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Problem Explanation:
This can be a tricky problem to understand. You need to find where in the array a number should be inserted by order, and return the index where it should go.
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Hint: 1
The first thing to do is sort the array from lower to bigger, just to make the code easier. This is where sort comes in, it needs a callback function so you have to create it.
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Hint: 2
Once the array is sorted, then just check for the first number that is bigger and return the index.
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Hint: 3
If there is no index for that number then you will have to deal with that case too.
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Spoiler Alert!
Solution ahead!
Basic Code Solution:
function getIndexToIns(arr, num) {
arr.sort(function(a, b) {
return a - b;
});
for (var a = 0; a < arr.length; a++) {
if (arr[a] >= num)
return a;
}
return arr.length;
}
Code Explanation:
- First I sort the array using
.sort(callbackFuntion)
to sort it by lowest to highest, from left to right. - Then I use a for loop to compare the items in the array starting from the smallest one. When an item on the array is greater than the number we are comparing against, then we return the index as an integer.
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Basic Code Solution:
function getIndexToIns(arr, num) {
// Find my place in this sorted array.
var times = arr.length; // runs the for loop once for each thing in the array
var count = 0;
for (i=0;i<times;i++){
if(num>arr[i]){count++;} } // counts how many array numbers are smaller than num
return count; // the above equals num's position in a sorted array
}
getIndexToIns([40, 60], 50);
Code Explanation:
- I do not sort the arr input array
- I run a for loop counting whenever the num input is bigger than an arr input number.
- This number is equivalent to what num’s position would be in a sorted array.
Basic Code Solution:
by @HarinaPana
function getIndexToIns(arr, num) {
arr.sort(function(a, b) {
return a - b;
});
var i = 0;
while (num > arr[i]) {
i++;
}
return i;
}
getIndexToIns([40, 60], 50);
Code Explanation:
- Sort existing array.
- Iterate through the array while checking if num is bigger.
- The loop will stop when num isn’t bigger than i and return the last element checked.
Intermediate Code Solution:
by @faustodc
function getIndexToIns(arr, num) {
arr.push(num);
arr.sort(function(a, b){return a-b});
return arr.indexOf(num);
}
Code Explanation:
- First we add the number
num
to the array usingpush()
which adds it as the last element of the array. - Then we use
sort()
with the callback functionfunction(a, b){return a-b}
to sort the numbers in ascending order. - Lastly we return the postion or index of
num
in the array with theindexOf()
function.
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Intermediate Code Solution:
Using .findIndex()
function getIndexToIns(arr, num) {
// sort and find right index
var index = arr.sort((curr, next) => curr > next)
.findIndex((currNum)=> num <= currNum);
// Returns proper answer
return index === -1 ? arr.length : index;
}
getIndexToIns([40, 60], 500);
Code Explanation:
- First sort the array in ascending order, this is currently done using array functions for minimal footprint.
- Once the array it is sorted, we directly apply the
.findIndex()
where we are going to compare every element in the array until we find wherenum <= currNum
meaning where the number we want to insert is less or equal to the current number number in the iteration. - Then we use ternary operations to check whether we got an index returned or
-1
. We only get-1
when the index was not found meaning when we get a false for all elements int he array, and for such case, it would mean thatnum
should be inserted at the end of the list hence why we usearr.length
.
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NOTES FOR CONTRIBUTIONS:
- DO NOT add solutions that are similar to any existing solutions. If you think it is similar but better, then try to merge (or replace) the existing similar solution.
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See
Wiki Challenge Solution Template
for reference.