K-th digit from the end of a number

28 Jan, 2022
Given an integer N, the task is to find the Kth digit from the end of an integer N. If the Kth digit is not present, then print -1.
Examples:
Input: N = 2354, K = 2Output: 5
Input: N = 1234, K = 1Output: 4
Naive Approach: This simplest approach to solve this problem is converting the integer N to string. Follow the steps below to solve this problem:
If K is less than equal to 0, then print -1 and return.
Convert N into string and store in temp.
If K is greater than size of temp, then print -1, otherwise print K character from last.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// c++ program for the above approach#include <bits/stdc++.h> using namespace std; // Function to find the kth digit// from last in an integer nvoid kthDigitFromLast(int n, int k){     // If k is less than equal to 0    if (k <= 0) {        cout << -1 << endl;        return;    }     // Convert integer into string    string temp = to_string(n);     // If k is greater than length of the    // string temp    if (k > temp.length()) {        cout << -1 << endl;    }    // Print the k digit from last    else {        cout << temp[temp.length() - k] - '0';    }} // Driver codeint main(){    // Given Input    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}
// Java program for the above approachimport java.io.*; class GFG{     // Function to find the kth digit// from last in an integer n   public static void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        System.out.println(-1);        return;    }     // Convert integer into string    String temp = Integer.toString(n);     // If k is greater than length of the    // string temp    if (k > temp.length())    {        System.out.println(-1);    }         // Print the k digit from last    else    {        int index = temp.length() - k;        int res = temp.charAt(index) - '0';        System.out.println(res);    }} // Driver codepublic static void main(String[] args){         // Given Input    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by Potta Lokesh
# Python3 program for the above approach # Function to find the kth digit# from last in an integer ndef kthDigitFromLast(n, k):     # If k is less than equal to 0    if (k <= 0):        print(-1)        return     # Convert integer into string    temp = str(n)     # If k is greater than length of the    # temp    if (k > len(temp)):        print(-1)             # Print the k digit from last    else:        print(ord(temp[len(temp) - k]) - ord('0')) # Driver codeif __name__ == '__main__':         # Given Input    n = 2354    k = 2     # Function call    kthDigitFromLast(n, k) # This code is contributed by mohit kumar 29
// C# program for the above approachusing System;using System.Collections.Generic; class GFG{     // Function to find the kth digit// from last in an integer nstatic void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        Console.Write(-1);        return;    }     // Convert integer into string    string temp = n.ToString();     // If k is greater than length of the    // string temp    if (k > temp.Length)    {        Console.WriteLine(-1);    }         // Print the k digit from last    else    {        Console.Write(temp[temp.Length - k] - 48);    }} // Driver codepublic static void Main(){         // Given Input    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by SURENDRA_GANGWAR
<script> // JavaScript program for the above approach // Function to find the kth digit// from last in an integer nfunction kthDigitFromLast(n, k){         // If k is less than equal to 0    if (k <= 0)    {        document.write(-1);        return;    }     // Convert integer into string    var temp = String(n);         // If k is greater than length of the    // string temp    if (k > temp.length)    {        document.write(-1);    }         // Print the k digit from last    else    {        var req = temp.charAt(temp.length - k)                 // Convert to number again        document.write(Number(req));    }} // Driver code // Given Inputvar n = 2354;var k = 2; // Function callkthDigitFromLast(n, k); // This code is contributed by Potta Lokesh </script>
5
Time complexity: O(d), where d is the number of digits in number N.Auxiliary Space: O(d)
Efficient Approach: This problem can be solved by iterating over the digits of the number N. Follow the steps below to solve this problem: 
If K is less than equal to 0, then print -1 and return.
Iterate while N and K-1 are greater than 0 :Update N as N/10 and decrement K by 1.
Update N as N/10 and decrement K by 1.
If N is 0, then print -1, otherwise print N%10.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// c++ program for the above approach#include <bits/stdc++.h> using namespace std; // Function to find the kth digit// from last in an integer nvoid kthDigitFromLast(int n, int k){    // If k is less than equal to 0    if (k <= 0) {        cout << -1 << endl;        return;    }     // Divide the number n by 10    // upto k-1 times    while ((k - 1) > 0 && n > 0) {        n = n / 10;        k--;    }     // If the number n is equal 0    if (n == 0) {        cout << -1 << endl;    }    // Print the right most digit    else {        cout << n % 10 << endl;    }} // Driver codeint main(){    // Given Input    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}
// Java program for the above approachimport java.io.*; class GFG{     // Function to find the kth digit// from last in an integer npublic static void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        System.out.println(-1);        return;    }     // Divide the number n by 10    // upto k-1 times    while ((k - 1) > 0 && n > 0)    {        n = n / 10;        k--;    }     // If the number n is equal 0    if (n == 0)    {        System.out.println(-1);    }         // Print the right most digit    else    {        System.out.println(n % 10);    }} // Driver codepublic static void main(String[] args){    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by Potta Lokesh
# Python program for the above approach# Function to find the kth digit# from last in an eger ndef kthDigitFromLast( n,  k):     # If k is less than equal to 0    if (k <= 0):        print("-1")        return         # Divide the number n by 10    # upto k-1 times    while ((k - 1) > 0 and n > 0):        n = n / 10        k -= 1         # If the number n is equal 0    if (n == 0):        print("-1")         # Pr the right most digit    else:        print(int(n % 10 ))     # Driver codeif __name__ == '__main__':         # Given Input    n = 2354    k = 2     # Function call    kthDigitFromLast(n, k)     # this code is contributed by shivanisinghss2110
// C# program for the above approachusing System;class GFG{     // Function to find the kth digit// from last in an integer npublic static void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        Console.Write(-1);        return;    }     // Divide the number n by 10    // upto k-1 times    while ((k - 1) > 0 && n > 0)    {        n = n / 10;        k--;    }     // If the number n is equal 0    if (n == 0)    {        Console.Write(-1);    }         // Print the right most digit    else    {        Console.Write(n % 10);    }} // Driver codepublic static void Main(String[] args){    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by shivanisinghss2110
<script> // JavaScript program for the above approach // Function to find the kth digit// from last in an integer nfunction kthDigitFromLast(n, k){         // If k is less than equal to 0    if (k <= 0)    {        document.write(-1);        return;    }         // Divide the number n by 10    // upto k-1 times    while ((k - 1) > 0 && n > 0)    {        n = n / 10;        k--;    }         // If the number n is equal 0    if (n == 0)    {        document.write(-1);    }         // Print the right most digit    else    {        document.write(parseInt(n % 10));    }} // Driver code // Given Inputvar n = 2354;var k = 2; // Function callkthDigitFromLast(n, k); // This code is contributed by Potta Lokesh </script>
5
Time complexity: O(d), where d is the number of digits in number N.Auxiliary Space: O(1)
Efficient Approach: This problem can be solved using power function. Follow the steps below to solve this problem:
If K is less than equal to 0, then print -1 and return.
Initialize a variable say, divisor as pow(10, K-1).
If divisor is greater than N, then print -1, otherwise print (N/divisor) %10.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// c++ program for the above approach#include <bits/stdc++.h> using namespace std; // Function to find the kth digit// from last in an integer nvoid kthDigitFromLast(int n, int k){    // If k is less than equal to 0    if (k <= 0) {        cout << -1 << endl;        return;    }     // Calculate kth digit using power function    int divisor = (int)pow(10, k - 1);     // If divisor is greater than n    if (divisor > n) {        cout << -1 << endl;    }    // Otherwise print kth digit from last    else {        cout << (n / divisor) % 10 << endl;    }} // Driver codeint main(){    // Given Input    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}
// Java program for the above approachimport java.io.*; class GFG{     // Function to find the kth digit// from last in an integer npublic static void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        System.out.println(-1);        return;    }     // Calculate kth digit using power function    int diviser = (int)Math.pow(10, k - 1);     // If diviser is greater than n    if (diviser > n)    {        System.out.println(-1);    }         // Otherwise print kth digit from last    else    {        System.out.println((n / diviser) % 10);    }} // Driver codepublic static void main(String[] args){    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by Potta Lokesh
# Python program for the above approach # Function to find the kth digit# from last in an integer ndef kthDigitFromLast(n, k):         # If k is less than equal to 0    if (k <= 0):        print("-1")        return         # Calculate kth digit using power function    divisor = int(pow(10, k - 1))     # If divisor is greater than n    if (divisor > n):        print("-1")         # Otherwise print kth digit from last    else:        print(int((n / divisor) % 10))     # Given Inputn = 2354;k = 2; # Function callkthDigitFromLast(n, k); # This code is contributed by SoumikMondal
// C# program for the above approachusing System; class GFG{     // Function to find the kth digit// from last in an integer npublic static void kthDigitFromLast(int n, int k){         // If k is less than equal to 0    if (k <= 0)    {        Console.Write(-1);        return;    }     // Calculate kth digit using power function    int diviser = (int)Math.Pow(10, k - 1);     // If diviser is greater than n    if (diviser > n)    {        Console.Write(-1);    }         // Otherwise print kth digit from last    else    {        Console.Write((n / diviser) % 10);    }} // Driver codepublic static void Main(String[] args){    int n = 2354;    int k = 2;     // Function call    kthDigitFromLast(n, k);}} // This code is contributed by shivanisinghss2110
<script> // JavaScript program for the above approachfunction kthDigitFromLast(n, k){         // If k is less than equal to 0    if (k <= 0)    {        document.write(-1);        return;    }     // Calculate kth digit using power function    var diviser = parseInt(Math.pow(10, k - 1));     // If diviser is greater than n    if (diviser > n)    {        document.write(-1);    }         // Otherwise print kth digit from last    else    {        document.write(parseInt((n / diviser) % 10));    }} // Driver code // Given Inputvar n = 2354;var k = 2; // Function callkthDigitFromLast(n, k); // This code is contributed by Potta Lokesh </script>
5
Time complexity: O(log(K)), where d is the number of digits in number N. This time complexity is due to the calculation of power of 10 using power function.Auxiliary Space: O(1)
 
mohit kumar 29
SURENDRA_GANGWAR
lokeshpotta20
shivanisinghss2110
SoumikMondal
sooda367
clintra
maths-power
number-digits
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