Find a number K having sum of numbers obtained by repeated removal of last digit of K is N
Given an integer N, the task is to find an integer K such that the sum of the numbers formed by repeated removal of the last digit of K is equal to N.
Examples:
Input: N = 136
Output: 123
Explanation:
The numbers formed by repeatedly removing the last digit of 123 are {123, 12, 1}.
Therefore, the sum of these numbers = 123 + 12 + 1 = 136( = N).
Input: N = 107
Output: 98
Explanation:
The numbers formed by repeatedly removing the last digit of 98 are {98, 9}.
Therefore, the sum of these numbers = 98 + 7 = 107( = N).
Approach: The approach is based on the following observations:
- Consider K = 123.
- The possible numbers formed from 123 are 1, 12, and 123.
- Now, 123 can be expressed as 100 + 20 + 3. If all the other numbers are expressed similarly, then the idea is to know the position and frequency of each digit in all the numbers combined, to get the total sum as N.
Digit Frequency of each digit Sum units tens hundreds 1 1 1 1 1*1 + 1*10 + 1*100 = 111 2 1 1 2*1 + 2*10 = 22 3 1 3*1 = 3
- Now, for the given number N of length L. Divide the number with L number of 1s to get the highest place digit.
- Calculate the remainder which will be our newly formed N.
- Again divide the newly formed N with (L – 1) number of 1s to get 2nd highest place digit and continue till the L becomes 0.
Follow the steps below to solve the problem:
- Let L be the count of digits in the given number N.
- Initialize string str as L numbers of 1s in it.
- Initialize a variable ans as zero that will store the resultant number K.
- Iterate until the string str is not empty and follow the steps below:
- Convert the string str to number using the function stoi() and store it in M.
- Divide N by M and update ans as:
ans = ans*10 + (N/M)
- Update N to N % M.
- Remove the last character from the string str.
- After the above steps, print the value stored in the ans which is the required value of K.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the value of Kint findK(int N, int l){ // Stores l number of 1s string ones = ""; while (l--) { // Storing 1's ones = ones + '1'; } // Stores the value of K int ans = 0; // Iterate until ones is empty while (ones != "") { // Convert ones to number int m = stoi(ones); // Update the ans ans = (ans * 10) + (N / m); // Update N to N%m N = N % m; // Removing last digit from ones ones.pop_back(); } // Return the value of K return ans;}// Driver Codeint main(){ // Given number N int N = 136; // Number of digits in N int L = to_string(N).length(); // Funtion Call cout << findK(N, L); return 0;} |
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Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to find the // value of Kstatic int findK(int N, int l){ // Stores l number of 1s String ones = ""; while (l-- > 0) { // Storing 1's ones += '1'; } // Stores the value of K int ans = 0; // Iterate until ones is empty while (!ones.equals("")) { // Convert ones to number int m = Integer.valueOf(ones); // Update the ans ans = (ans * 10) + (N / m); // Update N to N%m N = N % m; // Removing last digit from ones ones = ones.substring(0, ones.length() - 1); } // Return the value of K return ans;}// Driver Codepublic static void main(String[] args){ // Given number N int N = 136; // Number of digits in N int L = String.valueOf(N).length(); // Funtion Call System.out.print(findK(N, L));}}// This code is contributed by 29AjayKumar |
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Python3
# Python3 program for # the above approach# Function to find # the value of Kdef findK(N, l): # Stores l number of 1s ones = "" while (l): # Storing 1's ones = ones + '1' l -= 1 # Stores the value of K ans = 0 # Iterate until ones # is empty while (ones != ""): # Convert ones to number m = int(ones) # Update the ans ans = (ans * 10) + (N // m) # Update N to N%m N = N % m # Removing last digit from ones ones = ones.replace(ones[-1], "", 1) # Return the value of K return ans# Driver Codeif __name__ == "__main__": # Given number N N = 136 # Number of digits in N L = len(str(N)) # Funtion Call print (findK(N, L))# This code is contributed by Chitranayal |
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C#
// C# program for // the above approachusing System;class GFG{// Function to find the // value of Kstatic int findK(int N, int l){ // Stores l number of 1s String ones = ""; while (l-- > 0) { // Storing 1's ones += '1'; } // Stores the value of K int ans = 0; // Iterate until ones is empty while (!ones.Equals("")) { // Convert ones to number int m = Int32.Parse(ones); // Update the ans ans = (ans * 10) + (N / m); // Update N to N%m N = N % m; // Removing last digit from ones ones = ones.Substring(0, ones.Length - 1); } // Return the value of K return ans;}// Driver Codepublic static void Main(String[] args){ // Given number N int N = 136; // Number of digits in N int L = String.Join("", N).Length; // Funtion Call Console.Write(findK(N, L));}}// This code is contributed by Princi Singh |
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Output:
123
Time Complexity: O(log10N)
Auxiliary Space: O(1)
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