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 Codeif __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 otherfunction 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 Codelet 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.
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