Given a number n and a value k, the task is to find the smallest m(m>=n), such that m can be represented as a sum of distinct powers of k.
Examples:
Input: n = 5, k = 5
Output: 5
Explanation: 5 = 51Input: n = 29, k = 5
Output: 30
Explanation: 30 = 51 + 52
Approach:
- Store the k-nary(base k) representation of n. Then traverse through each element of the base k representation.
- If the base k representation of this position is 1 or 0, then continue. If it is >1, it means that the current power of k occurs more than once.
- In that case, that power is added k (k’s-position value) times, which makes it one power more (((k-1)+1).kx = k.kx = kx+1).
- Since the smallest number has to be found, after this step, all the lower powers of k are reduced to 0 as adding(k-b)kx (b=value at that position in the base k representation) has already made the bigger than the previous number.
- Finally, convert the number back to decimal form.
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;typedef long long ll;#define pb push_back// Function to find the smallest number// greater than or equal to n represented // as the sum of distinct powers of kvoid greaterK(ll n, ll k){ // Vector P to store the base k // representation of the number vector<ll> p; ll x = n; while (x) { p.pb(x % k); x /= k; } int idx = 0; for (ll i = 0; i < (ll)p.size() - 1; ++i) { if (p[i] >= 2) { // If the representation is >=2, then // this power of k has to be added // once again and then increase the // next power of k and make the // current power 0 p[i] = 0; p[i + 1]++; // Reduce all the lower power of // k to 0 for (int j = idx; j < i; ++j) { p[j] = 0; } idx = i + 1; } if (p[i] == k) { p[i] = 0; p[i + 1]++; } } ll j = (ll)p.size() - 1; // Check if the most significant // bit also satisfy the above // conditions if (p[j] >= 2) { for (auto& i : p) i = 0; p.pb(1); } ll ans = 0; // Converting back from the // k-nary representation to // decimal form. for (ll i = p.size() - 1; i >= 0; --i) { ans = ans * k + p[i]; } cout << ans << endl;}int main(){ ll n = 29, k = 7; greaterK(n, k); return 0;} |
Java
// Java implementation of the above approachclass GFG{// Function to find the smallest number// greater than or equal to n represented // as the sum of distinct powers of kstatic void greaterK(int n, int k){ // Vector P to store the base k // representation of the number int []p = new int[String.valueOf(n).length() + 2]; int index = 0; int x = n; while (x > 0) { p[index]=(int) (x % k); x /= k; index++; } int idx = 0; for (int i = 0; i < p.length - 1; ++i) { if (p[i] >= 2) { // If the representation is >=2, then // this power of k has to be added // once again and then increase the // next power of k and make the // current power 0 p[i] = 0; p[i + 1]++; // Reduce all the lower power of // k to 0 for (int j = idx; j < i; ++j) { p[j] = 0; } idx = i + 1; } if (p[i] == k) { p[i] = 0; p[i + 1]++; } } int j = p.length - 1; // Check if the most significant // bit also satisfy the above // conditions if (p[j] >= 2) { p[index] = 1; index++; } int ans = 0; // Converting back from the // k-nary representation to // decimal form. for (int i = p.length - 1; i >= 0; --i) { ans = ans * k + p[i]; } System.out.print(ans +"\n");}// Driver codepublic static void main(String[] args){ int n = 29, k = 7; greaterK(n, k);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the above approach# Function to find the smallest number# greater than or equal to n represented # as the sum of distinct powers of kdef greaterK(n, k): # Vector P to store the base k # representation of the number index = 0 p = [0 for i in range(n + 2)] x = n while (x > 0): p[index] = x % k x //= k index += 1 idx = 0 for i in range(0,len(p)-1, 1): if (p[i] >= 2): # If the representation is >=2, then # this power of k has to be added # once again and then increase the # next power of k and make the # current power 0 p[i] = 0 p[i + 1] += 1 # Reduce all the lower power of # k to 0 for j in range(idx, i, 1): p[j] = 0 idx = i + 1 if (p[i] == k): p[i] = 0 p[i + 1] += 1 j = len(p) - 1 # Check if the most significant # bit also satisfy the above # conditions if (p[j] >= 2): p[index] = 1 index += 1 ans = 0 # Converting back from the # k-nary representation to # decimal form. i = len(p)-1 while(i>= 0): ans = ans * k + p[i] i -= 1 print(ans)if __name__ == '__main__': n = 29 k = 7 greaterK(n, k)# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;class GFG{// Function to find the smallest number// greater than or equal to n represented // as the sum of distinct powers of kstatic void greaterK(int n, int k){ // List P to store the base k // representation of the number int []p = new int[String.Join("",n).Length + 2]; int index = 0; int x = n; int j; while (x > 0) { p[index] = (int) (x % k); x /= k; index++; } int idx = 0; for (int i = 0; i < p.Length - 1; ++i) { if (p[i] >= 2) { // If the representation is >=2, then // this power of k has to be added // once again and then increase the // next power of k and make the // current power 0 p[i] = 0; p[i + 1]++; // Reduce all the lower power of // k to 0 for (j = idx; j < i; ++j) { p[j] = 0; } idx = i + 1; } if (p[i] == k) { p[i] = 0; p[i + 1]++; } } j = p.Length - 1; // Check if the most significant // bit also satisfy the above // conditions if (p[j] >= 2) { p[index] = 1; index++; } int ans = 0; // Converting back from the // k-nary representation to // decimal form. for (int i = p.Length - 1; i >= 0; --i) { ans = ans * k + p[i]; } Console.Write(ans +"\n");}// Driver codepublic static void Main(String[] args){ int n = 29, k = 7; greaterK(n, k);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the above approach// Function to find the smallest number// greater than or equal to n represented// as the sum of distinct powers of kfunction greaterK(n, k) { // Vector P to store the base k // representation of the number let p = new Array(String(n).length + 2).fill(0); let index = 0; let x = n; while (x > 0) { p[index] = (x % k); x = Math.floor(x / k); index++; } let idx = 0; for (let i = 0; i < p.length - 1; ++i) { if (p[i] >= 2) { // If the representation is >=2, then // this power of k has to be added // once again and then increase the // next power of k and make the // current power 0 p[i] = 0; p[i + 1]++; // Reduce all the lower power of // k to 0 for (let j = idx; j < i; ++j) { p[j] = 0; } idx = i + 1; } if (p[i] == k) { p[i] = 0; p[i + 1]++; } } let j = p.length - 1; // Check if the most significant // bit also satisfy the above // conditions if (p[j] >= 2) { p[index] = 1; index++; } let ans = 0; // Converting back from the // k-nary representation to // decimal form. for (let i = p.length - 1; i >= 0; --i) { ans = ans * k + p[i]; } document.write(ans + "<br>");}// Driver codelet n = 29, k = 7;greaterK(n, k);// This code is contributed by gfgking</script> |
Output:
49
Time Complexity: O(logkn)
Auxiliary Space: O(logkn)
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