In a small town the population is p0 = 1000 at the beginning of a year. The population regularly increases by 2 percent per year and moreover 50 new inhabitants per year come to live in the town. How many years does the town need to see its population greater or equal to p = 1200 inhabitants?
At the end of the first year there will be:
1000 + 1000 * 0.02 + 50 => 1070 inhabitantsAt the end of the 2nd year there will be:
1070 + 1070 * 0.02 + 50 => 1141 inhabitants(** number of inhabitants is an integer **)At the end of the 3rd year there will be:
1141 + 1141 * 0.02 + 50 => 1213It will need 3 entire years.
More generally given parameters:
p0, percent, aug (inhabitants coming or leaving each year), p (population to surpass)
the function nb_year should return n number of entire years needed to get a population greater or equal to p.
aug is an integer, percent a positive or null floating number, p0 and p are positive integers (> 0)
Examples:
nb_year(1500, 5, 100, 5000) -> 15
nb_year(1500000, 2.5, 10000, 2000000) -> 10
Note:
Don’t forget to convert the per cent parameter as a percentage in the body of your function: if the parameter percent is 2 you have to convert it to 0.02.
Brute Force
procedure nbYear(p0, percent, aug, p)
1. years = 0
2. for years until p0 >= p
A. p0 += p0 * (percent / 100) + aug
3. return yearsJavaScript Implementation
Depending on the language, the implementation using for loop or recursive call might make difference in the performance of the algorithm.
Rehan
Not able to pass all test cases with these solution. You can implement below solution
function nbYear(p0, percent, aug, p) {
// your code
console.log(“args”, p0,percent,aug, p)
let sum =0;
let count =0;
while(sum < p){
sum = Math.floor(p0 + p0 * (percent/100) + aug)
p0 = sum;
count ++;
}
console.log("sum", sum)
console.log("count", count)
return count;
}
Moshe Kessler
Hi! which one of the solutions have you tried out?
Your solution looks valid as well!