Sum all numbers between two whole numbers
TARGET
Given two numbers in an array, sum all the numbers including (and between) both integers (eg [4,2] → 2 + 3 + 4 = 9 ).
I was able to resolve the question, but was wondering if there is a more elegant solution (especially using Math.max and Math.min) - see below for more questions ...
MY DECISION
//arrange array for lowest to highest number
function order(min,max) {
return min - max;
}
function sumAll(arr) {
var list = arr.sort(order);
var a = list[0]; //smallest number
var b = list[1]; //largest number
var c = 0;
while (a <= b) {
c = c + a; //add c to itself
a += 1; // increment a by one each time
}
return c;
}
sumAll([10, 5]);
MY QUESTION (S)
- Is there a more efficient way to do this?
- How do I use Math.max () and Math.min () for an array?
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4 answers
var array = [4, 2];
var max = Math.max.apply(Math, array); // 4
var min = Math.min.apply(Math, array); // 2
function sumSeries (smallest, largest) {
// The formulate to sum a series of integers is
// n * (max + min) / 2, where n is the length of the series.
var n = (largest - smallest + 1);
var sum = n * (smallest + largest) / 2; // note integer division
return sum;
}
var sum = sumSeries(min, max);
console.log(sum);
+7
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The sum of the first n integers (from 1 to n, inclusive) is given by the formula n (n + 1) / 2. It is also the nth triangular number.
S1 = 1 + 2 + ... + (a-1) + a + (a+1) + ... + (b-1) + b
= b(b+1)/2
S2 = 1 + 2 + ... + (a-1)
= (a-1)a/2
S1 - S2 = a + (a+1) + ... + (b-1) + b
= (b(b+1)-a(a-1))/2
We now have a general formula for calculating the sum. It will be much more efficient if we sum up a large range (for example, 1 to 2 million).
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Following is SumAll's one-time recursive software solution using es6.
const SumAll = (num, sum = 0) => num - 1 > 0 ? SumAll(num-1,sum += num) : sum+num;
console.log(SumAll(10));
Note :: Although the best example is using the Algorithm as mentioned above. However, if the above can be improved.
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