The question as per the practice course is :
Write a JavaScript program to find the maximum integer n such that (1 + 2 + ... + n <= given integer ) is true. For eg. If a given integer is 10, value of maximum integer n is 4 so that 1+2+3+4 <= 10 is true. Your output code should be in the format console.log("Value of n is ", variableName)
My code is :
var num = prompt("Enter a number");
function test(x) {
var sum = 1,
n = 1,
a = 0;
while (sum <= x) {
sum += n;
n = n + 1;
a += 1;
}
return a;
}
var output = test(num);
console.log("Result is :", output);
I'm getting the correct outputs as per the test cases I've entered(10-4,15-5,16-6,17-6) but the website says there is something wrong with the program.
What am i doing wrong?
8 Answers 8
Better answer than looping: exploit maths. Starting with Triangular number formula:
1 + 2 + ... + n = n * (n + 1) / 2
Thus, for input x, you need to find n such that
n * (n + 1) / 2 <= x
To solve this, we need to clean up the inequality, then use the quadratic equation formula:
n^2 + n <= 2x
n^2 + n - 2x <= 0
n <= (-1 + sqrt(1 + 8x)) / 2
as the final solution. e.g. for
x = 10: n <= (-1 + sqrt(81)) / 2; n <= 4
x = 16: n <= (-1 + sqrt(128)) / 2; n <= 5.156854249492381
Round the upper limit down, and you have the largest allowed integer. Translated into JavaScript:
function test(x) {
return Math.floor((Math.sqrt(8 * x + 1) - 1) / 2);
}
var num = prompt("Enter a number");
console.log("Result is :", test(num));
5 Comments
while in the question title. Using the mathematical approach almost always yields a quicker and more efficient result. I agree that applying it where possible will lead to a much more fruitful programming experience that is able to quickly tackle technical and theoretical problemsConsider if the passed value is 11. Then, the maximum integer n should be 4, because 1+2+3+4 < 11 is true, while 1+2+3+4+5 < 11 is false. Your current code outputs 5 for an input of 11, though, which is incorrect; your while loop is sometimes overshooting sum.
You also need to initialize sum to start at 0, not at 1.
Subtract one from a before returning it:
function test(x) {
var sum = 0,
n = 1,
a = 0;
while (sum <= x) {
sum += n;
n = n + 1;
a += 1;
console.log(a, sum);
}
return a - 1;
}
console.log(test(10));
console.log(test(11));
var num = prompt("Enter a number");
var output = test(num);
console.log("Result is :", output);
1 Comment
The code below should work for you. Basically, what I did was that if the input is 10, and your sum is 9, it will still go into the while loop. Then it will add n again and now your number is greater than your input (which is 10), but you still return it. Here what I did is that at the end of the while loop, if your sum is greater than your input, subtract one from a. That way it will still execute, but it will fix the problem.
Also another error I noticed was that sum started at 1, and n started at 1. You wanted 1+2+3+...+n, however using your previous method, you got 1+1+2+3+...+n.
var num = prompt("Enter a number");
function test(x) {
var sum = 0,
n = 1,
tempSum = 1,
a = 0;
while (sum <= x) {
sum += n;
n++;
a++;
if (sum > x) {
a--;
}
}
return a;
}
var output = test(num);
console.log("Result is :", output);
4 Comments
Your order of operation is a little funky; all you have to do is add the incrementor. The while false case will make sure the sum only passes over the number once. So when you return, reduce the number by one:
var num = prompt("Enter a number");
var output = test(num);
console.log("Result is :", output);
function test(num){
let sum = 0
let inc = 0
while(sum<=num)
sum+=++inc
return --inc;
}
Comments
This is a reduced version of your code, basically we increment first the number to add (n) in each iteration, and then we add it to the variable holding the sum. When the loop conditions evaluates to false you need to decrement one to n to get your value:
var num = prompt("Enter a number");
function test(x)
{
var sum = 0, n = 0;
while (sum <= x)
{
sum += (++n);
}
return --n;
}
var output = test(num);
console.log("Result is :", output);
Comments
I think this will work for you:
var num = prompt("Enter a number");
function test(x) {
var sum = 1,
n = 0;
while ((sum+n) <= x) {
n = n + 1;
sum += n;
}
return n;
}
var output = test(num);
console.log("Result is :", output);
Comments
Try below function to find max Number
function maxNumber(a){
var i=1,sum=0,maxNumber=0;
while(sum<=a) {
sum=sum+i;
if(sum<=a)
{
maxNumber=i;
}
i+=1;
}
return maxNumber;
}
doubled checked condition sum<=a to preserve the previous loop value and if condition not satisfied that means current loop value is not useful so returned preserved value of previous loop
Output tested :
Comments
Below will help you get the job done.
var num = prompt("Enter a number");
function findMaxNumber(num){
var sum = 0;
var counter = 0;
while(sum < num){
if(sum + counter > num){
break; // Exit loop
}
sum = sum + counter;
counter++;
}
return --counter; // Loop will cause this to be 1 higher than the max int.
}
console.log('Result is: ' + findMaxNumber(num));
4and you get3, but1+2+3 = 6.