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Daily Challenge #63- Two Sum

dev.to staff on September 10, 2019

Write a function that takes an array of numbers (integers for the tests) and a target number. It should find two different items in the array that,...

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Avalander • Sep 10 '19 • Edited

Haskell

A recursive solution that checks for each item x in the list if there is an item in the rest of the list that, added to x, results in the target value.

import Data.List (findIndex)

twoSum :: [Int] -> Int -> (Int, Int)
twoSum xs n = findSum (head xs) 0 (tail xs)
  where
    findSum x curr ys = case findIndex (isSum x) ys of
      Just(i) -> (curr, i + curr + 1)
      Nothing -> findSum (head ys) (curr + 1) (tail ys)

    isSum x y = x + y == n

Zohar Peled • Sep 11 '19 • Edited

The input will always be valid that's even worst than premature optimizations.

However, for the purpose of this challenge, I'll take that assumption, because while super important, input validation is boring.

My answer inspired by willsmart's answer.

C#, with comments on every line, so that everyone can understand it even if they don't speak the language:

The simple, O(n²) solution - using a nested loop.


    (int Index1, int Index2) TwoSum(int[] array, int target)
    {
        // iterate the array from 0 to length -2
        for (var i = 0; i < array.Length - 1; i++)
        {
            // iterate the array from i+1 to length -1
            for (var j = i + 1; j < array.Length; j++)
            {
                // if sum found
                if (array[i] + array[j] == target)
                {
                    // return both indexes
                    return (i, j);
                }
            }
        }
        // indicate no match found, required by the c# compier.
        return (-1, -1); 
    }

The sophisticated, O(n) solution - using a single loop and a helper hash map.

    int Index1, int Index2) TwoSumImproved(int[] array, int target)
    {
        // In .Net, A dictionary is basically a hash map. here the keys and values are both ints
        var map = new Dictionary<int, int>();

        // iterate the full length of the array.
        for (var i = 0; i < array.Length; i++)
        {
            // calculate the value needed to add to the current value to get the target sum
            var valueNeeded = target - array[i];

            // check if the map contains the value needed as it's key, If it does, return true and populate the int j
            // The TryGetValue time complexity is approaching O(1)
            if (map.TryGetValue(valueNeeded, out int j))
            {
                // return both indexes
                return (j, i);
            }
            // add to the map, where the key is the array value and the value is the index.
            map.Add(array[i], i);
        }
        return (-1, -1); // indicate no match found
    }

Try it online

Oleksii Filonenko • Sep 10 '19

Inspired by Michael Kohl's solution in F# (O(n) rules!), I did the same thing in Rust. All the credit to him :)

fn two_sum(ns: &[i32], sum: i32) -> Option<(usize, usize)> {
    let mut map = HashMap::new();
    for (i, &n) in ns.iter().enumerate() {
        let diff = sum - n;
        if let Some(&j) = map.get(&diff) {
            return Some((j, i));
        }
        map.insert(n, i);
    }
    None
}

Dev Prakash • Sep 12 '19


global sum_value
sum_value = int(input("Enter the sum value :"))
arr = [ int(i) for i in input("Enter array elements :" ).split( )]
def two_sum (arr) :
    for i in range(len(arr)) :
        for j in  range(i+1 , len(arr)) :
            if arr[j] + arr[i] == sum_value :
                return (i , j)
        return "NO SUCH PAIRS FOUND"

print (two_sum(arr))

Another one

global sum_value
sum_value = int(input("Enter the sum value :"))
arr = [ int(i) for i in input("Enter array elements :" ).split( )]
def two_sum (arr) :
    for i in range(len(arr)) :
        second_num = sum_value - arr[i]
        if second_num in arr :
            return (i , arr.index(second_num))
    return "NO SUCH PAIRS FOUND"
print (two_sum(arr))

Melvin Yeo • Sep 12 '19 • Edited

In Python, written with a O(n) solution with the given assumptions using a dictionary and enumeration to generate an index.

def twoSum(arr: list, target: int) -> (int, int):
    num_dict = dict()
    for index, num in enumerate(arr):
        diff = target - num
        if diff in num_dict:
            return (min(index, num_dict[diff]), max(index, num_dict[diff]))
        num_dict[num] = index

    raise IndexError

Craig McIlwrath • Sep 10 '19

Haskell:

import Control.Applicative
import Data.List (find) 

twoSum :: (Eq a, Num a) => [a] -> a -> Maybe (Int, Int) 
twoSum xs target = let withId = zip xs [0..]
                       sums = (\(x,a) (y,b) -> (x+y, (a, b))) <$> withId <*> withId
                   in find ((==target) . fst) sums >>= (return . snd)

I know the question said the input is always good, but I didn't want to ignore possible errors in when using the function...

I'm not sure of the speed of this code either. My guess is it only evaluates combinations until it finds a correct one? But I'm not an expert on lazy evaluation and I'm pretty sure this could be faster.

Amin • Sep 10 '19

JavaScript

Recursive solution.

"use strict";

const addWith = target => current => target + current;
const equalsTo = target => current => target === current;

function twoSum(numbers, sum, index = 0) {
    if (numbers.length === 0) {
        return null;
    }

    const [first, ...rest] = numbers;

    const foundIndex = numbers.map(addWith(first)).findIndex(equalsTo(sum));

    if (foundIndex > 0) {
        return [index, foundIndex + index];
    }

    return twoSum(rest, sum, index + 1);
}

console.log(twoSum([18, 7, 1, 6, 0, 15, 3, 12, 111, 1111], 4)); // [2, 6]

Try it online.

willsmart • Sep 10 '19

reduce is good for this one.
Here's a JS quickie with O(n) complexity.

twoSum = (numbers, target) => numbers.reduce(([result, partials], num, i) => ([
  result || (num in partials && [partials[num], i]), 
  Object.assign(partials, {[target - num]: i})
]), [undefined, {}])[0]

The partials map maps the index of each value to the value required to bring it up to the target. This can then be picked up by a later value to the get the answer.

E. Choroba • Sep 11 '19

Perl solution. Walk the array, for each element, store the expected counterpart into a hash. If the element already exists in the hash, we have the pair.

#!/usr/bin/perl
use warnings;
use strict;

sub two_sum {
    my ($arr, $sum) = @_;
    my %expect;
    for my $i (0 .. $#$arr) {
        return [ $expect{ $sum - $arr->[$i] }, $i ]
            if exists $expect{ $sum - $arr->[$i] };
        $expect{ $arr->[$i] } = $i;
    }
}

use Test::More tests => 2;
is_deeply two_sum([1, 2, 3], 4), [0, 2];
is_deeply two_sum([1 .. 99], 197), [97, 98];

peter279k • Jun 22 '20

Here is the Python solution:

def two_sum(numbers, target):
    num_index = 0
    for number in numbers:
        index = 1
        while index < len(numbers):
            if number + numbers[index] == target:
                return [num_index, index]
            index += 1
        num_index += 1

    return [0, len(numbers)-1]

Jérémie Astor • Sep 10 '19

In gwion:

fun <~int, int~>Tuple twosum(int data[], int total) {
  for(int i; i < data.size(); ++i) {
    for(i + 1 => int j; j < data.size(); ++j) {
      if((data[i]+data[j]) == total)
        return <(i, j);
    }
  }
  return null;
}

([1, 2, 3], 4) => twosum @=> >(index0, index1);
<<< index0, " ", index1 >>>;

The fun thing is that it undercovered a (now fixed) bug on tuple unpacking. 🍾

playwrightpo • Sep 10 '19 • Edited

Ruby

The initial solution comes to mind which obviously could be improved in performance purposes having more time

  def two_sum(array, target)
    array.each_with_index do |item, i|
      (i + 1).upto(array.size - 1) do |j|
        return [i, j] if item + array[j] == target
      end
    end

    nil
  end

ahmed1921 • Apr 30 '20

Swift Solution

class Solution {
static func twosum(numbers: [Double], target: Double) -> [Int] {
let result = numbers.filter({ return numbers.contains( target - $0 ) })

return [numbers.firstIndex(of: result[0])! , numbers.lastIndex(of: result [result.count - 1 ] )!] }
}

SRINU • Sep 11 '19

def twoSum(arr,target):
    for i in range(len(arr)-1):
        for j in range(i+1,len(arr)):
            if(arr[i]+arr[j]==target):
                return([i,j])
    else:
        return("no indices found with that sum")
if __name__=="__main__":
    arr=list(map(int,input().split()))
    target=int(input())
    print(twoSum(arr,target))