Given a positive integer N, the task is to check whether the given number N can be represented as the sum of the distinct powers of 3. If found to be true, then print “Yes”. Otherwise, “No”.
Examples:
Input: N = 28
Output: Yes
Explanation:
The number N(= 28) can be represented (1 + 7) = (30 + 33), which is a perfect power 2.Input: N = 6
Output: No
Approach: The simplest approach to solve the given problem is to generate all possible permutations of all distinct powers of 3 and if there exists any such combination whose sum is a perfect power of 3. As 315 > 107 so there are only 16 distinct powers i.e., [0, 15].
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to try all permutations// of distinct powersbool PermuteAndFind(vector<long> power, int idx, long SumSoFar, int target){ // Base Case if (idx == power.size()) { // If the distinct powers // sum is obtained if (SumSoFar == target) return true; // Otherwise return false; } // If current element not selected // in power[] bool notselect = PermuteAndFind(power, idx + 1, SumSoFar, target); // If current element selected in // power[] bool select = PermuteAndFind( power, idx + 1, SumSoFar + power[idx], target); // Return 1 if any permutation // found return (select || notselect);}// Function to check the N can be// represented as the sum of the// distinct powers of 3void DistinctPowersOf3(int N){ // Stores the all distincts powers // of three to [0, 15] vector<long> power(16); power[0] = 1; for (int i = 1; i < 16; i++) power[i] = 3 * power[i - 1]; // Function Call bool found = PermuteAndFind(power, 0, 0L, N); // print if (found == true) { cout << "Yes"; } else { cout << "No"; }}// Driven Codeint main(){ int N = 91; DistinctPowersOf3(N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to try all permutations // of distinct powers public static boolean PermuteAndFind(int[] power, int idx, int SumSoFar, int target) { // Base Case if (idx == power.length) { // If the distinct powers // sum is obtained if (SumSoFar == target) return true; // Otherwise return false; } // If current element not selected // in power[] boolean notselect = PermuteAndFind(power, idx + 1, SumSoFar, target); // If current element selected in // power[] boolean select = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target); // Return 1 if any permutation // found return (select || notselect); } // Function to check the N can be // represented as the sum of the // distinct powers of 3 public static void DistinctPowersOf3(int N) { // Stores the all distincts powers // of three to [0, 15] int[] power = new int[16]; power[0] = 1; for (int i = 1; i < 16; i++) power[i] = 3 * power[i - 1]; // Function Call boolean found = PermuteAndFind(power, 0, 0, N); // print if (found == true) { System.out.println("Yes"); } else { System.out.println("No"); } } // Driven Code public static void main(String args[]) { int N = 91; DistinctPowersOf3(N); }}// This code is contributed by _saurabh_jaiswal. |
Python3
# Python3 program for the above approach# Function to try all permutations# of distinct powersdef PermuteAndFind(power, idx, SumSoFar, target): # Base Case if (idx == len(power)): # If the distinct powers # sum is obtained if (SumSoFar == target): return True # Otherwise return False # If current element not selected # in power[] select = PermuteAndFind(power, idx + 1, SumSoFar, target) # If current element selected in # power[] notselect = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target) # Return 1 if any permutation # found return(select or notselect)# Function to check the N can be# represented as the sum of the# distinct powers of 3def DistinctPowersOf3(N): # Stores the all distincts powers # of three to[0, 15] power = [0 for x in range(16)] power[0] = 1 for i in range(1, 16): power[i] = 3 * power[i - 1] # Function Call found = PermuteAndFind(power, 0, 0, N) # Print if (found == True): print("Yes") else: print("No")# Driver CodeN = 91DistinctPowersOf3(N)# This code is contributed by amreshkumar3 |
C#
// C# program for the above approachusing System;class GFG{ // Function to try all permutations // of distinct powers public static bool PermuteAndFind(int[] power, int idx, int SumSoFar, int target) { // Base Case if (idx == power.Length) { // If the distinct powers // sum is obtained if (SumSoFar == target) return true; // Otherwise return false; } // If current element not selected // in power[] bool select = PermuteAndFind(power, idx + 1, SumSoFar, target); // If current element selected in // power[] bool notselect = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target); // Return 1 if any permutation // found return (select || notselect); } // Function to check the N can be // represented as the sum of the // distinct powers of 3 public static void DistinctPowersOf3(int N) { // Stores the all distincts powers // of three to [0, 15] int[] power = new int[16]; power[0] = 1; for (int i = 1; i < 16; i++) power[i] = 3 * power[i - 1]; // Function Call bool found = PermuteAndFind(power, 0, 0, N); // print if (found == true) { Console.Write("Yes"); } else { Console.Write("No"); } }// Driver Codepublic static void Main(){ int N = 91; DistinctPowersOf3(N);}}// This code is contributed by avijitmondal1998. |
Javascript
<script>// Javascript program for the above approach// Function to try all permutations// of distinct powersfunction PermuteAndFind(power, idx, SumSoFar, target) { // Base Case if (idx == power.length) { // If the distinct powers // sum is obtained if (SumSoFar == target) return true; // Otherwise return false; } // If current element not selected // in power[] let select = PermuteAndFind(power, idx + 1, SumSoFar, target); // If current element selected in // power[] let notselect = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target); // Return 1 if any permutation // found return select || notselect;}// Function to check the N can be// represented as the sum of the// distinct powers of 3function DistinctPowersOf3(N) { // Stores the all distincts powers // of three to [0, 15] let power = new Array(16); power[0] = 1; for (let i = 1; i < 16; i++) power[i] = 3 * power[i - 1]; // Function Call let found = PermuteAndFind(power, 0, 0, N); // print if (found == true) { document.write("Yes"); } else { document.write("No"); }}// Driven Codelet N = 91;DistinctPowersOf3(N);</script> |
Output
Yes
Time Complexity: O(2N)
Auxiliary Space: O(2N)
Another Approach: The above approach can also be optimized by observing the fact that N in ternary form (Base 3). Each digit is 3i, and the ternary number(N) is the sum of them.
To have distinct powers of 3, to sum up to N, in ternary form the ith digit can be either 0,1 or 2(In Binary, it is 0 and 1). Thus, there can be three cases for each digit at ith position:
- Digit can be 0 i.e. there is No 3i number in N.
- Digit can be 1 i.e, there is One 3i number in N.
- Digit cannot be 2 because then there are 2 of 3i, therefore, not distinct.
Follow the below steps to solve the problem:
- Iterate until N becomes 0 and perform the following steps:
- If the value of N%3 is 2, then print “No”.
- Otherwise, divide N by 3.
- After completing the above steps, if the value of N is 0 then print “Yes”. Otherwise, “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check whether the given// N can be represented as the sum of// the distinct powers of 3void DistinctPowersOf3(int N){ // Iterate until N is non-zero while (N > 0) { // Termination Condition if (N % 3 == 2) { cout << "No"; return; } // Right shift ternary bits // by 1 for the next digit N /= 3; } // If N can be expressed as the // sum of perfect powers of 3 cout << "Yes";}// Driver Codeint main(){ int N = 91; DistinctPowersOf3(N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to check whether the given// N can be represented as the sum of// the distinct powers of 3public static void DistinctPowersOf3(int N){ // Iterate until N is non-zero while (N > 0) { // Termination Condition if (N % 3 == 2) { System.out.println("No"); return; } // Right shift ternary bits // by 1 for the next digit N /= 3; } // If N can be expressed as the // sum of perfect powers of 3 System.out.println("Yes");}// Driver Codepublic static void main(String args[]){ int N = 91; DistinctPowersOf3(N);}}// This code is contributed by _saurabh_jaiswal. |
Python3
# Python3 program for the above approach# Function to check whether the given# N can be represented as the sum of# the distinct powers of 3def DistinctPowersOf3(N): # Iterate until N is non-zero while (N > 0): # Termination Condition if (N % 3 == 2): cout << "No" return # Right shift ternary bits # by 1 for the next digit N //= 3 # If N can be expressed as the # sum of perfect powers of 3 print ("Yes")# Driver Codeif __name__ == '__main__': N = 91 DistinctPowersOf3(N)# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;class GFG{ // Function to check whether the given// N can be represented as the sum of// the distinct powers of 3static void DistinctPowersOf3(int N){ // Iterate until N is non-zero while (N > 0) { // Termination Condition if (N % 3 == 2) { Console.Write("No"); return; } // Right shift ternary bits // by 1 for the next digit N /= 3; } // If N can be expressed as the // sum of perfect powers of 3 Console.Write("Yes");}// Driver Codepublic static void Main(){ int N = 91; DistinctPowersOf3(N);}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript program for the above approach// Function to check whether the given// N can be represented as the sum of// the distinct powers of 3function DistinctPowersOf3(N){ // Iterate until N is non-zero while (N > 0) { // Termination Condition if (N % 3 == 2) { document.write("No"); return; } // Right shift ternary bits // by 1 for the next digit N = Math.floor(N/ 3); } // If N can be expressed as the // sum of perfect powers of 3 document.write("Yes");}// Driver Codelet N = 91;DistinctPowersOf3(N);// This code is contributed by patel2127</script> |
Output
Yes
Time Complexity: O(log3N)
Auxiliary Space: O(1)
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