Daily Challenge #63- Two Sum

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, when added together, give the target value. The indices of these items should then be returned in a tuple like so: (index1, index2).

For the purposes of this kata, some tests may have multiple answers; any valid solutions will be accepted.

The input will always be valid (numbers will be an array of length 2 or greater, and all of the items will be numbers; target will always be the sum of two different items from that array).

Based on: https://leetcode.com/problems/two-sum/

twoSum [1, 2, 3] 4 === (0, 2)

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Thank you to wthit56 and CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[peter279k]

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]

[ahmed1921]

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 ] )!] }
}

[ahmed1921]

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 ] )!] }
}

[Melvin Yeo]

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

[Dev Prakash]

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))

[E. Choroba]

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];

[Zohar Peled]

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

[SRINU]

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))

[Jérémie Astor]

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. 🍾

[willsmart]

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.