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 = 136Output:123Explanation:

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 = 107Output:98Explanation:

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**.

DigitFrequency of each digitSumunitstenshundreds1 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**1**s 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**1**s 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**1**s 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:

- Convert the string

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 K` `int` `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 Code` `int` `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;` `}` |

## Java

`// Java program for` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the` `// value of K` `static` `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 Code` `public` `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` |

## Python3

`# Python3 program for` `# the above approach` `# Function to find` `# the value of K` `def` `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 Code` `if` `__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` |

## C#

`// C# program for` `// the above approach` `using` `System;` `class` `GFG{` `// Function to find the` `// value of K` `static` `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 Code` `public` `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` |

## Javascript

`<script>` ` ` `// JavaScript program for` ` ` `// the above approach` ` ` `// Function to find the` ` ` `// value of K` ` ` `function` `findK(N, l) {` ` ` `// Stores l number of 1s` ` ` `var` `ones = ` `""` `;` ` ` `while` `(l) {` ` ` `// Storing 1's` ` ` `ones += ` `"1"` `;` ` ` `l -= 1;` ` ` `}` ` ` `// Stores the value of K` ` ` `var` `ans = 0;` ` ` `// Iterate until ones is empty` ` ` `while` `(ones !== ` `""` `) {` ` ` `// Convert ones to number` ` ` `var` `m = parseInt(ones);` ` ` `// Update the ans` ` ` `ans = parseInt(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 Code` ` ` `// Given number N` ` ` `var` `N = 136;` ` ` `// Number of digits in N` ` ` `var` `L = N.toString().length;` ` ` `// Funtion Call` ` ` `document.write(findK(N, L));` ` ` `</script>` |

**Output:**

123

**Time Complexity:** O(log_{10}N)**Auxiliary Space:** O(1)

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