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‎courses/recommend.md‎

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import courses from '@site/src/database/courses';
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<Courses courses={courses} />
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<Courses courses={courses} />

‎dsa-solutions/gfg-solutions/Easy problems/Lucky-Numbers.md‎

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---
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id: lucky-number
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title: Nth Fibonacci Number
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sidebar_label: Nth Fibonacci Number
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tags:
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- Easy
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- Dynamic Programming
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- Math
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description: "This tutorial covers the solution to the Nth Fibonacci Number problem from the GeeksforGeeks."
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---
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## Problem Description
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Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers,
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Take the set of integers
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`1`, `2`, `3`, `4`, `5`, `6`, `7`, `8`, `9`, `10`, `11`, `12`, `13`, `14`, `15`, `16`, `17`, `18`, `19`,……
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First, delete every second number, we get following reduced set.
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`1`, `3`, `5`, `7`, `9`, `11`, `13`, `15`, `17`, `19`,…………
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Now, delete every third number, we get
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`1`, `3`, `7`, `9`, `13`, `15`, `19`,….….
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Continue this process indefinitely……
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Any number that does NOT get deleted due to above process is called “lucky”.
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You are given a number N, you need to tell whether the number is lucky or not. If the number is lucky return 1 otherwise 0.
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## Examples
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**Example 1:**
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```
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Input:
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N = 5
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Output: 0
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Explanation: 5 is not a lucky number
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as it gets deleted in the second
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iteration.
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```
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**Example 2:**
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```
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Input:
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N = 19
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Output: 1
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Explanation: 19 is a lucky number because
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it does not get deleted throughout the process.
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```
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## Your Task
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You don't need to read input or print anything. You only need to complete the function isLucky() that takes N as parameter and returns either False if the N is not lucky else True.
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Expected Time Complexity: $O(sqrt(n))$
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Expected Auxiliary Space: $O(1)$ for iterative approach.
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## Constraints
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* `1 ≤ n ≤ 10^5`
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## Problem Explanation
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Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers,
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Take the set of integers
64-
`1`, `2`, `3`, `4`, `5`, `6`, `7`, `8`, `9`, `10`, `11`, `12`, `13`, `14`, `15`, `16`, `17`, `18`, `19`,……
65-
First, delete every second number, we get following reduced set.
66-
`1`, `3`, `5`, `7`, `9`, `11`, `13`, `15`, `17`, `19`,…………
67-
Now, delete every third number, we get
68-
`1`, `3`, `7`, `9`, `13`, `15`, `19`,….….
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Continue this process indefinitely……
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Any number that does NOT get deleted due to above process is called “lucky”.
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You are given a number N, you need to tell whether the number is lucky or not. If the number is lucky return 1 otherwise 0.
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## Code Implementation
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<Tabs>
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<TabItem value="Python" label="Python" default>
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```py
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def is_lucky(N):
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lucky_numbers = list(range(1, N + 1))
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delete_step = 2
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while delete_step <= len(lucky_numbers):
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lucky_numbers = [num for i, num in enumerate(lucky_numbers) if (i + 1) % delete_step != 0]
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delete_step += 1
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return 1 if N in lucky_numbers else 0
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```
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</TabItem>
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<TabItem value="C++" label="C++">
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```cpp
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int isLucky(int N) {
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vector<int> luckyNumbers;
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for (int i = 1; i <= N; i++) {
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luckyNumbers.push_back(i);
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}
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int deleteStep = 2;
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while (deleteStep <= luckyNumbers.size()) {
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for (int i = deleteStep - 1; i < luckyNumbers.size(); i += deleteStep) {
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luckyNumbers.erase(luckyNumbers.begin() + i);
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}
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deleteStep++;
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}
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for (int num : luckyNumbers) {
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if (num == N) {
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return 1;
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}
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}
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return 0;
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}
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```
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</TabItem>
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<TabItem value="Javascript" label="Javascript" default>
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```javascript
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function isLucky(N) {
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let luckyNumbers = Array.from({ length: N }, (_, i) => i + 1);
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let deleteStep = 2;
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while (deleteStep <= luckyNumbers.length) {
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luckyNumbers = luckyNumbers.filter((_, i) => (i + 1) % deleteStep !== 0);
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deleteStep++;
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}
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return luckyNumbers.includes(N) ? 1 : 0;
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}
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```
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</TabItem>
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<TabItem value="Typescript" label="Typescript" default>
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```typescript
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function isLucky(N) {
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let luckyNumbers = Array.from({ length: N }, (_, i) => i + 1);
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let deleteStep = 2;
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while (deleteStep <= luckyNumbers.length) {
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luckyNumbers = luckyNumbers.filter((_, i) => (i + 1) % deleteStep !== 0);
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deleteStep++;
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}
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return luckyNumbers.includes(N) ? 1 : 0;
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}
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```
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</TabItem>
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<TabItem value="Java" label="Java" default>
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```java
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public int isLucky(int N) {
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List<Integer> luckyNumbers = new ArrayList<>();
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for (int i = 1; i <= N; i++) {
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luckyNumbers.add(i);
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}
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int deleteStep = 2;
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while (deleteStep <= luckyNumbers.size()) {
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for (int i = deleteStep - 1; i < luckyNumbers.size(); i += deleteStep) {
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luckyNumbers.remove(i);
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}
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deleteStep++;
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}
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for (int num : luckyNumbers) {
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if (num == N) {
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return 1;
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}
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}
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return 0;
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}
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```
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</TabItem>
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</Tabs>
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## Time Complexity
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* The iterative approach has a time complexity of $O(n^2)$.
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## Space Complexity
194-
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* The space complexity is $O(n)$ since we are using only a fixed amount of extra space.
1+
---
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id: lucky-number
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title: Nth Fibonacci Number
4+
sidebar_label: Nth Fibonacci Number
5+
tags:
6+
- Easy
7+
- Dynamic Programming
8+
- Math
9+
description: "This tutorial covers the solution to the Nth Fibonacci Number problem from the GeeksforGeeks."
10+
---
11+
12+
## Problem Description
13+
14+
Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers,
15+
Take the set of integers
16+
`1`, `2`, `3`, `4`, `5`, `6`, `7`, `8`, `9`, `10`, `11`, `12`, `13`, `14`, `15`, `16`, `17`, `18`, `19`,……
17+
First, delete every second number, we get following reduced set.
18+
`1`, `3`, `5`, `7`, `9`, `11`, `13`, `15`, `17`, `19`,…………
19+
Now, delete every third number, we get
20+
`1`, `3`, `7`, `9`, `13`, `15`, `19`,….….
21+
Continue this process indefinitely……
22+
Any number that does NOT get deleted due to above process is called “lucky”.
23+
24+
You are given a number N, you need to tell whether the number is lucky or not. If the number is lucky return 1 otherwise 0.
25+
26+
## Examples
27+
28+
**Example 1:**
29+
30+
```
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Input:
32+
N = 5
33+
Output: 0
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Explanation: 5 is not a lucky number
35+
as it gets deleted in the second
36+
iteration.
37+
```
38+
39+
**Example 2:**
40+
41+
```
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Input:
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N = 19
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Output: 1
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Explanation: 19 is a lucky number because
46+
it does not get deleted throughout the process.
47+
```
48+
49+
## Your Task
50+
51+
You don't need to read input or print anything. You only need to complete the function isLucky() that takes N as parameter and returns either False if the N is not lucky else True.
52+
53+
Expected Time Complexity: $O(sqrt(n))$
54+
55+
Expected Auxiliary Space: $O(1)$ for iterative approach.
56+
57+
## Constraints
58+
59+
- `1 ≤ n ≤ 10^5`
60+
61+
## Problem Explanation
62+
63+
Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers,
64+
Take the set of integers
65+
`1`, `2`, `3`, `4`, `5`, `6`, `7`, `8`, `9`, `10`, `11`, `12`, `13`, `14`, `15`, `16`, `17`, `18`, `19`,……
66+
First, delete every second number, we get following reduced set.
67+
`1`, `3`, `5`, `7`, `9`, `11`, `13`, `15`, `17`, `19`,…………
68+
Now, delete every third number, we get
69+
`1`, `3`, `7`, `9`, `13`, `15`, `19`,….….
70+
Continue this process indefinitely……
71+
Any number that does NOT get deleted due to above process is called “lucky”.
72+
73+
You are given a number N, you need to tell whether the number is lucky or not. If the number is lucky return 1 otherwise 0.
74+
75+
## Code Implementation
76+
77+
<Tabs>
78+
<TabItem value="Python" label="Python" default>
79+
<SolutionAuthor name="@Ishitamukherjee2004"/>
80+
81+
```py
82+
def is_lucky(N):
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lucky_numbers = list(range(1, N + 1))
84+
delete_step = 2
85+
while delete_step <= len(lucky_numbers):
86+
lucky_numbers = [num for i, num in enumerate(lucky_numbers) if (i + 1) % delete_step != 0]
87+
delete_step += 1
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return 1 if N in lucky_numbers else 0
89+
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```
91+
92+
</TabItem>
93+
<TabItem value="C++" label="C++">
94+
<SolutionAuthor name="@Ishitamukherjee2004"/>
95+
96+
```cpp
97+
int isLucky(int N) {
98+
vector<int> luckyNumbers;
99+
for (int i = 1; i <= N; i++) {
100+
luckyNumbers.push_back(i);
101+
}
102+
int deleteStep = 2;
103+
while (deleteStep <= luckyNumbers.size()) {
104+
for (int i = deleteStep - 1; i < luckyNumbers.size(); i += deleteStep) {
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luckyNumbers.erase(luckyNumbers.begin() + i);
106+
}
107+
deleteStep++;
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}
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for (int num : luckyNumbers) {
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if (num == N) {
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return 1;
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}
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}
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return 0;
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}
116+
117+
```
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</TabItem>
120+
121+
<TabItem value="Javascript" label="Javascript" default>
122+
<SolutionAuthor name="@Ishitamukherjee2004"/>
123+
124+
```javascript
125+
function isLucky(N) {
126+
let luckyNumbers = Array.from({ length: N }, (_, i) => i + 1);
127+
let deleteStep = 2;
128+
while (deleteStep <= luckyNumbers.length) {
129+
luckyNumbers = luckyNumbers.filter((_, i) => (i + 1) % deleteStep !== 0);
130+
deleteStep++;
131+
}
132+
return luckyNumbers.includes(N) ? 1 : 0;
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}
134+
```
135+
136+
</TabItem>
137+
138+
<TabItem value="Typescript" label="Typescript" default>
139+
<SolutionAuthor name="@Ishitamukherjee2004"/>
140+
141+
```typescript
142+
function isLucky(N) {
143+
let luckyNumbers = Array.from({ length: N }, (_, i) => i + 1);
144+
let deleteStep = 2;
145+
while (deleteStep <= luckyNumbers.length) {
146+
luckyNumbers = luckyNumbers.filter((_, i) => (i + 1) % deleteStep !== 0);
147+
deleteStep++;
148+
}
149+
return luckyNumbers.includes(N) ? 1 : 0;
150+
}
151+
```
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153+
</TabItem>
154+
155+
<TabItem value="Java" label="Java" default>
156+
<SolutionAuthor name="@Ishitamukherjee2004"/>
157+
158+
```java
159+
public int isLucky(int N) {
160+
List<Integer> luckyNumbers = new ArrayList<>();
161+
for (int i = 1; i <= N; i++) {
162+
luckyNumbers.add(i);
163+
}
164+
int deleteStep = 2;
165+
while (deleteStep <= luckyNumbers.size()) {
166+
for (int i = deleteStep - 1; i < luckyNumbers.size(); i += deleteStep) {
167+
luckyNumbers.remove(i);
168+
}
169+
deleteStep++;
170+
}
171+
for (int num : luckyNumbers) {
172+
if (num == N) {
173+
return 1;
174+
}
175+
}
176+
return 0;
177+
}
178+
179+
180+
```
181+
182+
</TabItem>
183+
</Tabs>
184+
185+
## Time Complexity
186+
187+
- The iterative approach has a time complexity of $O(n^2)$.
188+
189+
## Space Complexity
190+
191+
- The space complexity is $O(n)$ since we are using only a fixed amount of extra space.

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