Given two integers N and X. The task is to print the absolute difference between the first X and last X digits in N. Considering the number of digits is atleast 2*x.
Examples:
Input: N = 21546, X = 2 Output: 25 The first two digit in 21546 is 21. The last two digit in 21546 is 46. The absolute difference of 21 and 46 is 25. Input: N = 351684617, X = 3 Output: 266
Simple Approach:
- Store the last x digits of the number in last.
- Find the number of digits in the number.
- Remove all the digits except the first x.
- Store the first x integers of the number in first.
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// number of digits in the integerlong long digitsCount(long long n){ int len = 0; while (n > 0) { len++; n /= 10; } return len;}// Function to find the absolute differencelong long absoluteFirstLast(long long n, int x){ // Store the last x digits in last int i = 0, mod = 1; while (i < x) { mod *= 10; i++; } int last = n % mod; // Count the no. of digits in N long long len = digitsCount(n); // Remove the digits except the first x while (len != x) { n /= 10; len--; } // Store the first x digits in first int first = n; // Return the absolute difference between // the first and last return abs(first - last);}// Driver codeint main(){ long long n = 21546, x = 2; cout << absoluteFirstLast(n, x); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{// Function to find the// number of digits in the integerstatic int digitsCount(int n){ int len = 0; while (n > 0) { len++; n /= 10; } return len;}// Function to find the absolute differencestatic int absoluteFirstLast(int n, int x){ // Store the last x digits in last int i = 0, mod = 1; while (i < x) { mod *= 10; i++; } int last = n % mod; // Count the no. of digits in N int len = digitsCount(n); // Remove the digits except the first x while (len != x) { n /= 10; len--; } // Store the first x digits in first int first = n; // Return the absolute difference between // the first and last return Math.abs(first - last);}// Driver codepublic static void main(String args[]){ int n = 21546, x = 2; System.out.println(absoluteFirstLast(n, x));}}// This code is contributed by// Surendra_Gangwar |
Python3
# Python3 implementation of the above approach # Function to find the # number of digits in the integer def digitsCount(n) : length = 0; while (n > 0) : length += 1; n //= 10; return length; # Function to find the absolute difference def absoluteFirstLast(n, x) : # Store the last x digits in last i = 0 ; mod = 1; while (i < x) : mod *= 10; i += 1; last = n % mod; # Count the no. of digits in N length = digitsCount(n); # Remove the digits except the first x while (length != x) : n //= 10; length -= 1; # Store the first x digits in first first = n; # Return the absolute difference between # the first and last return abs(first - last); # Driver code if __name__ == "__main__" : n = 21546 ; x = 2; print(absoluteFirstLast(n, x)); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System;class GFG { // Function to find the // number of digits in the integer static int digitsCount(int n) { int len = 0; while (n > 0) { len++; n /= 10; } return len; } // Function to find the absolute difference static int absoluteFirstLast(int n, int x) { // Store the last x digits in last int i = 0, mod = 1; while (i < x) { mod *= 10; i++; } int last = n % mod; // Count the no. of digits in N int len = digitsCount(n); // Remove the digits except the first x while (len != x) { n /= 10; len--; } // Store the first x digits in first int first = n; // Return the absolute difference between // the first and last return Math.Abs(first - last); } // Driver code public static void Main(String []args) { int n = 21546, x = 2; Console.Write(absoluteFirstLast(n, x)); } } // This code has been contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the above approach// Function to find the number of // digits in the integerfunction digitsCount($n){ $len = 0; while ($n > 0) { $len++; $n = (int)($n / 10); } return $len;}// Function to find the absolute differencefunction absoluteFirstLast($n, $x){ // Store the last x digits in last $i = 0; $mod = 1; while ($i < $x) { $mod *= 10; $i++; } $last = $n % $mod; // Count the no. of digits in N $len = digitsCount($n); // Remove the digits except the first x while ($len != $x) { $n = (int)($n / 10); $len--; } // Store the first x digits in first $first = $n; // Return the absolute difference // between the first and last return abs($first - $last);}// Driver code$n = 21546;$x = 2;echo absoluteFirstLast($n, $x);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the above approach// Function to find the// number of digits in the integerfunction digitsCount(n){ let len = 0; while (n > 0) { len++; n = Math.floor(n / 10); } return len;}// Function to find the absolute differencefunction absoluteFirstLast(n, x){ // Store the last x digits in last let i = 0, mod = 1; while (i < x) { mod *= 10; i++; } let last = n % mod; // Count the no. of digits in N let len = digitsCount(n); // Remove the digits except the first x while (len != x) { n = Math.floor(n / 10); len--; } // Store the first x digits in first let first = n; // Return the absolute difference between // the first and last return Math.abs(first - last);}// driver program let n = 21546, x = 2; document.write(absoluteFirstLast(n, x)); </script> |
Output:
25
Time Complexity: O(X+k) where X is the given integer and k is the number of digits in n.
Auxiliary Space: O(1), since no extra space has been taken.
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