Minimum inverting factor in an array

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 arr_i and arr_j 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).

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 <bits/stdc++.h> 
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[0]); 
  
    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

<?php 
// PHP implementation of the above approach 
  
// Function to find the minimum inverting factor 
function findMinimumInvertingFactor($arr, $N) 
{ 
    // ans stores the minimum inverting factor 
    $ans = PHP_INT_MAX; 
  
    // Iterate over the loop and convert each 
    // array element into its reversed form 
    for ($i = 0; $i < $N; $i++) 
    { 
        $s=""; 
        $num = $arr[$i]; 
  
        // Extract each digit of the number and 
        // store it in reverse order 
        while ($num > 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 
?> 

Output:

Time Complexity : O(N * logN)

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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