'Kyu 8 Code Wars - Finding the Sum of an array after removing the highest and lowest values
I am practising on code wars and am currently stuck on a kyu 8 question, all the tests seem to pass bar the last one. I will add my code and the tests below plus the output I get below.
function sumArray(array) {
if (array == null || array.length <= 2) {
return 0
} else {
let largestInt = Math.max.apply(null, array)
let smallestInt = Math.min.apply(null, array)
let indexSmallest = array.indexOf(largestInt)
let indexLargest = array.indexOf(smallestInt)
array.splice(indexSmallest, 1)
array.splice(indexLargest, 1)
let sum = 0
for (let i = 0; i < array.length; i++) {
sum += array[I]
}
return sum
}
}
The tests:
const {
assert
} = require("chai");
it("example tests", () => {
assert.strictEqual(sumArray(null), 0);
assert.strictEqual(sumArray([]), 0);
assert.strictEqual(sumArray([3]), 0);
assert.strictEqual(sumArray([3, 5]), 0);
assert.strictEqual(sumArray([6, 2, 1, 8, 10]), 16);
assert.strictEqual(sumArray([0, 1, 6, 10, 10]), 17);
assert.strictEqual(sumArray([-6, -20, -1, -10, -12]), -28);
assert.strictEqual(sumArray([-6, 20, -1, 10, -13]), 3);
});
The output:
Test Results: example tests expected -10 to equal 3
Solution 1:[1]
function total(array) {
// always assure at least an empty array.
array = Array.from(array ?? []);
// sort array values ascending.
array.sort((a, b) => a - b);
array.pop(); // remove last/higest value.
array.shift(); // remove first/lowest value.
// for any to be reduced/summed-up (empty) array
// the initial value of zero always assures the
// minimum expected result of zero.
return array
.reduce((total, value) => total + value, 0);
}
const testEntries = [
[ null, 0 ],
[ [ ], 0 ],
[ [ 3 ], 0 ],
[ [ 3, 5 ], 0 ],
[ [ 6, 2, 1, 8, 10 ], 16 ],
[ [ 0, 1, 6, 10, 10 ], 17 ],
[ [ -6, -20, -1, -10, -12 ], -28 ],
[ [ -6, 20, -1, 10, -13 ], 3 ],
];
console.log(
testEntries
.map(([value, result]) =>
`(total([${ value }]) === ${ result }) ... ${ total(value) === result }`
)
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
