Return a new array where each element is the number of smaller elements to the right of that element in the original input The 2019 Stack Overflow Developer Survey Results Are InFunction to find the shortest word in an array, where not every element is a stringCodility “PermMissingElem” SolutionFor an array, check if an index exists where the sum of the elements to the left of it is equal to the sum of the elements right of itCount of Smaller Numbers After SelfLeetcode Count of Smaller Numbers After Self solutionProducts excluding each element of the arrayFind the length of the longest consecutive elementsDeepest pit of an arrayFind the elements that appear only onceFind the majority element, which appears more than half the time
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Return a new array where each element is the number of smaller elements to the right of that element in the original input
The 2019 Stack Overflow Developer Survey Results Are InFunction to find the shortest word in an array, where not every element is a stringCodility “PermMissingElem” SolutionFor an array, check if an index exists where the sum of the elements to the left of it is equal to the sum of the elements right of itCount of Smaller Numbers After SelfLeetcode Count of Smaller Numbers After Self solutionProducts excluding each element of the arrayFind the length of the longest consecutive elementsDeepest pit of an arrayFind the elements that appear only onceFind the majority element, which appears more than half the time
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The task:
Given an array of integers, return a new array where each element in
the new array is the number of smaller elements to the right of that
element in the original input array.
For example, given the array [3, 4, 9, 6, 1], return [1, 1, 2, 1, 0],
since:
- There is 1 smaller element to the right of 3
- There is 1 smaller element to the right of 4
- There are 2 smaller elements to the right of 9
- There is 1 smaller element to the right of 6
- There are no smaller elements to the right of 1
const lst = [3, 4, 9, 6, 1];
My solutions:
const numberOfSmallerElem = lst => lst.map((x,i) => lst.slice(i + 1).reduce((acc,y) => y < x ? ++acc : acc, 0));
console.log(numberOfSmallerElem(lst));
function numberOfSmallerElem2(lst)
for (let i in lst)
lst[i] = lst.slice(i).reduce((acc,y) => y < lst[i] ? ++acc : acc, 0);
return lst;
console.log(numberOfSmallerElem2(lst));
function numberOfSmallerElem3(lst)
const ret = [];
for (let i = 0, len = lst.length; i < len - 1; i++)
const reference = lst[i];
let counter = 0;
for (let j = i + 1; j < len; j++)
if(lst[j] < reference) counter++;
ret.push(counter);
ret.push(0);
return ret;
console.log(numberOfSmallerElem3(lst));
javascript algorithm programming-challenge functional-programming
asked 8 mins ago
thadeuszlaythadeuszlay
823516
$endgroup$
add a comment |
$begingroup$
The task:
Given an array of integers, return a new array where each element in
the new array is the number of smaller elements to the right of that
element in the original input array.
For example, given the array [3, 4, 9, 6, 1], return [1, 1, 2, 1, 0],
since:
- There is 1 smaller element to the right of 3
- There is 1 smaller element to the right of 4
- There are 2 smaller elements to the right of 9
- There is 1 smaller element to the right of 6
- There are no smaller elements to the right of 1
const lst = [3, 4, 9, 6, 1];
My solutions:
const numberOfSmallerElem = lst => lst.map((x,i) => lst.slice(i + 1).reduce((acc,y) => y < x ? ++acc : acc, 0));
console.log(numberOfSmallerElem(lst));
function numberOfSmallerElem2(lst)
for (let i in lst)
lst[i] = lst.slice(i).reduce((acc,y) => y < lst[i] ? ++acc : acc, 0);
return lst;
console.log(numberOfSmallerElem2(lst));
function numberOfSmallerElem3(lst)
const ret = [];
for (let i = 0, len = lst.length; i < len - 1; i++)
const reference = lst[i];
let counter = 0;
for (let j = i + 1; j < len; j++)
if(lst[j] < reference) counter++;
ret.push(counter);
ret.push(0);
return ret;
console.log(numberOfSmallerElem3(lst));
javascript algorithm programming-challenge functional-programming
asked 8 mins ago
thadeuszlaythadeuszlay
823516
$endgroup$
add a comment |
$begingroup$
The task:
Given an array of integers, return a new array where each element in
the new array is the number of smaller elements to the right of that
element in the original input array.
For example, given the array [3, 4, 9, 6, 1], return [1, 1, 2, 1, 0],
since:
- There is 1 smaller element to the right of 3
- There is 1 smaller element to the right of 4
- There are 2 smaller elements to the right of 9
- There is 1 smaller element to the right of 6
- There are no smaller elements to the right of 1
const lst = [3, 4, 9, 6, 1];
My solutions:
const numberOfSmallerElem = lst => lst.map((x,i) => lst.slice(i + 1).reduce((acc,y) => y < x ? ++acc : acc, 0));
console.log(numberOfSmallerElem(lst));
function numberOfSmallerElem2(lst)
for (let i in lst)
lst[i] = lst.slice(i).reduce((acc,y) => y < lst[i] ? ++acc : acc, 0);
return lst;
console.log(numberOfSmallerElem2(lst));
function numberOfSmallerElem3(lst)
const ret = [];
for (let i = 0, len = lst.length; i < len - 1; i++)
const reference = lst[i];
let counter = 0;
for (let j = i + 1; j < len; j++)
if(lst[j] < reference) counter++;
ret.push(counter);
ret.push(0);
return ret;
console.log(numberOfSmallerElem3(lst));
javascript algorithm programming-challenge functional-programming
asked 8 mins ago
thadeuszlaythadeuszlay
823516
$endgroup$
The task:
Given an array of integers, return a new array where each element in
the new array is the number of smaller elements to the right of that
element in the original input array.
For example, given the array [3, 4, 9, 6, 1], return [1, 1, 2, 1, 0],
since:
- There is 1 smaller element to the right of 3
- There is 1 smaller element to the right of 4
- There are 2 smaller elements to the right of 9
- There is 1 smaller element to the right of 6
- There are no smaller elements to the right of 1
const lst = [3, 4, 9, 6, 1];
My solutions:
const numberOfSmallerElem = lst => lst.map((x,i) => lst.slice(i + 1).reduce((acc,y) => y < x ? ++acc : acc, 0));
console.log(numberOfSmallerElem(lst));
function numberOfSmallerElem2(lst)
for (let i in lst)
lst[i] = lst.slice(i).reduce((acc,y) => y < lst[i] ? ++acc : acc, 0);
return lst;
console.log(numberOfSmallerElem2(lst));
function numberOfSmallerElem3(lst)
const ret = [];
for (let i = 0, len = lst.length; i < len - 1; i++)
const reference = lst[i];
let counter = 0;
for (let j = i + 1; j < len; j++)
if(lst[j] < reference) counter++;
ret.push(counter);
ret.push(0);
return ret;
console.log(numberOfSmallerElem3(lst));
javascript algorithm programming-challenge functional-programming
javascript algorithm programming-challenge functional-programming
asked 8 mins ago
thadeuszlaythadeuszlay
823516
asked 8 mins ago
thadeuszlaythadeuszlay
823516
asked 8 mins ago
thadeuszlaythadeuszlay
823516
asked 8 mins ago
thadeuszlaythadeuszlay
823516
asked 8 mins ago
thadeuszlaythadeuszlay
823516
823516
add a comment |
add a comment |
0
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