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Minimum inverting factor in an array

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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:
The minimum inverting factor is 9, of the pair (56, 47).

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

A naive approach is to iterate over two loops to find all possible pairs. Reverse both 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 Steps to solve this problem:

  •  Initialize ans to the maximum possible value of an integer (INT_MAX).
  •  Iterate over the loop 0 to N-1:
    • Initialize an empty string s
    • Extract each digit of the number arr[i] and store it in reverse order in s.
    • Find the position pos upto which trailing zeroes occur in s.
    • Form the reversed number num by iterating from pos to the end of s, multiplying the existing num by 10 and adding the digit at the current index.
    • Update arr[i] with the reversed number num.
  • Sort the array arr in ascending order using sort() function.
  • Iterate over loop 1 to N-1.
    • Compute the absolute difference between adjacent elements in arr.
    • Update ans with the minimum of the current value of ans and the computed difference.
  •  Return ans.

Below is the implementation of the 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
?>


Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to find the minimum inverting factor
    function findMinimumInvertingFactor(arr, N)
    {
     
        // ans stores the minimum inverting factor
        let ans = Number.MAX_VALUE;
 
        // Iterate over the loop and convert each
        // array element into its reversed form
        for (let i = 0; i < N; i++)
        {
            let s = "";
            let num = arr[i];
 
            // Extract each digit of the number and
            // store it in reverse order
            while (num > 0)
            {
                s+=String.fromCharCode((num % 10) + '0'.charCodeAt());
                num = parseInt(num / 10, 10);
            }
 
            // Find the position upto which trailing
            // zeroes occur
            let pos;
            for (pos = 0; pos < s.length;pos++)
                if (s[pos] != 0)
                    break;
 
            // Form the reversed number
            num = 0;
            for (let j = pos; j < s.length; j++)
                num = num * 10 + (s[j].charCodeAt() - '0'.charCodeAt());
            arr[i] = num;
        }
        arr.sort(function(a, b){return a - b});
 
        // Consider all adjacent pairs and update the
        // answer accordingly
        for (let i = 1; i < N; i++)
            ans = Math.min(ans, Math.abs(arr[i] - arr[i - 1]));
 
        return ans;
    }
     
    let arr = [ 56, 20, 47, 93, 45 ];
    let N = arr.length;
   
    document.write(findMinimumInvertingFactor(arr, N));
     
    // This code is contributed by decode2207.
</script>


Output

9

Time Complexity: O(N * logN)
Auxiliary Space: O(len), where len is the digits of the maximum number present in the array.



Last Updated : 01 Mar, 2023
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