Given that a zoo has a single chick. A chick gives birth to 2 chicks every day and the life expectancy of a chick is 6 days. The task is to find the number of chicks on the Nth day.
Examples:
Input: N = 3
Output: 9
First day: 1 chick
Second day: 1 + 2 = 3
Third day: 3 + 6 = 9Input: N = 12
Output: 173988
Simple approach: It is given that the life expectancy of a chick is 6 days, so no chick dies till the sixth day. The 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 newborn 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 dayll 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 codeint main()
{ int n = 7;
cout << getChicks(n);
return 0;
} |
Java
// Java implementation of the approachimport java.util.*;
public class GFG {
// Function to return the number// of chicks on the nth daystatic 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+1];
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 codepublic static void main(String[] args) {
int n = 7;
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 daydef getChicks(n):
# Size of dp[] has to be
# at least 6 (1-based indexing)
size = max(n, 7);
dp = [0]*(size+1);
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[7] = 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 coden = 7;
print(getChicks(n));
# This code is contributed by Princi Singh |
C#
// C# implementation of the approachusing System;
class GFG
{ // Function to return the number// of chicks on the nth daystatic 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+1];
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 codestatic public void Main ()
{ int n = 7;
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+1);
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[7] = 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 = 7;
document.write(getChicks(n));
</script> |
Output
726
Time Complexity: O(n)
Auxiliary Space: O(max(n, 7)), where n is the given input.
An efficient approach:
The problem would be more easy if there was no life expectancy of chicks. But here we have to take care of that also. We will maintain an array to keep count of how many chicks expires in each day.
We will maintain an array to keep count of how many chicks expire each day. We can take this array of size 42(minimum) as the maximum length of N is 35 and we have to take more extra 6 days as the life expectancy of a chick is 6 days.
Follow the steps below to implement the above approach:
- Initially, maintain a variable called cnt to keep the count of live chicks. Initialize cnt=1 as there was only 1 chick on day 1.
- Declare an array named expires[] of size greater than or equal to 42.
- assign expires[0]=6 as first chick will expire on day 6.
- Run a loop from i=1 to n
- In each iteration, reduce cnt by expires[i] as those chicks will expire on that day.
- Add 2*cnt to expires[i+6] as 2*cnt will expire on (i+6)th day.
- Add 2*cnt to cnt as 2*cnt chicks are born on each day
- At the end, return cnt after the loop ends.
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 daylong long int NoOfChicks(int n)
{ long long cnt = 1ll;
vector<long long> expires(50, 0);
expires[6] = 1;
for (int i = 1; i < n; i++) {
cnt -= expires[i];
expires[i + 6] += 2 * cnt;
cnt += (2 * cnt);
}
return cnt;
}// Driver codeint main()
{ int n = 7;
cout << NoOfChicks(n);
return 0;
} |
Java
// Java implementation of the approachimport java.util.*;
public class GFG {
// Function to return the number
// of chicks on the nth day
public static long NoOfChicks(int N)
{
// Code here
long temp = 1;
long ans = 1;
long arr[] = new long[N];
boolean flag = false;
arr[0] = ans;
int ind = 0;
for (int i = 1; i < N; i++) {
temp++;
if (temp % 7 == 0 || flag) {
flag = true;
ans -= arr[ind++];
}
long val = ans * 2;
arr[i] = val;
ans += val;
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int n = 7;
System.out.println(NoOfChicks(n));
}
} |
Python3
# Python implementation of the approach# Function to return the number# of chicks on the nth daydef NoOfChicks(N):
arr = [0]*(N+1)
actual = [0]*(N+1)
arr[1] = 1
actual[1] = 1
for i in range(2, N+1):
curr = i-6
add = arr[i-1]
actual[i] = add*2
arr[i] = add*2
arr[i] += arr[i-1]
if curr >= 1:
arr[i] -= 3*actual[curr]
actual[i] -= 2*actual[curr]
return arr[N]
# Driver coden = 7
print(NoOfChicks(n))
|
C#
// C# implementation of the approachusing System;
public class GFG {
// Function to return the number
// of chicks on the nth day
public static long NoOfChicks(int N)
{
// Code here
long temp = 1;
long ans = 1;
long[] arr = new long[N];
bool flag = false;
arr[0] = ans;
int ind = 0;
for (int i = 1; i < N; i++) {
temp++;
if (temp % 7 == 0 || flag) {
flag = true;
ans -= arr[ind++];
}
long val = ans * 2;
arr[i] = val;
ans += val;
}
return ans;
}
static public void Main()
{
// Code
int n = 7;
Console.WriteLine(NoOfChicks(n));
}
}// This code is contributed by lokesh |
Javascript
// Function to return the number
// of chicks on the nth day
function NoOfChicks(N)
{
// Code here
let temp = 1;
let ans = 1;
let arr = [];
let flag = false;
arr[0] = ans;
let ind = 0;
for (let i = 1; i < N; i++) {
temp++;
if (temp % 7 == 0 || flag) {
flag = true;
ans -= arr[ind++];
}
let val = ans * 2;
arr[i] = val;
ans += val;
}
return ans;
}
let n = 7;
console.log(NoOfChicks(n));
// This code is contributed by utkarshshirode02 |
Output
726
Time Complexity: O(N)
Auxiliary Space: O(N)
An constant space O(1) approach:
As you know the life expectancy of chicks is 6 days, so in order to get the number of chicks alive today (lets say ith day), you have to subtract the newBornChick at i – 6th day from ith day and double it, described above. This method of creating vector of size 50 can be reduced by only size 7 or constant space by simply creating a findIndex() method, which takes the ith element as input and return an index by doing a modulo division with 6. if i = 7 then findIndex() will return 1 and we will store the value (newly born chicks on that day) to newBornChicks[i], and if the method return 0, we will store the value at index 6(means this method will return 6).
Here, we will maintain a vector/array of size 7 which will store the number of newly born chicks every day.
Follow the steps below to implement the above idea:
- Declare an array newBornChicks[] and totalChicks[] of size 7
- Initialise newBornChicks[1] and totalChicks = 1; giving the initial values.
- Run a for loop from 1 to N (inclusive).
- check, if i is less than equal 6,
- If true, store double of totalChicks at newBornChicks[i].
- update totalChicks with totalChicks + newBornChicks[i].
- if i is greater than 6,
- first find the index by performing modulus division operation
- subtract the number of dead chicks which are born i-6th day, totalChicks -= newBornChicks[index].
- now push the today’s new born chicks to at that index.
- and update the total chicks by totalChicks with today’s newBornChicks
- Return the totalChicks.
Below is the implementation of the above approach:
C++
/// C++ implementation #include<bits/stdc++.h>using namespace std;
class GFG
{ int findIndex (int k)
{
if(k%6 == 0)
{
///returning the end location which is 6;
return 6;
}
else
{
/// returning the mod of 6 to maintain the array location from 1 to 5.
return (k%6);
}
}
public:
long long int NoOfChicks(int N)
{
/// if n == 1
if(N == 1)
{
return 1;
}
/// declearing newBornChicks array and totalChicks
long long int newBorn[7], totalChicks;
/// initialising by 1
newBorn[1] = totalChicks = 1;
/// loop will run from 2 to n (inclusive).
for(int i = 2; i <= N; i++)
{
if(i <= 6)
{
/// instead of multiply by two, using leftShift by 1.
/// storing today's newBornChicks at index i.
newBorn[i] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[i];
}
else
{
int index = findIndex(i);
/// subtracting dead chicks from total chicks present today.
totalChicks -= newBorn[index];
/// instead of multiply by two, using leftShift by 1.
newBorn[index] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[index];
}
}
return totalChicks;
}
};//{ Driver Code Starts.int main()
{ int N = 7;
GFG obj;
cout << obj.NoOfChicks(N) <<"\n";
return 0;
}// } Driver Code Ends |
Java
/// Java implementation public class GFG
{ private static int findIndex (int k)
{
if(k%6 == 0)
{
///returning the end location which is 6;
return 6;
}
else
{
/// returning the mod of 6 to maintain the array location from 1 to 5.
return (k%6);
}
}
public static long NoOfChicks(int N)
{
if(N == 1)
{
return 1;
}
/// declearing newBornChicks array and totalChicks
long []newBorn = new long[7];
long totalChicks;
/// initialising by 1
newBorn[1] = totalChicks = 1;
for(int i = 2; i <= N; i++)
{
if(i <= 6)
{
/// instead of multiply by two, using leftShift by 1.
/// storing today's newBornChicks at index i.
newBorn[i] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[i];
}
else
{
int index = findIndex(i);
/// subtracting dead chicks from total chicks present today.
totalChicks -= newBorn[index];
/// instead of multiply by two, using leftShift by 1.
newBorn[index] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[index];
}
}
return totalChicks;
}
/// main class
public static void main(String[] args)
{
int n = 7;
System.out.println(NoOfChicks(n));
}
} |
Python
# Python implementationdef findIndex(k):
if(k % 6 == 0):
# returning the end location which is 6;
return 6
else:
# returning the mod of 6 to maintain the array location from 1 to 5.
return (k % 6)
def NoOfChicks(N):
# if n == 1
if(N == 1):
return 1
# ndeclearing newBornChicks array and totalChicks
newBorn = []
totalChicks = 0
for i in range(0, 7):
newBorn.append(0)
# initialising by 1
newBorn[1] = totalChicks = 1
# loop will run from 2 to n (inclusive)
for i in range(2, N + 1):
if(i <= 6):
# instead of multiply by two, using leftShift by 1.
# storing today's newBornChicks at index i.
newBorn[i] = totalChicks << 1
# updating today total alive chicks
totalChicks += newBorn[i]
else:
index = findIndex(i)
# subtracting dead chicks from total chicks present today.
totalChicks -= newBorn[index]
# instead of multiply by two, using leftShift by 1.
newBorn[index] = totalChicks << 1
# updating today total alive chicks
totalChicks += newBorn[index]
return totalChicks
# Driver CodeN = 7
print(NoOfChicks(N))
# This code is contributed by Samim Hossain Mondal. |
C#
// Include namespace systemusing System;
// / C# implementationpublic class GFG
{ private static int findIndex(int k)
{
if (k % 6 == 0)
{
// /returning the end location which is 6;
return 6;
}
else
{
// / returning the mod of 6 to maintain the array location from 1 to 5.
return (k % 6);
}
}
public static long NoOfChicks(int N)
{
if (N == 1)
{
return 1;
}
// / declearing newBornChicks array and totalChicks
long[] newBorn = new long[7];
long totalChicks;
// / initialising by 1
newBorn[1] = totalChicks = 1;
for (int i = 2; i <= N; i++)
{
if (i <= 6)
{
// / instead of multiply by two, using leftShift by 1.
// / storing today's newBornChicks at index i.
newBorn[i] = totalChicks << 1;
// / updating today total alive chicks
totalChicks += newBorn[i];
}
else
{
var index = GFG.findIndex(i);
// / subtracting dead chicks from total chicks present today.
totalChicks -= newBorn[index];
// / instead of multiply by two, using leftShift by 1.
newBorn[index] = totalChicks << 1;
// / updating today total alive chicks
totalChicks += newBorn[index];
}
}
return totalChicks;
}
// / main class
public static void Main(String[] args)
{
var n = 7;
Console.WriteLine(GFG.NoOfChicks(n));
}
}// This code is contributed by utkarshshirode02 |
Javascript
// Javascript implementation function findIndex (k)
{
if(k%6 == 0)
{
///returning the end location which is 6;
return 6;
}
else
{
/// returning the mod of 6 to maintain the array location from 1 to 5.
return (k%6);
}
}
function NoOfChicks( N)
{
/// if n == 1
if(N == 1)
{
return 1;
}
/// declearing newBornChicks array and totalChicks
let newBorn =[], totalChicks;
for(let i = 0; i < 7; i++)
newBorn.push(0);
/// initialising by 1
newBorn[1] = totalChicks = 1;
/// loop will run from 2 to n (inclusive).
for(let i = 2; i <= N; i++)
{
if(i <= 6)
{
/// instead of multiply by two, using leftShift by 1.
/// storing today's newBornChicks at index i.
newBorn[i] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[i];
}
else
{
let index = findIndex(i);
/// subtracting dead chicks from total chicks present today.
totalChicks -= newBorn[index];
/// instead of multiply by two, using leftShift by 1.
newBorn[index] = totalChicks<<1;
/// updating today total alive chicks
totalChicks += newBorn[index];
}
}
return totalChicks;
}
let N = 7;
console.log(NoOfChicks(N));
// This code is contributed by garg28harsh.
|
Output
726
Time Complexity: O(N).
Auxiliary Space: O(1).