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).
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 factorint 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 Codeint 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 approachimport java.util.*;class GFG{ // Function to find the minimum inverting factorstatic 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 Codepublic 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 approachimport 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 approachusing System; class GFG{ // Function to find the minimum inverting factorstatic 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 Codepublic 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 factorfunction 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.
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