K-th digit in 'a' raised to power 'b'

Try it on GfG Practice

Given three numbers a, b and k, find k-th digit in a^b from right side

Examples: 

Input : a = 3, b = 3, k = 1
Output : 7
Explanation: 3^3 = 27 for k = 1. First digit is 7 in 27

Input : a = 5, b = 2,  k = 2
Output : 2
Explanation: 5^2 = 25 for k = 2. First digit is 2 in 25

The approach is simple.

• Computes the value of a^b
• Then iterates through the digits of the result. Starting from the rightmost digit, it extracts each digit one by one by using the modulus operation (p % 10).
• The extracted digit is compared with the desired position k. It continues removing the last digit (using integer division by 10) until it reaches the k-th digit, which is then returned.

#include <bits/stdc++.h>
using namespace std;

// Modular exponentiation to compute (a^b) % (10^k)
int kthDigit(int a, int b, int k){
    
    long long mod = pow(10LL, k); 
    long long res = 1;
    long long base = a;

    while (b > 0) {
        if (b & 1) {
            res = (res * base) % mod;
        }
        base = (base * base) % mod;
        b >>= 1;
    }
    
    for (int i = 1; i < k; i++)
            res /= 10;

    return (int)(res);
}

// Driver code
int main(){
    
    int a = 5, b = 2;
    int k = 1;
    cout << kthDigit(a, b, k); 
    return 0;
}

public class Main {
    static int kthDigit(int a, int b, int k) {
        long mod = (long) Math.pow(10, k);
        long res = 1;
        long base = a;

        while (b > 0) {
            if ((b & 1) == 1) {
                res = (res * base) % mod;
            }
            base = (base * base) % mod;
            b >>= 1;
        }
        for (int i = 1; i < k; i++)
            res /= 10;

        return (int)res;
    }

    public static void main(String[] args) {
        int a = 5, b = 2, k = 1;
        System.out.println(kthDigit(a, b, k));
    }
}

def kthDigit(a, b, k):
    mod = 10 ** k
    res = 1
    base = a

    while b > 0:
        if b & 1:
            res = (res * base) % mod
        base = (base * base) % mod
        b >>= 1

    for i in range(1,k):
        res //= 10
    return res

# Driver code
if __name__ == "__main__":
    a = 5
    b = 2
    k = 1
    print(kthDigit(a, b, k))

using System;

class GfG{
    
    static int kthDigit(int a, int b, int k){

        long mod = (long)Math.Pow(10, k);
        long res = 1;
        long baseVal = a;

        while (b > 0){
            
            if ((b & 1) == 1){
                
                res = (res * baseVal) % mod;
            }
            baseVal = (baseVal * baseVal) % mod;
            b >>= 1;
        }
        for (int i = 1; i < k; i++)
            res /= 10;

        return (int)(res);
    }

    static void Main(){
        
        int a = 5, b = 2, k = 1;
        Console.WriteLine(kthDigit(a, b, k));
    }
}

function kthDigit(a, b, k) {
    let mod = 10 ** k;
    let res = 1;
    let base = a % mod;

    while (b > 0) {
        if (b & 1) {
            res = (res * base) % mod;
        }
        base = (base * base) % mod;
        b >>= 1;
    }

    for (let i = 1; i < k; i++) {
        res = Math.floor(res / 10);
    }

    return res;
}

// Example usage:
let a = 5, b = 2, k = 1;
console.log(kthDigit(a, b, k));  // Output: 5

5

Time Complexity: O(log b)
Auxiliary Space: O(1)