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Posted on Jun 10, 2020

Daily Challenge #256 - How Many Are Smaller Than I?

#challenge

Write function smaller(arr) that given an array arr, you have to return the amount of numbers that are smaller to the right of that number in the index.

For example:
smaller([5, 4, 3, 2, 1]) === [4, 3, 2, 1, 0]
smaller([1, 2, 0]) === [1, 1, 0]

Tests:
smaller([5, 4, 3, 2, 1])
smaller([1, 2, 3])
smaller([1, 1, -1, 0, 0])
smaller([5, 4, 7, 9, 2, 4, 4, 5, 6])

Good luck!

This challenge comes from joh_pot on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (8)

JP Antunes • Jun 10 '20

Simple JS solution.

const smaller = arr => arr.reduce((acc, val, idx) => {
    acc[idx] = arr.filter((e, i) => e < val && i > idx).length;
    return acc;
},[]);

Amin • Jun 11 '20 • Edited

TypeScript

const smaller = (items: number[]): Uint32Array => {
    const length: number = items.length;
    const output: Uint32Array = new Uint32Array(length);

    for (let current: number = 0; current < length; current++) {
        output[current] = 0;

        for (let next: number = current + 1; next < length; next++) {
            if (items[next] < items[current]) {
                output[current]++;
            }
        }
    }

    return output;
};

David Leger • Jun 12 '20

const smaller = (arr: number[]): number[] => {
  // loop over items
  return arr.map((item, i) => {
    // get array of remaining items
    const rest = arr.slice(i + 1, arr.length);

    // filter for items less than the current item
    const lessThan = rest.filter(curr => curr < item);

    // count the items less than
    return lessThan.length;
  })
}

tawfeeq khalilia • Oct 9 '22 • Edited

Here is a python solution

def smaller(arr):
    lst = []
    for ch in range(len(arr)):
        lst.append(sum([1 for ch1 in range(ch , len(arr)) if arr[ch1]<arr[ch]]))
    return lst

Vidit Sarkar • Jun 10 '20

Here is a C++ solution,

vector<int> smaller(vector<int> arr){
    map<int, int> m;
    vector<int> ans(arr.size(), 0);

    for(int i = arr.size()-1; i >= 0; i--){
        auto it = m.insert({arr[i], m[arr[i]]++});

        for(auto itr = m.begin(); itr != it.first; itr++) ans[i] += (*itr).second;
    }

    return ans;
}

Test cases,

smaller([5, 4, 3, 2, 1]) => [4, 3, 2, 1, 0]
smaller([1, 2, 0]) => [1, 1, 0]
smaller([1, 2, 3]) => [0, 0, 0]
smaller([1, 1, -1, 0, 0]) => [3, 3, 0, 0, 0]
smaller([5, 4, 7, 9, 2, 4, 4, 5, 6]) => [4, 1, 5, 5, 0, 0, 0, 0, 0]

Timon van Spronsen • Jun 15 '20

function smaller(arr) {
  return arr.map((x, i) => arr.slice(i).filter((y) => y < x).length);
}

Mike Talbot ⭐ • Jun 10 '20


const smaller = arr => arr.reduce((c,a,i)=>{
   for(let j =0; j < i; j++) { 
       c[j] = (c[j] || 0) + arr[j] > a ? 1 : 0 
   }
   return c
}, [])

saswat • Jun 16 '20 • Edited

This is a trivial python solution.

DEV Community

Daily Challenge #256 - How Many Are Smaller Than I?

Top comments (8)

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