Tell us what’s happening:
My solution finds the largest numbers in the array by eliminating all but the largest in each sub-array. The problem is that the numbers are still in sub arrays, like this:
[[27],[5],[39],[1001]]
The working solution requires a single array with 4 numbers like this:
[27, 5, 19, 1001]
To achieve this, I’ve created a blank array (largestArray) so that each remaining number is sliced in. However, only the final number is added (1001, in this case).
I suspected that the while loop was running several times before the second ‘if’ ran once, meaning that i==3 before a value was sliced, so I tried rewriting the code, seen below as ALT CODE, with the same problem.
Since ‘i’ is only incremented when a value is pushed, it leads me to believe that the problem is not the loop, but the method of adding a value to the array (slice). Now I believe that the slice function is overwriting each value so that only the most recent remains. However, using methods like ‘push’ or even ‘unshift’ (unsuitable anyway) return ‘1’, not even [1], and I have no idea why.
Short version: Only the value from the final sub-array is added to a new array, suspect slicing is the cause. All thoughts welcome.
Your code so far
function largestOfFour(arr) {
var largestArray = {};
for(var i = 0; i < arr.length; i++){
while(arr[i].length > 1){
if(arr[i][0] < arr[i][1]) {
arr[i].splice(0, 1);
}
else{
arr[i].splice(1, 1);
}
}
if(arr[i].length == 1) {
largestArray = arr[i].slice();
}
}
return largestArray;
}
largestOfFour([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
ALT CODE:
function largestOfFour(arr) {
var largestArray = {};
var i = 0;
while(i < arr.length){
if(arr[i][0] < arr[i][1] && arr[i].length > 1) {
arr[i].splice(0, 1);
}
else if (arr[i][0] > arr[i][1] && arr[i].length > 1){
arr[i].splice(1, 1);
}
else if(arr[i].length == 1) {
largestArray = arr[i].slice();
i++;
}
}
return largestArray;
}
largestOfFour([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
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Link to the challenge: