Given that a zoo has a single chick. A chick gives birth to 2 chicks everyday and the life expectancy of a chick is 6 days. The task is to find the number of chicks on the **N ^{th}** day.

**Examples:**

Input:N = 3Output:9

First day: 1 chick

Second day: 1 + 2 = 3

Third day: 3 + 6 = 9Input:N = 12Output:173988

**Simple approach:** It is given that the life expectancy of a chick is 6 days, so no chick dies till the sixth day. Everyday population of the current day will be 3 times of the previous day. One more thing is to note that the chick born on ith day is not counted on that day, it will be counted in the next day and the changes begin from the seventh day. So main calculation starts from the seventh day onwards. **On Seventh Day:** Chicks from the 1st day die so based on manual calculation it will be 726. **On Eighth Day:** Two new born chicks born on (8-6)th i.e 2nd day dies. This will affect the current population by 2/3. This population needs to be get deducted from the previous day’s population because today i.e 8th day more newborns will be born, so we cannot deduct directly from today’s population. This will then be multiplied by three times because of newborns born on that day.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define ll long long int` `// Function to return the number` `// of chicks on the nth day` `ll getChicks(` `int` `n)` `{` ` ` `// Size of dp[] has to be` ` ` `// at least 6 (1-based indexing)` ` ` `int` `size = max(n, 7);` ` ` `ll dp[size];` ` ` `dp[0] = 0;` ` ` `dp[1] = 1;` ` ` `// Every day current population` ` ` `// will be three times of the previous day` ` ` `for` `(` `int` `i = 2; i <= 6; i++) {` ` ` `dp[i] = dp[i - 1] * 3;` ` ` `}` ` ` `// Manually calculated value` ` ` `dp[7] = 726;` ` ` `// From 8th day onwards` ` ` `for` `(` `int` `i = 8; i <= n; i++) {` ` ` `// Chick population decreases by 2/3 everyday.` ` ` `// For 8th day on [i-6] i.e 2nd day population` ` ` `// was 3 and so 2 new born die on the 6th day` ` ` `// and so on for the upcoming days` ` ` `dp[i] = (dp[i - 1] - (2 * dp[i - 6] / 3)) * 3;` ` ` `}` ` ` `return` `dp[n];` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3;` ` ` `cout << getChicks(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `public` `class` `GFG {` `// Function to return the number` `// of chicks on the nth day` `static` `long` `getChicks(` `int` `n)` `{` ` ` `// Size of dp[] has to be` ` ` `// at least 6 (1-based indexing)` ` ` `int` `size = Math.max(n, ` `7` `);` ` ` `long` `[]dp = ` `new` `long` `[size];` ` ` `dp[` `0` `] = ` `0` `;` ` ` `dp[` `1` `] = ` `1` `;` ` ` `// Every day current population` ` ` `// will be three times of the previous day` ` ` `for` `(` `int` `i = ` `2` `; i < ` `6` `; i++) {` ` ` `dp[i] = dp[i - ` `1` `] * ` `3` `;` ` ` `}` ` ` `// Manually calculated value` ` ` `dp[` `6` `] = ` `726` `;` ` ` `// From 8th day onwards` ` ` `for` `(` `int` `i = ` `8` `; i <= n; i++) {` ` ` `// Chick population decreases by 2/3 everyday.` ` ` `// For 8th day on [i-6] i.e 2nd day population` ` ` `// was 3 and so 2 new born die on the 6th day` ` ` `// and so on for the upcoming days` ` ` `dp[i] = (dp[i - ` `1` `] - (` `2` `* dp[i - ` `6` `] / ` `3` `)) * ` `3` `;` ` ` `}` ` ` `return` `dp[n];` `}` `// Driver code` `public` `static` `void` `main(String[] args) {` `int` `n = ` `3` `;` ` ` `System.out.println(getChicks(n));` ` ` `}` `}` `// This code has been contributed by 29AjayKumar` |

## Python3

` ` `# Python implementation of the approach` ` ` `# Function to return the number` `# of chicks on the nth day` `def` `getChicks(n):` ` ` ` ` `# Size of dp[] has to be` ` ` `# at least 6 (1-based indexing)` ` ` `size ` `=` `max` `(n, ` `7` `);` ` ` `dp ` `=` `[` `0` `]` `*` `size;` ` ` ` ` `dp[` `0` `] ` `=` `0` `;` ` ` `dp[` `1` `] ` `=` `1` `;` ` ` ` ` `# Every day current population` ` ` `# will be three times of the previous day` ` ` `for` `i ` `in` `range` `(` `2` `,` `7` `):` ` ` `dp[i] ` `=` `dp[i ` `-` `1` `] ` `*` `3` `;` ` ` ` ` `# Manually calculated value` ` ` `dp[` `6` `] ` `=` `726` `;` ` ` ` ` `# From 8th day onwards` ` ` `for` `i ` `in` `range` `(` `8` `,n` `+` `1` `):` ` ` ` ` `# Chick population decreases by 2/3 everyday.` ` ` `# For 8th day on [i-6] i.e 2nd day population` ` ` `# was 3 and so 2 new born die on the 6th day` ` ` `# and so on for the upcoming days` ` ` `dp[i] ` `=` `(dp[i ` `-` `1` `] ` `-` `(` `2` `*` `dp[i ` `-` `6` `] ` `/` `/` `3` `)) ` `*` `3` `;` ` ` ` ` `return` `dp[n];` ` ` `# Driver code` `n ` `=` `3` `;` ` ` `print` `(getChicks(n));` `# This code is contributed by Princi Singh` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the number` `// of chicks on the nth day` `static` `long` `getChicks(` `int` `n)` `{` ` ` `// Size of dp[] has to be` ` ` `// at least 6 (1-based indexing)` ` ` `int` `size = Math.Max(n, 7);` ` ` `long` `[]dp = ` `new` `long` `[size];` ` ` `dp[0] = 0;` ` ` `dp[1] = 1;` ` ` `// Every day current population` ` ` `// will be three times of the previous day` ` ` `for` `(` `int` `i = 2; i < 6; i++)` ` ` `{` ` ` `dp[i] = dp[i - 1] * 3;` ` ` `}` ` ` `// Manually calculated value` ` ` `dp[6] = 726;` ` ` `// From 8th day onwards` ` ` `for` `(` `int` `i = 8; i <= n; i++)` ` ` `{` ` ` `// Chick population decreases by 2/3 everyday.` ` ` `// For 8th day on [i-6] i.e 2nd day population` ` ` `// was 3 and so 2 new born die on the 6th day` ` ` `// and so on for the upcoming days` ` ` `dp[i] = (dp[i - 1] - (2 * dp[i - 6] / 3)) * 3;` ` ` `}` ` ` `return` `dp[n];` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` ` ` `int` `n = 3;` ` ` `Console.WriteLine(getChicks(n));` `}` `}` `// This code has been contributed by @Tushil..` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` ` ` ` ` `// Function to return the number` ` ` `// of chicks on the nth day` ` ` `function` `getChicks(n)` ` ` `{` ` ` `// Size of dp[] has to be` ` ` `// at least 6 (1-based indexing)` ` ` `let size = Math.max(n, 7);` ` ` `let dp = ` `new` `Array(size);` ` ` `dp.fill(0);` ` ` `dp[0] = 0;` ` ` `dp[1] = 1;` ` ` `// Every day current population` ` ` `// will be three times of the previous day` ` ` `for` `(let i = 2; i < 6; i++)` ` ` `{` ` ` `dp[i] = dp[i - 1] * 3;` ` ` `}` ` ` `// Manually calculated value` ` ` `dp[6] = 726;` ` ` `// From 8th day onwards` ` ` `for` `(let i = 8; i <= n; i++)` ` ` `{` ` ` `// Chick population decreases by 2/3 everyday.` ` ` `// For 8th day on [i-6] i.e 2nd day population` ` ` `// was 3 and so 2 new born die on the 6th day` ` ` `// and so on for the upcoming days` ` ` `dp[i] = (dp[i - 1] -` ` ` `(2 * parseInt(dp[i - 6] / 3, 10))) * 3;` ` ` `}` ` ` `return` `dp[n];` ` ` `}` ` ` ` ` `let n = 3;` ` ` `document.write(getChicks(n));` `</script>` |

**Output:**

9

**Efficient approach:** If you look closely, you can observe a pattern that is a number of chicks for Nth day in the zoo can be calculated directly using the formula **pow(3, N – 1)**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define ll long long int` `// Function to return the number` `// of chicks on the nth day` `ll getChicks(` `int` `n)` `{` ` ` `ll chicks = (ll)` `pow` `(3, n - 1);` ` ` `return` `chicks;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3;` ` ` `cout << getChicks(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function to return the number` `// of chicks on the nth day` `static` `int` `getChicks(` `int` `n)` `{` ` ` `int` `chicks = (` `int` `)Math.pow(` `3` `, n - ` `1` `);` ` ` `return` `chicks;` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `n = ` `3` `;` ` ` `System.out.println (getChicks(n));` `}` `}` `// This code is contributed by Tushil.` |

## Python 3

`# Python 3 implementation of the approach` `# Function to return the number` `# of chicks on the nth day` `def` `getChicks( n):` ` ` `chicks ` `=` `pow` `(` `3` `, n ` `-` `1` `)` ` ` `return` `chicks` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `n ` `=` `3` ` ` `print` `( getChicks(n))` `# This code is contributed by ChitraNayal` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function to return the number` ` ` `// of chicks on the nth day` ` ` `static` `int` `getChicks(` `int` `n)` ` ` `{` ` ` ` ` `int` `chicks = (` `int` `)Math.Pow(3, n - 1);` ` ` ` ` `return` `chicks;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` ` ` `int` `n = 3;` ` ` `Console.WriteLine(getChicks(n));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` ` ` ` ` `// Function to return the number` ` ` `// of chicks on the nth day` ` ` `function` `getChicks(n)` ` ` `{` ` ` ` ` `let chicks = Math.pow(3, n - 1);` ` ` ` ` `return` `chicks;` ` ` `}` ` ` ` ` `let n = 3;` ` ` `document.write(getChicks(n));` `</script>` |

**Output:**

9

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