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

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

Examples:

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

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

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; 
} 

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 

OUTPUT

YES
NO

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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