 0
The minimum inverting factor is 0, of the pair (15, 150).

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A naive approach is to iterate over two loops to find all possible pairs. Reverse both the numbers individually and find their absolute difference. Update the inverting factor (minimum absolute difference) at every step. Time Complexity would be O(N2).
An efficient approach would be to precompute the reverse of each array element and store it in its reversed form only (considering the case of trailing zeroes also). Now, to find the minimum inverting factor, sort the array in non-decreasing order. Since the array is sorted, minimum absolute difference always occurs among any two adjacent numbers.
Below is the implementation of above approach .

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum inverting factor``int` `findMinimumInvertingFactor(``int` `arr[], ``int` `N)``{``    ``// ans stores the minimum inverting factor``    ``int` `ans = INT_MAX;` `    ``// Iterate over the loop and convert each``    ``// array element into its reversed form``    ``for` `(``int` `i = 0; i < N; i++) {``        ``string s;``        ``int` `num = arr[i];` `        ``// Extract each digit of the number and``        ``// store it in reverse order``        ``while` `(num > 0) {``            ``s.push_back(num % 10 + ``'0'``);``            ``num /= 10;``        ``}` `        ``// Find the position upto which trailing``        ``// zeroes occur``        ``int` `pos;``        ``for` `(pos = 0; pos < s.size(); pos++)``            ``if` `(s[pos] != 0)``                ``break``;` `        ``// Form the reversed number``        ``num = 0;``        ``for` `(``int` `j = pos; j < s.size(); j++)``            ``num = num * 10 + (s[j] - ``'0'``);``        ``arr[i] = num;``    ``}``    ``sort(arr, arr + N);` `    ``// Consider all adjacent pairs and update the``    ``// answer accordingly``    ``for` `(``int` `i = 1; i < N; i++)``        ``ans = min(ans, ``abs``(arr[i] - arr[i - 1]));` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 56, 20, 47, 93, 45 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << findMinimumInvertingFactor(arr, N) << endl;``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;` `class` `GFG``{``    ` `// Function to find the minimum inverting factor``static` `int` `findMinimumInvertingFactor(``int` `arr[], ``int` `N)``{``    ``// ans stores the minimum inverting factor``    ``int` `ans = Integer.MAX_VALUE;` `    ``// Iterate over the loop and convert each``    ``// array element into its reversed form``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``String s = ``""``;``        ``int` `num = arr[i];` `        ``// Extract each digit of the number and``        ``// store it in reverse order``        ``while` `(num > ``0``)``        ``{``            ``s+=(``char``)((num % ``10``) + ``'0'``);``            ``num /= ``10``;``        ``}` `        ``// Find the position upto which trailing``        ``// zeroes occur``        ``int` `pos;``        ``for` `(pos = ``0``; pos < s.length(); pos++)``            ``if` `(s.charAt(pos) != ``0``)``                ``break``;` `        ``// Form the reversed number``        ``num = ``0``;``        ``for` `(``int` `j = pos; j < s.length(); j++)``            ``num = num * ``10` `+ (s.charAt(j) - ``'0'``);``        ``arr[i] = num;``    ``}``    ``Arrays.sort(arr);` `    ``// Consider all adjacent pairs and update the``    ``// answer accordingly``    ``for` `(``int` `i = ``1``; i < N; i++)``        ``ans = Math.min(ans, Math.abs(arr[i] - arr[i - ``1``]));` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``56``, ``20``, ``47``, ``93``, ``45` `};``    ``int` `N = arr.length;` `    ``System.out.println(findMinimumInvertingFactor(arr, N));``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the above approach` `import` `sys` `# Function to find the minimum inverting factor``def` `findMinimumInvertingFactor(arr, N) :``    ` `    ``# ans stores the minimum inverting factor``    ``ans ``=` `sys.maxsize` `    ``# Iterate over the loop and convert each``    ``# array element into its reversed form``    ``for` `i ``in` `range``(N) :``        ``num ``=` `arr[i]``        ``s ``=` `""` `        ``# Extract each digit of the number and``        ``# store it in reverse order``        ``while` `(num > ``0``) :``            ``s ``+``=` `str``(num ``%` `10``)``            ``num ``/``/``=` `10``         `  `        ``# Find the position upto which trailing``        ``# zeroes occur``        ``for` `pos ``in` `range``(``len``(s)) :``            ``if` `(s[pos] !``=` `"0"``) :``                ``break``;` `        ``# Form the reversed number``        ``num ``=` `0``        ``for` `j ``in` `range``(pos, ``len``(s)) :``            ``num ``=` `num ``*` `10` `+` `(``ord``(s[j]) ``-` `ord``(``"0"``))``        ``arr[i] ``=` `num``     ` `    ``arr.sort()``    ``# Consider all adjacent pairs and update the``    ``# answer accordingly``    ``for` `i ``in` `range``(N) :``        ``ans ``=` `min``(ans, ``abs``(arr[i] ``-` `arr[i ``-` `1``]))` `    ``return` `ans`` `  `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr``=` `[ ``56``, ``20``, ``47``, ``93``, ``45` `]``    ``N ``=` `len``(arr)``    ` `    ``print``(findMinimumInvertingFactor(arr, N))` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to find the minimum inverting factor``static` `int` `findMinimumInvertingFactor(``int` `[]arr, ``int` `N)``{``    ``// ans stores the minimum inverting factor``    ``int` `ans = ``int``.MaxValue;` `    ``// Iterate over the loop and convert each``    ``// array element into its reversed form``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``String s = ``""``;``        ``int` `num = arr[i];` `        ``// Extract each digit of the number and``        ``// store it in reverse order``        ``while` `(num > 0)``        ``{``            ``s+=(``char``)((num % 10) + ``'0'``);``            ``num /= 10;``        ``}` `        ``// Find the position upto which trailing``        ``// zeroes occur``        ``int` `pos;``        ``for` `(pos = 0; pos < s.Length;pos++)``            ``if` `(s[pos] != 0)``                ``break``;` `        ``// Form the reversed number``        ``num = 0;``        ``for` `(``int` `j = pos; j < s.Length; j++)``            ``num = num * 10 + (s[j] - ``'0'``);``        ``arr[i] = num;``    ``}``    ``Array.Sort(arr);` `    ``// Consider all adjacent pairs and update the``    ``// answer accordingly``    ``for` `(``int` `i = 1; i < N; i++)``        ``ans = Math.Min(ans, Math.Abs(arr[i] - arr[i - 1]));` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 56, 20, 47, 93, 45 };``    ``int` `N = arr.Length;` `    ``Console.WriteLine(findMinimumInvertingFactor(arr, N));``}``}` `/* This code contributed by PrinciRaj1992 */`

## PHP

 ` 0)``        ``{``            ``\$s``.=``chr``(``\$num` `% 10 + ord(``'0'``));``            ``\$num` `=(int)(``\$num``/10);``        ``}` `        ``// Find the position upto which trailing``        ``// zeroes occur``        ``for` `(``\$pos` `= 0; ``\$pos` `< ``strlen``(``\$s``); ``\$pos``++)``            ``if` `(``\$s``[``\$pos``] != 0)``                ``break``;` `        ``// Form the reversed number``        ``\$num` `= 0;``        ``for` `(``\$j` `= ``\$pos``; ``\$j` `< ``strlen``(``\$s``); ``\$j``++)``            ``\$num` `= ``\$num` `* 10 + (ord(``\$s``[``\$j``]) - ord(``'0'``));``        ``\$arr``[``\$i``] = ``\$num``;``    ``}``    ``sort(``\$arr``);` `    ``// Consider all adjacent pairs and update the``    ``// answer accordingly``    ``for` `(``\$i` `= 1; ``\$i` `< ``\$N``; ``\$i``++)``        ``\$ans` `= min(``\$ans``, ``abs``(``\$arr``[``\$i``] - ``\$arr``[``\$i` `- 1]));` `    ``return` `\$ans``;``}` `// Driver Code` `\$arr` `= ``array``( 56, 20, 47, 93, 45 );``\$N` `= ``count``(``\$arr``);` `echo` `findMinimumInvertingFactor(``\$arr``, ``\$N``).``"\n"``;` `// This code is contributed by mits``?>`

## Javascript

 ``
Output:
`9`

Time Complexity : O(N * logN)

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