# Smallest number greater than or equal to N using only digits 1 to K

Given a number N and an integer K, the task is to find the smallest number greater than or equal to N, formed using only first K non-zero digits( 1, 2, …, K-1, K).
Examples:

Input: N = 124, K = 3
Output: 131
Explanation:
The smallest number greater than or equal to 124, which is only made of digits 1, 2, 3 is 131.

Input: N = 325242, K = 4
Output: 331111

Naive Approach:
The simplest solution is to start a for loop from N + 1 and find the first number made up of the first K digits.

Efficient Solution:

• To obtain an efficient solution, we need to understand the fact that a maximum of 9-digit numbers can be formed up-to 1010. So, we will iterate over digits of the number in reverse and check:
1. If current digit >= K then, make that digit = 1.
2. If the current digit < K and there is no digit greater than K after this, then increment the digit by 1 and copy all the remaining digits as it is.
• Once we have iterated over all the digits and still haven’t found any digit which is less than K then we need to add a digit (1) to the answer.

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the smallest` `// number greater than or equal` `// to N which is made up of` `// first K digits` `#include ` `using` `namespace` `std;`   `// Function to count the` `// digits greater than K` `int` `CountGreater(``int` `n, ``int` `k)` `{` `    ``int` `a = 0;` `    ``while` `(n) {` `        ``if` `((n % 10) > k) {` `            ``a++;` `        ``}` `        ``n = n / 10;` `    ``}` `    ``return` `a;` `}`   `// Function to print the list` `int` `PrintList(list<``int``> ans)` `{` `    ``for` `(``auto` `it = ans.begin();` `         ``it != ans.end(); it++)` `        ``cout << *it;` `}`   `// Function to find the number` `// greater than or equal to n,` `// which is only made of first` `// k digits` `void` `getNumber(``int` `n, ``int` `k)` `{` `    ``int` `count = CountGreater(n, k);`   `    ``// If the number itself` `    ``// satisfy the conditions` `    ``if` `(count == 0) {` `        ``cout << n;` `        ``return``;` `    ``}`   `    ``list<``int``> ans;` `    ``bool` `changed = ``false``;`   `    ``// Check digit from back` `    ``while` `(n > 0) {` `        ``int` `digit = n % 10;` `        ``if` `(changed == ``true``) {` `            ``ans.push_front(digit);` `        ``}` `        ``else` `{` `            ``// If digit > K is` `            ``// present previously and` `            ``// current digit is less` `            ``// than K` `            ``if` `(count == 0 && digit < k) {` `                ``ans.push_front(digit + 1);` `                ``changed = ``true``;` `            ``}` `            ``else` `{` `                ``ans.push_front(1);` `                ``// If current digit is` `                ``// greater than K` `                ``if` `(digit > k) {` `                    ``count--;` `                ``}` `            ``}` `        ``}` `        ``n = n / 10;` `    ``}`   `    ``// If an extra digit needs` `    ``// to be added` `    ``if` `(changed == ``false``) {` `        ``ans.push_front(1);` `    ``}`   `    ``// Print the number` `    ``PrintList(ans);`   `    ``return``;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 51234;` `    ``int` `K = 4;` `    ``getNumber(N, K);` `    ``return` `0;` `}`

## Java

 `// Java program to find the smallest` `// number greater than or equal` `// to N which is made up of` `// first K digits` `import` `java.util.*;`   `class` `GFG{`   `// Function to count the` `// digits greater than K` `static` `int` `CountGreater(``int` `n, ``int` `k)` `{` `    ``int` `a = ``0``;` `    `  `    ``while` `(n > ``0``) ` `    ``{` `        ``if` `((n % ``10``) > k) ` `        ``{` `            ``a++;` `        ``}` `        ``n = n / ``10``;` `    ``}` `    ``return` `a;` `}`   `// Function to print the list` `static` `void` `PrintList(List ans)` `{` `    ``for``(``int` `it : ans)` `        ``System.out.print(it);` `}`   `// Function to find the number` `// greater than or equal to n,` `// which is only made of first` `// k digits` `static` `void` `getNumber(``int` `n, ``int` `k)` `{` `    ``int` `count = CountGreater(n, k);`   `    ``// If the number itself` `    ``// satisfy the conditions` `    ``if` `(count == ``0``)` `    ``{` `        ``System.out.print(n);` `        ``return``;` `    ``}`   `    ``List ans = ``new` `LinkedList<>();` `    ``boolean` `changed = ``false``;`   `    ``// Check digit from back` `    ``while` `(n > ``0``)` `    ``{` `        ``int` `digit = n % ``10``;` `        `  `        ``if` `(changed == ``true``) ` `        ``{` `            ``ans.add(``0``, digit);` `        ``}` `        ``else` `        ``{` `            `  `            ``// If digit > K is` `            ``// present previously and` `            ``// current digit is less` `            ``// than K` `            ``if` `(count == ``0` `&& digit < k)` `            ``{` `                ``ans.add(``0``, digit + ``1``);` `                ``changed = ``true``;` `            ``}` `            ``else` `            ``{` `                ``ans.add(``0``, ``1``);` `                `  `                ``// If current digit is` `                ``// greater than K` `                ``if` `(digit > k)` `                ``{` `                    ``count--;` `                ``}` `            ``}` `        ``}` `        ``n = n / ``10``;` `    ``}`   `    ``// If an extra digit needs` `    ``// to be added` `    ``if` `(changed == ``false``)` `    ``{` `        ``ans.add(``0``, ``1``);` `    ``}`   `    ``// Print the number` `    ``PrintList(ans);`   `    ``return``;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``51234``;` `    ``int` `K = ``4``;` `    `  `    ``getNumber(N, K);` `}` `}`   `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program to find the smallest ` `# number greater than or equal ` `# to N which is made up of ` `# first K digits `   `# Function to count the ` `# digits greater than K ` `def` `CountGreater(n, k):`   `    ``a ``=` `0` `    ``while` `(n > ``0``):` `        ``if` `((n ``%` `10``) > k):` `            ``a ``+``=` `1` `            `  `        ``n ``=` `n ``/``/` `10`   `    ``return` `a`   `# Function to print the list` `def` `PrintList (ans):`   `    ``for` `i ``in` `ans:` `        ``print``(i, end ``=` `'')`   `# Function to find the number ` `# greater than or equal to n, ` `# which is only made of first ` `# k digits ` `def` `getNumber(n, k):`   `    ``count ``=` `CountGreater(n, k)` `    `  `    ``# If the number itself ` `    ``# satisfy the conditions ` `    ``if` `(count ``=``=` `0``):` `        ``print``(n)` `        ``return`   `    ``ans ``=` `[]` `    ``changed ``=` `False`   `    ``# Check digit from back` `    ``while` `(n > ``0``):` `        ``digit ``=` `n ``%` `10` `        ``if` `(changed ``=``=` `True``):` `            ``ans.insert(``0``, digit)`   `        ``else``:` `            `  `            ``# If digit > K is ` `            ``# present previously and ` `            ``# current digit is less ` `            ``# than K` `            ``if` `(count ``=``=` `0` `and` `digit < k):` `                ``ans.insert(``0``, digit ``+` `1``)` `                ``changed ``=` `True`   `            ``else``:` `                ``ans.insert(``0``, ``1``)` `                `  `                ``# If current digit is ` `                ``# greater than K` `                ``if` `(digit > k):` `                    ``count ``-``=` `1`   `        ``n ``=` `n ``/``/` `10`   `    ``# If an extra digit needs ` `    ``# to be added ` `    ``if` `(changed ``=``=` `False``):` `        ``ans.insert(``0``, ``1``)`   `    ``# Print the number` `    ``PrintList(ans)` `    ``return`   `# Driver Code` `N ``=` `51234` `K ``=` `4`   `getNumber(N, K)`   `# This code is contributed by himanshu77`

## C#

 `// C# program to find the smallest` `// number greater than or equal` `// to N which is made up of` `// first K digits` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{`   `// Function to count the` `// digits greater than K` `static` `int` `CountGreater(``int` `n, ``int` `k)` `{` `  ``int` `a = 0;`   `  ``while` `(n > 0) ` `  ``{` `    ``if` `((n % 10) > k) ` `    ``{` `      ``a++;` `    ``}` `    ``n = n / 10;` `  ``}` `  ``return` `a;` `}`   `// Function to print the list` `static` `void` `PrintList(List<``int``> ans)` `{` `  ``foreach``(``int` `it ``in` `ans)` `    ``Console.Write(it);` `}`   `// Function to find the number` `// greater than or equal to n,` `// which is only made of first` `// k digits` `static` `void` `getNumber(``int` `n, ``int` `k)` `{` `  ``int` `count = CountGreater(n, k);`   `  ``// If the number itself` `  ``// satisfy the conditions` `  ``if` `(count == 0)` `  ``{` `    ``Console.Write(n);` `    ``return``;` `  ``}`   `  ``List<``int``> ans = ``new` `List<``int``>();` `  ``bool` `changed = ``false``;`   `  ``// Check digit from back` `  ``while` `(n > 0)` `  ``{` `    ``int` `digit = n % 10;`   `    ``if` `(changed == ``true``) ` `    ``{` `      ``ans.Insert(0, digit);` `    ``}` `    ``else` `    ``{` `      ``// If digit > K is` `      ``// present previously and` `      ``// current digit is less` `      ``// than K` `      ``if` `(count == 0 && digit < k)` `      ``{` `        ``ans.Insert(0, digit + 1);` `        ``changed = ``true``;` `      ``}` `      ``else` `      ``{` `        ``ans.Insert(0, 1);`   `        ``// If current digit is` `        ``// greater than K` `        ``if` `(digit > k)` `        ``{` `          ``count--;` `        ``}` `      ``}` `    ``}` `    ``n = n / 10;` `  ``}`   `  ``// If an extra digit needs` `  ``// to be added` `  ``if` `(changed == ``false``)` `  ``{` `    ``ans.Insert(0, 1);` `  ``}`   `  ``// Print the number` `  ``PrintList(ans);`   `  ``return``;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `  ``int` `N = 51234;` `  ``int` `K = 4;` `  ``getNumber(N, K);` `}` `}`   `// This code is contributed by Princi Singh`

## Javascript

 `// JS Program to find the smallest` `// number greater than or equal` `// to N which is made up of` `// first K digits`   `// Function to count the` `// digits greater than K` `function` `CountGreater(n, k)` `{` `    ``let a = 0;` `    ``while` `(n > 0) {` `        ``if` `((n % 10) > k) {` `            ``a++;` `        ``}` `        ``n = Math.floor(n / 10);` `    ``}` `    ``return` `a;` `}`   `// Function to print the list` `function` `PrintList(ans)` `{` `    ``console.log(ans.join(``""``))` `}`   `// Function to find the number` `// greater than or equal to n,` `// which is only made of first` `// k digits` `function` `getNumber(n, k)` `{` `    ``let count = CountGreater(n, k);`   `    ``// If the number itself` `    ``// satisfy the conditions` `    ``if` `(count == 0) {` `        ``process.stdout.write(``""` `+ n);` `        ``return``;` `    ``}`   `    ``let ans = [];` `    ``let changed = ``false``;`   `    ``// Check digit from back` `    ``while` `(n > 0) {` `        ``let digit = n % 10;` `        ``if` `(changed == ``true``) {` `            ``ans.unshift(digit);` `        ``}` `        ``else` `{` `            ``// If digit > K is` `            ``// present previously and` `            ``// current digit is less` `            ``// than K` `            ``if` `(count == 0 && digit < k) {` `                ``ans.unshift(digit + 1);` `                ``changed = ``true``;` `            ``}` `            ``else` `{` `                ``ans.unshift(1);` `                ``// If current digit is` `                ``// greater than K` `                ``if` `(digit > k) {` `                    ``count--;` `                ``}` `            ``}` `        ``}` `        ``n = Math.floor(n / 10);` `    ``}`   `    ``// If an extra digit needs` `    ``// to be added` `    ``if` `(changed == ``false``) {` `        ``ans.unshift(1);` `    ``}`   `    ``// Print the number` `    ``PrintList(ans);`   `    ``return``;` `}`   `// Driver Code` `let N = 51234;` `let K = 4;` `getNumber(N, K);`   `// This code is contributed by phasing17.`

Output:

`111111`

Time Complexity: O(log10N)

Space Complexity: O(N) as ans list has been created.

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