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Coderbyte

A common coding interview question

elisabethgross profile image elisabethgross Updated on ・2 min read

Hey everyone! Welcome back to Code Review, a series of interview challenges released weekly that are completely free for the community. This week we’ll be working on a common, relatively straightforward question that I personally have been asked multiple times in interviews. I chose this challenge because there are multiple ways to approach the problem, each with various time and space trade-offs.

The Challenge:

Write a function, FindIntersection, that reads an array of strings which will contain two elements: the first element will represent a list of comma-separated numbers sorted in ascending order, the second element will represent a second list of comma-separated numbers (also sorted). Your goal is to return a string of numbers that occur in both elements of the input array in sorted order. If there is no intersection, return the string "false".

For example: if the input array is ["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"] the output string should be "1, 4, 13" because those numbers appear in both strings. The array given will not be empty, and each string inside the array will be of numbers sorted in ascending order and may contain negative numbers.

The Brute Force Solution:

A brute-force solution is to loop over the numbers of the first string, and for each number in the first string, loop over all numbers of the other string, looking for a match. If a match is found, concat that value to a result string.

function FindIntersection (strArr) {
  const inBothStrings = []
  const arr1 = strArr[0].split(', ')
  const arr2 = strArr[1].split(', ')
  arr1.forEach(elementArr1 => {
    const numArr1 = parseInt(elementArr1)
    arr2.forEach(elementArr2 => {
      const numArr2 = parseInt(elementArr2)
      if (numArr1 === numArr2) {
        inBothStrings.push(numArr1)
      }
    })
  })
  return inBothStrings.join(',')
}

Although this will work, it is not the most optimal solution. The worst case scenario (if there are no matches) is that for every element in the first string, we will have to iterate through every element in the second string. This has a time complexity of O(nm) where n and m are the size of the given strings.

If you haven’t heard of Big O notation, check out this great article which goes into all the details. Understanding Big O notation and being able to communicate how optimal your solution is to an interviewer is an extremely important part of any technical interview.

Try it Out:

Head on over to Coderbyte and try and solve the problem in a more optimized way. Remember to come back next week where I’ll discuss some other solutions and common pitfalls to this problem. Good luck coding :)

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Discussion

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citizen428 profile image
Michael Kohl

I don't usually do JS, but it has became pretty decent in the last few years:

function FindIntersection(strArr) {
  const [s1, s2] = strArr.map(a => new Set(a.split(", ")));
  const intersection = [...s1].filter(x => s2.has(x));
  return intersection.length === 0 ? "false": intersection.join(",");
}
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thorstenhirsch profile image
Thorsten Hirsch

But [...s1].filter(x => s2.has(x)) is still O(n²), isn't it?

  • filter() will step through s1 (=outer forEach)
  • s2.has() will check against each element in s2 (=inner forEach), probably exiting early

I'm pretty sure there is a O(s1) + O(s2) = O(n) solution.

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citizen428 profile image
Michael Kohl

Membership tests on Sets are O(1) not O(n), in other words there is no inner forEach.

Thread Thread
citizen428 profile image
Michael Kohl

To clarify: you don't need to iterate over each element in a set for a membership test. They can be implemented as e.g. hash maps. If a key exists, the element is in the set. Hash map lookups are generally O(1).

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thorstenhirsch profile image
Thorsten Hirsch

Yes, I've looked up how Set works and you're right. Thanks for clarifying.

Another cool solution (but not as readable as yours) however is this "destructive" one on SO.

Thread Thread
citizen428 profile image
Michael Kohl

FWIW filter + has is also how the MDN docs define intersection.

developer.mozilla.org/en-US/docs/W...

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sleepingnapster profile image
sleepingnapster

Good answer but you lost the ordering when using sets.

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citizen428 profile image
Michael Kohl

According to the docs sets will be iterated in insertion order. So the solution will propably work, but rely on an implementation detail to do so, which is not ideal.

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elisabethgross profile image
elisabethgross Author

Nice solution! Love your use of Sets. Very concise and clear.

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dwilmer profile image
Daan Wilmer
function findIntersection($numbersAsStrings) {
    // parse the strings into numbers
    $numbersAsArrays = array_map(function($string) {
        return explode(', ', $string);
    }, $numbersAsStrings);

    // find the intersections
    $intersection = array_intersect(...$numbersAsArrays);

    // if empty, return string "false"; otherwise, return the numbers as a string
    if (empty($intersection)) {
        return 'false';
    } else {
        return implode(', ', $intersection);
    }
}

because I'd argue that letting your language work for you, and therefore optimizing developer time, is better than optimizing execution time when not strictly necessary.

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dwilmer profile image
Daan Wilmer

To follow up: I made three versions of this code. The first version is as above, the second version has the naive approach to finding the intersection:

// find the intersections
$arrA = $numbersAsArrays[0];
$arrB = $numbersAsArrays[1];
$intersection = [];
foreach ($arrA as $num) {
    if (in_array($num, $arrB)) {
        $intersection[] = $num;
    }
}

and a third approach uses a map for lookup:

// find the intersections
$mapA = array_combine($numbersAsArrays[0], $numbersAsArrays[0]);
$mapB = array_combine($numbersAsArrays[1], $numbersAsArrays[1]);

$intersection = [];
foreach ($mapA as $num) {
    if (isset($mapB[$num])) {
        $intersection[] = $num;
    }
}

Performing some timed tests shows that using the built-in function consistently takes about two to three times as long as the self-built map-based function, growing at roughly O(n) using a random array of size 1000 and 10000. Only in exceptional cases would I not use the built-in function.
The naive approach, on the other hand, grows with O(n²) and takes significantly longer at larger array sizes. However, with 26ms with an array size of 1000, it depends on the use case whether I would start optimizing this (if this isn't easily replaced with a built-in function).

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elisabethgross profile image
elisabethgross Author

Nice work!! Always always interesting to test out a language's built in methods and know whats what when it comes to time complexity.

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elisabethgross profile image
elisabethgross Author

When the language solves the problem for you... touché

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cxsttlx profile image
Juan Castillo

this is my solution, Im no pretty good at JS but I want to be a master on it :)

const FindIntersection  = (str) => {
  const arr = str.flatMap(i => i.split(', '))
  const itr = [...new Set(arr.filter((unique, index) => arr.indexOf(unique) !== index))]
  return itr.length ? itr : false
}
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cxsttlx profile image
Juan Castillo

ok, i have no idea to make my code look like editor

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elisabethgross profile image
elisabethgross Author

Nice job! And try surrounding your code with triple backticks to format the code ;)

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cxsttlx profile image
Juan Castillo

thanks I made it :D i'll wait for more

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gabgab2003 profile image
DownloadPizza

also you can put javascript directly after your backticks to make it have syntax highlighting (maybe its js, dont know, also works for other languages)

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elisabethgross profile image
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devtronic profile image
Julian Finkler
function FindIntersection(strArr) { 

  const allNumbers = strArr.reduce((c, a) => c = [...c, ...a.split(',').map(n => Number(n.trim()))], []);
  const numberCounts = {};
  const duplicates = [];
  allNumbers.forEach((n) => {
    if(!numberCounts.hasOwnProperty(n)){
      numberCounts[n] = 0;
    }
    if(++numberCounts[n] == strArr.length){
      duplicates.push(n);
    }
  })

  return duplicates.join(','); 

}
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idanarye profile image
Idan Arye

Rust solution - works for any number of strings:

fn find_intersection(strings: &[&str]) -> String {
    if strings.is_empty() {
        // NOTE: mathematically speaking, the result here should be **all** the numbers - but we
        // obviously can't return that...
        return "".to_owned(); // no arrays to intersect
    }
    let mut iterators: Vec<_> = strings.iter().map(
        |s| s.split(',').map(|n| n.trim().parse::<isize>().unwrap())
    ).collect();
    let mut current_values = Vec::with_capacity(iterators.len());
    for it in iterators.iter_mut() {
        if let Some(value) = it.next() {
            current_values.push(value);
        } else {
            return "".to_owned(); // at least one array is empty => intersection is empty
        }
    }
    let mut result = Vec::<String>::new();

    loop {
        let (min_value, min_count) = {
            let mut min_value = current_values[0];
            let mut min_count = 1;
            for value in current_values.iter().skip(1) {
                use std::cmp::Ordering::*;
                match value.cmp(&min_value) {
                    Less => {
                        min_value = *value;
                        min_count = 1;
                    },
                    Equal => {
                        min_count += 1;
                    },
                    Greater => {},
                }
            };
            (min_value, min_count)
        };
        if min_count == current_values.len() {
            result.push(min_value.to_string());
        }
        for (current_value, it) in current_values.iter_mut().zip(iterators.iter_mut()) {
            if *current_value == min_value {
                if let Some(value) = it.next() {
                    assert!(*current_value <= value, "numbers must be ordered");
                    *current_value = value;
                } else {
                    return result.join(", ");
                }
            }
        }
    }
}
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gabgab2003 profile image
DownloadPizza

Is there any difference between

String::from("")

and

"".to_owned()

?

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idanarye profile image
Idan Arye

None that I'm aware of. I prefer "".to_owned() over String::from("") or "".to_string() to avoid the confusion of "why are you converting a string to a string?" that I've seen in some newbie Rust threads.

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firee80 profile image
Firee80

Hi! I wrote a solution that should be able to handle n-number of lines.
Also it does not care if there are duplicate numbers or they are out of order.
It just uses .map and .reduce.. probably it still could be optimized by using Set.

const lines = ['1,2,3,4,5,6', '2,4,6,7,8,9,10', '1,2,3,4,5,6,8']

const mapArrayToHashTable = items => {
  return items.reduce((last, item) => ({...last, [''+item]: item}), {}) 
}

const getHashTable = line => {
  const textNumbers = line.split(',')
  const numbers = textNumbers.map(number => parseInt(number))
  return mapArrayToHashTable(numbers)
}

const [firstTable, ...otherTables] = lines.map(getHashTable)

const result = otherTables.reduce((last, table) => {
  const keyObjs = Object.keys(table).map(key => last[key] ? {[key]: last[key]} : {})
  return keyObjs.reduce((last, keyObj) => ({...last, ...keyObj}), {})
}, firstTable)

console.log(Object.values(result)) // returns [2,4,6]
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firee80 profile image
Firee80

simplified the solution by using Sets and corrected the answer to correspond the assignment

const lines = ['1,2,3,4,5,6', '2,4,6,7,8,9,10', '1,2,3,4,5,6,8']

function FindIntersection(lines) {
  const getSet = line => new Set(line.split(',').map(number => parseInt(number)))
  const [firstSet, ...otherSets] = lines.map(getSet)
  const result = [...otherSets.reduce((lastSet, set) => new Set([...set].filter(number => lastSet.has(number))), firstSet)]
  return result.length > 0 ? result : 'false'
}

console.log(FindIntersection(lines)) // outputs: [2,4,6]
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firee80 profile image
Firee80

couldn't help myself.. improved the performance by sorting sets and checking result set. Changed the api from array input to n-amount of params. Also input is filtered from not a number

function FindIntersection(...lines) {  
  const getSet = (line = '') => {
    const numbers = line.split(',').map(number => parseInt(number))
    return new Set(numbers.filter(number => !isNaN(number)))
  }
  const sets = lines.map(getSet)
  const [shortestSet, ...otherSets] = [...sets].sort((setA, setB) => setA.size - setB.size)
  const defaultSet = lines.length > 1 ? shortestSet : new Set([])
  const resultSet = otherSets.reduce((resultSet, set) => {
    const commonNumbers = [...resultSet].filter(number => set.has(number))
    return new Set(commonNumbers)
  }, defaultSet)
  const result = [...resultSet]
  return result.length > 0 ? result : 'false'
}

console.log(FindIntersection()) // outputs: false
console.log(FindIntersection('1')) // outputs: false
console.log(FindIntersection('2', '3')) // outputs: false
console.log(FindIntersection('-1, 0', '0')) // outputs: [0]
console.log(FindIntersection('a, 3', 'b, c, 3')) // outputs: [3]
console.log(FindIntersection('-2, 3', '-4, -2')) // outputs: [-2]
console.log(FindIntersection('1,2,3,4', '2,3,4,5', '3,4,5,6')) // outputs: [3, 4]
console.log(FindIntersection('1,2,3,4,5,6', '2,4,6,7,8,9,10', '1,2,3,4,5,6,8')) // outputs: [2, 4, 6]
console.log(FindIntersection(...['1, 3, 4, 7, 13', '1, 2, 4, 13, 15'])) // outputs: [1, 4, 13]
Thread Thread
firee80 profile image
Firee80

Decided to have a one more improvement for performance (when one Set has size = 0, so no need to reduce other Sets. Also result from reduce is directly an array of numbers).
Also my apologies to @elisabethgross as I noticed the tag #codenewbie in the challenge and my answer isn't really for newbies.

function FindIntersection(...lines) {  
  const getSet = (line = '') => {
    const numbers = line.split(',').map(number => parseInt(number))
    return new Set(numbers.filter(number => !isNaN(number)))
  }
  const sets = lines.map(getSet)
  const [shortestSet, ...otherSets] = [...sets].sort((setA, setB) => setA.size - setB.size)
  const setsToReduce = shortestSet && shortestSet.size > 0 ? otherSets : [] 
  const defaultCommonNumbers = lines.length > 1 ? [...shortestSet] : []
  const resultCommonNumbers = setsToReduce.reduce((commonNumbers, set) => 
    commonNumbers.filter(number => set.has(number)), defaultCommonNumbers)
  return resultCommonNumbers.length > 0 ? resultCommonNumbers : 'false'
}

console.log(FindIntersection()) // outputs: false
console.log(FindIntersection('1')) // outputs: false
console.log(FindIntersection('2', '3')) // outputs: false
console.log(FindIntersection('-1, 0', '0')) // outputs: [0]
console.log(FindIntersection('a, 3', 'b, c, 3')) // outputs: [3]
console.log(FindIntersection('-2, 3', '-4, -2')) // outputs: [-2]
console.log(FindIntersection('1,2,3,4', '2,3,4,5', '3,4,5,6')) // outputs: [3, 4]
console.log(FindIntersection('1,2,3,4,5,6', '2,4,6,7,8,9,10', '1,2,3,4,5,6,8')) // outputs: [2, 4, 6]
console.log(FindIntersection(...['1, 3, 4, 7, 13', '1, 2, 4, 13, 15'])) // outputs: [1, 4, 13]
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artezan profile image
artezan

I used for loop so that improve the performance


const input = ["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"]
function FindIntersection(strArr) {
  const inBothStrings = []
   const [arr1,arr2] = strArr.map(item => item.split(", "))

  for (let index = 0; index < arr1.length; index++) {
    const element1 = arr1[index];
    const indexFound= arr2.findIndex(element2 => element2 === element1)

    if (indexFound !== -1) {
      inBothStrings.push(element1)
      arr2.splice(indexFound, 1)
    }
  }
  return inBothStrings.length ? inBothStrings.join(',') : false 
}
console.log(FindIntersection(input))

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jkodev profile image
Jonathan Jasson Huamani Cuadros

This is my approach.

function intersection(stringArray) {
    const parseArray = (array) => [...new Set(array.split(',').map(item => item.trim()))];
    const [a, b] = stringArray;
    const aParsed = parseArray(a);
    const bParsed = parseArray(b);
    const mergedData = [...aParsed, ...bParsed];
    const counter = {};
    mergedData.forEach(item => {
        counter[item] = (counter[item] || 0) + 1;
    });
    return Object.entries(counter).filter(item => item[1] > 1).map(item => item[0]).join(',');
}
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elisabethgross profile image
elisabethgross Author

Nice custom parseArray function!! And I like that use of Object.entries().

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assafsch profile image
Assaf Schwartz

Not a JS guy, so I decided to do it in Python.
As an interviewer, for these kind of question I'm not a fan of leveraging too much of the language built-ins (such as sets), as it masks algorithmic thinking, which I believe is an important part of the interview.

My solution uses a dictionary to count the number of occurrences for every item.

from collections import defaultdict

def find_intersect(strings):
    a, b = strings
    a = a.split(",")
    b = b.split(",")
    counter = defaultdict(int)
    intersect = list()
    for idx in range(max(len(a), len(b))):
        if idx < len(a):
            item = int(a[idx])
            counter[item]=counter[item]+1
            if counter[item] > 1:
                intersect.append(item)
        if idx < len(b):
            item = int(b[idx])
            counter[item]=counter[item]+1
            if counter[item] > 1:
                intersect.append(item)
    return intersect if intersect else "false"
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elisabethgross profile image
elisabethgross Author

I agree about not leveraging too many language built-ins, or, making sure to talk about the time complexity of those built-ins!

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lisabenmore profile image
Lisa Benmore

fwiw, another js option:

jsfiddle.net/4usxLhyo/1/

sorry, first post, not sure how to get the js syntax highlighting...
Edit: thanks @elisabethgross for the syntax highlighting help!

console.clear()

const one = '1,3,5,5,7,9,9';
const two = '2,4,5,6,8,9,9';
const three = '1,3,5,7,9';
const four = '2,4,6,8';

function findThings (str1, str2) {
  const arr1 = str1.split(',');
  const arr2 = str2.split(',');
  const result = [];

  for (const i of arr1) {
    if (arr2.indexOf(i) > -1) {
      result.push(i);
      arr2.splice(arr2.indexOf(i), 1);
    }
  }

  return result.length ? result : false;
}

console.log(findThings(one, two));
console.log(findThings(three, four )); ```


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elisabethgross profile image
elisabethgross Author

Nice! If you use the triple backticks, you can write javascript next to the top triple backticks. Check out this helpful markdown cheat sheet!

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philipgpm profile image
Philip Georgiev

coderbyte.com/results/philipgg:Fin...

I submitted a 0(2n-a) where a is the number of repetitive elements.

The idea is to use that the arrays are sorted so always increment from where we are.

if the elements are matching, save and then increment both counters.

Else increment the count of the array which has the lower number, to try to find the bigger one. (this is auto-switching between the two.)

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elisabethgross profile image
elisabethgross Author

Nice! Was waiting for someone to come up with this one!

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ferceg profile image
ferceg

My solution was the same.
When it comes to challenges I prefer algorithms that can be written in pseudocode and don't use many special (or language-specific) data types.

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teaglebuilt profile image
dillan teagle
import pytest



def findIntersection(arr):
    unique = list()
    duplicates = []
    for string in arr: 
        for char in string.split(','):
            if int("".join(char)) in unique:
                duplicates.append(int("".join(char)))
            else:
                unique.append(int("".join(char)))

    return duplicates



@pytest.mark.parametrize(
    ("arr_s", "expected"),
    (["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"], [1, 4, 13]),
)
def test(arr_s, expected):
    assert findIntersection(arr_s) == expected




if __name__ == "__main__":
    x = ["1, 3, 4, 7, 13", "1, 2, 4, 13, 15"]
    print(findIntersection(x))
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vaibhavyadav1998 profile image
Vaibhav Yadav

In JavaScript.

function findIntersection(strArr) {
  const arr = strArr.join(",").replace(/\s+/g, "").split(",");
  const count = {};
  const common = [];

  arr.map(i => count[i] ? count[i]++ : count[i] = 1);

  for (const i in count) {
    if (count[i] > 1) {
      common.push(parseInt(i, 10));
    }
  }

  return common.length ? common.sort((a, b) => a > b).join(",") : false;
}
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sohammondal profile image
Soham Mondal

This is my O(n) solution. We don't need to parse any of the strings to an integer.


const findIntersection = (arr) => {

  // n1 - larger array
  // n2 - smaller array

  let n1, n2;

  if (arr[0].split(',').length > arr[1].split(',').length) {
    n1 = arr[0].split(',');
    n2 = arr[1].split(',');
  } else {
    n1 = arr[1].split(',');
    n2 = arr[0].split(',');
  }

  let len = n1.length;

  // store the common items
  let common = [];

  for (let i = 0; i < len; i++) {

    // check if an item of one array (larger one) exists in the other array (smaller one)
    if (n2.includes(n1[i])) {
      common.push(n1[i]);
    }


  }

  return common.length ? common.join() : "false" ;
}

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stoiandan profile image
Stoian Dan

Python solution, not in a nice format, and assuming the input are two integer arrays:


def common_arr(first, second):
    mi, ma = (first,second) if len(first) <= len(second) else (second,first)
    common = []
    i = 0
    j = 0
    while(i < len(mi)):
        if(mi[i] == ma[j]):
                common.append(mi[i])
                i += 1
                if j < len(ma)-1:
                    j += 1
                else:
                    break
        elif mi[i] < ma[j]:
            i += 1
        else:
            if j < len(ma)-1:
                j += 1
            else:
                break

    return common

arr = common_arr([1,2,3,4,5,9],[3,5,6,8,9])
print(arr)

So idea:
You iterate over the smallest array, at each step, you are in one of 3 situations

arr1[i] == arra2[j] // you increment i and j because of the invariant, that says the arrays are sorted, so if you found a match you can safely increment.

arr1[i] < arr2[j] // since at index i we have the smallest element, we increment that one because being sorter we ca safely increment until we hit what is at index j or greater

arr1[i] < arr2[j] // same as above only for j

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leoalipazaga profile image
Leonardo Alipazaga
function FindIntersection(strArr) {
  return
    strArr
    .map(e => e.split(','))
    .reduce((listA, listB) => {
      const listFiltered = listA.filter(itemA => listB.includes(itemA));
      return listFiltered.length ? listFiltered : false;
    });
}
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gabgab2003 profile image
DownloadPizza
use std::cmp::Ordering;

fn main() {
    let a = find_intersections([
        String::from("-2, -1, 1, 1, 1, 2, 3, 6"),
        String::from("-1, 1, 1, 3, 4, 5, 6")
    ]);
    println!("{}", a);
}

fn find_intersections(arr: [String; 2]) -> String {
    let mut inter : Vec<isize> = Vec::new();
    let mut a = arr[0].split(",").map(
        |e| e.trim().parse::<isize>().unwrap()
    );
    let mut b = arr[1].split(",").map(
        |e| e.trim().parse::<isize>().unwrap()
    );

    let mut tmp_a = a.next();
    let mut tmp_b = b.next();

    while tmp_a.is_some() && tmp_b.is_some() {
        match tmp_a.unwrap().partial_cmp(&tmp_b.unwrap()).expect("NAN is evil") {
            Ordering::Less => tmp_a = a.next(),
            Ordering::Equal => {inter.push(tmp_a.unwrap()); tmp_a = a.next();tmp_b = b.next()},
            Ordering::Greater => tmp_b = b.next(),
        }
    }

    if inter.is_empty() {
        return String::from("false")
    }
    return inter.into_iter().map(|x| x.to_string()).collect::<Vec<String>>().join(", ")
}

A rust solution, im quite new to rust, so this is probably quite ugly

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sxync profile image
SaiKumar Immadi

A simple and straightforward O(m+n) solution.

const FindIntersection = array => {
  const commonNumbers = [];

  const array1 = array[0].split(", ").map(num => Number(num));
  const array2 = array[1].split(", ").map(num => Number(num));

  let index1 = 0;
  let index2 = 0;

  while (index1 < array1.length && index2 < array2.length) {
    if (array1[index1] < array2[index2]) {
      index1 += 1;
    } else if (array2[index2] < array1[index1]) {
      index2 += 1;
    } else {
      commonNumbers.push(array1[index1]);
      index1 += 1;
      index2 += 1;
    }
  }

  return commonNumbers.join(", ");
};

EDIT: I just checked the next article and it has the same solution listed. Sorry I did not check that before posting this here.

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hariadon profile image
hariadon

// I believe key here is that both arrays are sorted
function FindIntersection (strArr) {
const inBothStrings = []
const arr1 = strArr[0].split(', ')
const arr2 = strArr[1].split(', ')

let i=0,j=0;
while (i<arr1.length&&j<arr2.length){

//casting to numbers
let a = arr1[i]|0;
let b = arr2[j]|0;

if(a==b){
inBothStrings.push(a)
i++;
j++;
}
else if(a>b){

j++;
}
else if(b>a){
i++;
}

}

return inBothStrings.join(',')
}

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elisabethgross profile image
elisabethgross Author

Bingo! Nice work!

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orionaegis profile image
Orion Aegis
def sortedIntersection([a,b]):
  return any([map(set,a.split(', ')).intersection(b.split(', '))])

Would be the easiest to implement (I'm on my phone so I can't check it, also, python)

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ajinkabeer profile image
Ajin Kabeer

Thank you!

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anonymous999 profile image
Anonymous
const FindIntersection = arr => {
  return arr[0].match(/\d+/g).filter(v => (
    arr[1].match(/\d+/g).includes(v)
  )).join`,` || false
}
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gabgab2003 profile image
DownloadPizza

Kotlin (any number of strings): pl.kotl.in/Lgfcz5KnV?theme=darcula

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marko911 profile image
Marko Bilal

I dont understand why anyone would ask you this in an interview ? A massive red flag for me.

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gabgab2003 profile image
DownloadPizza

Clarification question: can the lists have duplicate values?

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elisabethgross profile image
elisabethgross Author

Good question! Nope, the lists will not have duplicates in themselves.