Intermediate Solution states:
But is this true? Calling arr.sort() actually sorts the array in place, does it not? The variable sortedArr is initialized but never used–the original arr is used instead.
What I’m driving at is the sortedArr variable is extraneous and doesn’t need to be declared.
Can someone tell me how ineffective / effective my recursive solution is:
function sumAll(arr) {
var min = Math.min(arr[0], arr[1]),
max = Math.max(arr[0], arr[1]);
if (min === max)
return (min);
return max + sumAll([min, max - 1]);
}
sumAll([1, 4]);
And generally, how can one measure code efficiency in js ?
Cheers!
en… why is no one using Math.abs()?
function sumAll(arr) {
return (arr[0]+arr[1])*(Math.abs(arr[0]-arr[1])+1)/2;
}
My solution is similar, except instead of using the newer spread syntax, I opted for the classical apply function.
function sumAll(arr) {
var n1 = Math.min.apply(null, arr);
var n2 = Math.max.apply(null, arr);
return ((n2 + n1) * (n2 - n1 + 1)) / 2;
}I also thought the reduce method should be used. I agree the code examples here without it are simpler hence a good practice. But it was fun learning more about the reduce method:
function sumAll(arr) {
var newArr = [];
newArr[0] = Math.min(…arr);
var i = Math.min(…arr);
var j=0;
while (i<Math.max(arr[0],arr[1])){
newArr[++j] = ++i;
}
return newArr.reduce(function(acc, val){
return acc + val;
},0);
}
This is my solution :
function sumAll(arr) {
var max = Math.max(arr[0], arr[1]);
var min = Math.min(arr[0], arr[1]);
var newArr = [];
for (i=min; i<=max; i++) {
newArr.push(i);
}
return newArr.reduce(function(a, b) {
return a + b;
});
}
sumAll([5, 10]);
This is my code for this freecodecamp exercise:
I used reduce and it helped a lot.
function sumAll(arr) {
var max = arr.reduce(function(a, b) {
return Math.max(a, b);
});
var min = arr.reduce(function(a, b) {
return Math.min(a, b);
});
var temp = 0;
for(var i = min; i < max + 1; i++) {
temp += i;
console.log(temp);
}
return temp;
}
sumAll([4, 1]);
You are not alone 
Look at mine
My solution:
function sumAll(arr) {
//sort array so lowest comes first
arr.sort(function(a,b){
return b < a;
});
//create sum variable and assyign total sum of arr to it
var sum = arr.reduce(function(acc, num){
return acc + num;
});
//check if we need to add the sum of number between first and last element of an array
if(arr[0] !== arr[1]){
//create second array to store numbers between first and last element of an array
var arr2 = [];
var index = arr[0]+1;//set the index
while(index < arr[1]){
arr2.push(index);//add between numbers to arr2
index++;
}
//assyign sum of arr to sum of arr2
sum += arr2.reduce(function(acc, num){
return acc + num;
});
}
return sum;
}
//test
sumAll([1, 4]);Here I am my code using the .sort method and a for loop.
function sumAll(arr) {
var sumArr = 0, sortArr = arr.sort(function(a,b){
return a - b;
});
for(var i = sortArr[0]; i < sortArr[1]; i += 2){
sumArr += i + (i + 1);
}
return sumArr;
}
sumAll([10, 5]);
But sincerely , if the suggestion to using reduce() is for learning then solution should include it. Remember its a leanring environment . As one who teaches at a university I cannot give my students a hint then not use it. It causes confusion and frustrates learners.
I went for
function sumAll(arr) {
var max = arr.reduce(function(a, b) {
return Math.max(a, b);
});
var min = arr.reduce(function(a, b) {
return Math.min(a, b);
});
var result = 0;
for (i = min; i <= max; i++){
result += i;
}
return result;
}A solution without using Math.max() and Math.min()
function sumAll(arr) {
var max;
var min;
var result = 0;
if (arr[0] > arr[1]) {
max = arr[0];
min = arr[1];
} else {
max = arr[1];
min = arr[0];
}
while (min <= max) {
result = result + min;
min++;
}
return result;
}
Hehe, same code here. Congrats. btw. i < max + 1 is the same as i <= max; 
My basic solution:
function sumAll(arr) {
arr.sort(function(a, b) {
return a - b;
});for(var i = arr[0]+1; i < arr[1]; i++) {
arr.push(i);
}return arr.reduce(function(val1, val2) {
return val1 + val2;
}, 0);
}
sumAll([5, 2]);
I “.sort()” the array. Loop and push new value in array. And, finally, I reduce.
You are using ES6 which does not work in the FC’s editor.
oh i see what you did there
Why not use some good old mathematics?
function sumAll(arr) {
max=Math.max(arr[0],arr[1]);
min=Math.min(arr[0],arr[1]);
return (max+min)*(max-min+1)/2;
}function sumAll(arr) {
return arr.sort(function(a,b){return a-b;}).reduce(function(a,b){
return ((b*(b+1))/2)-((a*(a-1)/2));
});
}
Is this code okay ?? It worked fine … But how efficient is this ??
More or less came up with verbatim the advanced code solution
function sumAll(arr) {
var result = 0;
for (let i=Math.min(...arr); i<=Math.max(...arr); i++){
result+=i;
}
return result;
}
sumAll([1, 4]);