Given two numbers as strings, find if one is a power of other

Last Updated : 19 Aug, 2022

Given two large numbers as strings, find if one is the power of another.

Examples:

Input : a = "374747", b = "52627712618930723"
Output : YES 
Explanation : 374747^3 = 52627712618930723

Input : a = "2", b = "4099"
Output : NO

Prerequisite: Multiply two large numbers represented as string The approach is simple. First, find smaller of the two strings and then multiply it with itself until it becomes equal to or greater than the larger string. If they are equal, return true, else return false. Below is the implementation of above approach

Implementation:

C++

// CPP program to check if one number is
// power of other
#include <bits/stdc++.h>
using namespace std;
 
bool isGreaterThanEqualTo(string s1, string s2)
{
    if (s1.size() > s2.size())
        return true;
 
    return (s1 == s2);
}
 
string multiply(string s1, string s2)
{
    int n = s1.size();
    int m = s2.size();
 
    vector<int> result(n + m, 0);
 
    // Multiply the numbers. It multiplies 
    // each digit of second string to each
    // digit of first and stores the result.
    for (int i = n - 1; i >= 0; i--) 
        for (int j = m - 1; j >= 0; j--) 
            result[i + j + 1] += 
               (s1[i] - '0') * (s2[j] - '0');
 
    // If the digit exceeds 9, add the 
    // cumulative carry to previous digit.
    int size = result.size();
    for (int i = size - 1; i > 0; i--) {
        if (result[i] >= 10) {
            result[i - 1] += result[i] / 10;
            result[i] = result[i] % 10;
        }
    }
 
    int i = 0;
    while (i < size && result[i] == 0)
        i++;
 
    // if all zeroes, return "0".
    if (i == size)
        return "0";
 
    string temp;
 
    // Remove starting zeroes.
    while (i < size) {
        temp += (result[i] + '0');
        i++;
    }
    return temp;
}
 
// Removes Extra zeroes from front of a string.
string removeLeadingZeores(string s)
{
    int n = s.size();
 
    int i = 0;
    while (i < n && s[i] == '0')
        i++;
 
    if (i == n)
        return "0";
 
    string temp;
    while (i < n)
        temp += s[i++];
 
    return temp;
}
 
bool isPower(string s1, string s2)
{
    // Make sure there are no leading zeroes 
    // in the string.
    s1 = removeLeadingZeores(s1);
    s2 = removeLeadingZeores(s2);
 
    if (s1 == "0" || s2 == "0")
        return false;
 
    if (s1 == "1" && s2 == "1")
        return true;
 
    if (s1 == "1" || s2 == "1")
        return true;
 
    // Making sure that s1 is smaller.
    // If it is greater, we recur we
    // reversed parameters.
    if (s1.size() > s2.size())
        return isPower(s2, s1);
 
    string temp = s1;
    while (!isGreaterThanEqualTo(s1, s2))
        s1 = multiply(s1, temp);
 
    return s1 == s2;
}
 
int main()
{
    string s1 = "374747", s2 = "52627712618930723";
    cout << (isPower(s1, s2) ? "YES\n" : "NO\n");
 
    s1 = "4099", s2 = "2";
    cout << (isPower(s1, s2) ? "YES\n" : "NO\n");
 
    return 0;
}

Java

// Java program to check if one number is 
// power of other 
import java.util.*;
 
class new_file
{
    static boolean isGreaterThanEqual(String s1, 
                                      String s2) 
    {
        if (s1.length() > s2.length())
            return true;
        if (s1.compareTo(s2) == 0)
            return true;
        return false;
    }
 
    static String multiply(String s1, String s2) 
    {
        int n = s1.length();
        int m = s2.length();
 
        int[] result = new int[n + m];
 
        // Multiply the numbers. It multiplies
        // each digit of second string to each
        // digit of first and stores the result.
        for (int i = n - 1; i >= 0; i--)
            for (int j = m - 1; j >= 0; j--)
                result[i + j + 1] += (s1.charAt(i) - '0') *
                                     (s2.charAt(j) - '0');
 
        // If the digit exceeds 9, add the
        // cumulative carry to previous digit.
        int size = result.length;
        for (int i = size - 1; i > 0; i--) 
        {
            if (result[i] >= 10)
            {
                result[i - 1] += result[i] / 10;
                result[i] = result[i] % 10;
            }
        }
 
        int i = 0;
        while (i < size && result[i] == 0)
            i++;
 
        // if all zeroes, return "0".
        if (i == size)
            return "0";
 
        String temp = "";
 
        // Remove starting zeroes.
        while (i < size)
        {
            temp += Integer.toString(result[i]);
            i++;
        }
        return temp;
    }
 
    // Removes Extra zeroes from front of a string.
    static String removeLeadingZeores(String s) 
    {
        int n = s.length();
        int i = 0;
        while (i < n && s.charAt(i) == '0')
            i++;
        if (i == n)
            return "0";
        String temp = "";
        while (i < n) 
        {
            temp += s.charAt(i);
            i++;
        }
 
        return temp;
    }
 
    static boolean isPower(String s1, String s2) 
    {
 
        // Make sure there are no leading zeroes
        // in the string.
        s1 = removeLeadingZeores(s1);
        s2 = removeLeadingZeores(s2);
        if (s1 == "0" || s2 == "0")
            return false;
        if (s1 == "1" && s2 == "1")
            return true;
        if (s1 == "1" || s2 == "1")
            return true;
 
        // Making sure that s1 is smaller.
        // If it is greater, we recur we
        // reversed parameters.
        if (s1.length() > s2.length())
            return isPower(s2, s1);
 
        String temp = new String(s1);
        while (!isGreaterThanEqual(s1, s2))
            s1 = multiply(s1, temp);
 
        if (s1.compareTo(s2) == 0)
            return true;
        return false;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String s1 = "374747", s2 = "52627712618930723";
        if (isPower(s1, s2))
            System.out.println("Yes");
        else
            System.out.println("No");
 
        s1 = "4099";
        s2 = "2";
 
        if (isPower(s1, s2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python3 program to check if one 
# number is a power of other 
def isGreaterThanEqualTo(s1, s2): 
 
    if len(s1) > len(s2): 
        return True
 
    return s1 == s2 
 
def multiply(s1, s2): 
 
    n, m = len(s1), len(s2)
    result = [0] * (n + m) 
 
    # Multiply the numbers. It multiplies 
    # each digit of second string to each 
    # digit of first and stores the result. 
    for i in range(n - 1, -1, -1): 
        for j in range(m - 1, -1, -1): 
            result[i + j + 1] += (int(s1[i]) *
                                  int(s2[j]))
 
    # If the digit exceeds 9, add the 
    # cumulative carry to previous digit. 
    size = len(result) 
    for i in range(size - 1, 0, -1): 
        if result[i] >= 10: 
            result[i - 1] += result[i] // 10
            result[i] = result[i] % 10
 
    i = 0
    while i < size and result[i] == 0: 
        i += 1
 
    # If all zeroes, return "0". 
    if i == size: 
        return "0"
 
    temp = "" 
     
    # Remove starting zeroes. 
    while i < size: 
        temp += str(result[i]) 
        i += 1
     
    return temp 
 
# Removes Extra zeroes from front of a string. 
def removeLeadingZeores(s): 
 
    n, i = len(s), 0
 
    while i < n and s[i] == '0': 
        i += 1
 
    if i == n: 
        return "0"
 
    temp = "" 
    while i < n: 
        temp += s[i]
        i += 1
         
    return temp 
 
def isPower(s1, s2): 
 
    # Make sure there are no leading 
    # zeroes in the string. 
    s1 = removeLeadingZeores(s1) 
    s2 = removeLeadingZeores(s2) 
 
    if s1 == "0" or s2 == "0": 
        return False
 
    if s1 == "1" and s2 == "1": 
        return True
 
    if s1 == "1" and s2 == "1": 
        return True
 
    # Making sure that s1 is smaller. 
    # If it is greater, we recur we 
    # reversed parameters. 
    if len(s1) > len(s2): 
        return isPower(s2, s1) 
 
    temp = s1 
    while not isGreaterThanEqualTo(s1, s2): 
        s1 = multiply(s1, temp) 
 
    return s1 == s2 
 
# Driver Code
if __name__ == "__main__":
 
    s1, s2 = "374747", "52627712618930723"
    if isPower(s1, s2):
        print("YES")
    else:
        print("NO")
         
    s1, s2 = "4099", "2"
    if isPower(s1, s2):
        print("YES")
    else:
        print("NO")
     
# This code is contributed by Rituraj Jain

C#

// C# program to check if one number is 
// power of other 
using System;
 
public class new_file
{
    static bool isGreaterThanEqual(String s1, 
                                    String s2) 
    {
        if (s1.Length > s2.Length)
            return true;
        if (s1.CompareTo(s2) == 0)
            return true;
        return false;
    }
 
    static String multiply(String s1, String s2) 
    {
        int n = s1.Length;
        int m = s2.Length;
        int i = 0;
        int[] result = new int[n + m];
 
        // Multiply the numbers. It multiplies
        // each digit of second string to each
        // digit of first and stores the result.
        for (i = n - 1; i >= 0; i--)
            for (int j = m - 1; j >= 0; j--)
                result[i + j + 1] += (s1[i] - '0') *
                                    (s2[j] - '0');
 
        // If the digit exceeds 9, add the
        // cumulative carry to previous digit.
        int size = result.Length;
        for (i = size - 1; i > 0; i--) 
        {
            if (result[i] >= 10)
            {
                result[i - 1] += result[i] / 10;
                result[i] = result[i] % 10;
            }
        }
 
        i = 0;
        while (i < size && result[i] == 0)
            i++;
 
        // if all zeroes, return "0".
        if (i == size)
            return "0";
 
        String temp = "";
 
        // Remove starting zeroes.
        while (i < size)
        {
            temp += (result[i]).ToString();
            i++;
        }
        return temp;
    }
 
    // Removes Extra zeroes from front of a string.
    static String removeLeadingZeores(String s) 
    {
        int n = s.Length;
        int i = 0;
        while (i < n && s[i] == '0')
            i++;
        if (i == n)
            return "0";
        String temp = "";
        while (i < n) 
        {
            temp += s[i];
            i++;
        }
 
        return temp;
    }
 
    static bool isPower(String s1, String s2) 
    {
 
        // Make sure there are no leading zeroes
        // in the string.
        s1 = removeLeadingZeores(s1);
        s2 = removeLeadingZeores(s2);
        if (s1 == "0" || s2 == "0")
            return false;
        if (s1 == "1" && s2 == "1")
            return true;
        if (s1 == "1" || s2 == "1")
            return true;
 
        // Making sure that s1 is smaller.
        // If it is greater, we recur we
        // reversed parameters.
        if (s1.Length > s2.Length)
            return isPower(s2, s1);
 
        String temp = s1;
        while (!isGreaterThanEqual(s1, s2))
            s1 = multiply(s1, temp);
 
        if (s1.CompareTo(s2) == 0)
            return true;
        return false;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        String s1 = "374747", s2 = "52627712618930723";
        if (isPower(s1, s2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
 
        s1 = "4099";
        s2 = "2";
 
        if (isPower(s1, s2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Python3 program to check if one
// number is a power of other
function isGreaterThanEqualTo(s1, s2){
 
    if(s1.length > s2.length)
        return true
 
    return s1 == s2
}
 
function multiply(s1, s2){
 
    let n  = s1.length,m = s2.length
    let result = new Array(n + m).fill(0)
 
    // Multiply the numbers. It multiplies
    // each digit of second string to each
    // digit of first && stores the result.
    for(let i=n - 1;i>=0;i--){
        for(let j=m-1;j>=0;j--)
            result[i + j + 1] += (parseInt(s1[i]) * parseInt(s2[j]))
    }
 
    // If the digit exceeds 9, add the
    // cumulative carry to previous digit.
    let size = result.length;
    for(let i=size - 1;i> 0;i--){
        if(result[i] >= 10){
            result[i - 1] += Math.floor(result[i] / 10)
            result[i] = result[i] % 10
        }
    }
 
    let i = 0
    while(i < size && result[i] == 0)
        i++
 
    // If all zeroes, return "0".
    if(i == size)
        return "0"
 
    let temp = ""
     
    // Remove starting zeroes.
    while(i < size){
        temp += (result[i])
        i += 1
    }
     
    return temp
}
 
// Removes Extra zeroes from front of a string.
function removeLeadingZeores(s){
 
    let n = s.length,i = 0
 
    while(i < n && s[i] == '0')
        i += 1
 
    if(i == n)
        return "0"
 
    let temp = ""
    while(i < n){
        temp += s[i]
        i += 1
    }
         
    return temp
}
 
function isPower(s1, s2){
 
    // Make sure there are no leading
    // zeroes in the string.
    s1 = removeLeadingZeores(s1)
    s2 = removeLeadingZeores(s2)
 
    if(s1 == "0" || s2 == "0")
        return false
 
    if(s1 == "1" && s2 == "1")
        return true
 
    if(s1 == "1" && s2 == "1")
        return true
 
    // Making sure that s1 is smaller.
    // If it is greater, we recur we
    // reversed parameters.
    if(s1.length > s2.length)
        return isPower(s2, s1)
 
    let temp = s1
    while(!isGreaterThanEqualTo(s1, s2))
        s1 = multiply(s1, temp)
 
    return s1 == s2
}
 
// Driver Code
 
let s1 = "374747", s2 = "52627712618930723"
if(isPower(s1, s2))
    document.write("YES","</br>")
else
    document.write("NO","</br>")
         
s1 = "4099",s2 = "2"
if(isPower(s1, s2))
    document.write("YES","</br>")
else
    document.write("NO","</br>")
     
// This code is contributed by shinjanpatra
 
</script>

Output

YES
NO

Complexity Analysis:

Time complexity : O(n*m) where n and m are the sizes of the respective input strings.
Auxiliary Space : O(n+m) where n and m are the sizes of the respective input strings.

Suggest improvement

Express 2^N in exactly N+1 powers of 2

Python | Count all prefixes in given string with greatest frequency

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Given two numbers as strings, find if one is a power of other

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