CODEWITHSUNDEEP
javaarraysalgorithmsortingdata-structures
Abhijit Sarkar
Abhijit Sarkar

Reputation: 24637

How to find 2-sum in linear time?

I enrolled in the Algorithms, Part II course on Coursera, and one of the interview questions (ungraded) is as follows:

2-sum. Given an array a of n 64-bit integers and a target value T, determine whether there are two distinct integers i and j such that a[i] + a[j] = T. Your algorithm should run in linear time in the worst case.

Hint: sort the array in linear time.

I can think of solving it in several ways:

  1. Insert the elements in a hash table in one pass. Then make a 2nd pass, looking for T - a[i] in the hash table. Space complexity is O(n) for the hash table, and O(n) for the 2 passes. This solution meets the time requirement stated in the question.

  2. Sort the array, and then run 2 pointers i and j from the beginning and the end, respectively, looking for a[i] + a[j] = T. If a[i] + a[j] < T, increment i, else decrement j. Space complexity is dependent on the sorting algorithm; assuming, Quick Sort, no additional space necessary. Time complexity, nlogn, so it fails the time requirement stated in the question.

  3. Considering that the question is given after the Radix Sort lecture, I'm assuming the intent is to use one of the Radix Sorts. Since the question specifies 64-bit integers, using binary representation of long, and using In-place MSD radix sort, the array can be sorted in-place in linear time. This seems like the best approach.

Other ideas?

P.S. I've seen this question, but it assumes a sorted array, and all the answers there use some kind of hashing.

I've also seen this question but it seems overly complicated for the specific case of 2-sum.

Upvotes: 4

Views: 2993

Answers (3)

Abhijit Sarkar
Abhijit Sarkar

Reputation: 24637

Answering my own question, here's the working solution in Java:

public class RadixSortsInterviewQuestions {
    private static final int MSB = 64;

    static Map.Entry<Integer, Integer> twoSum(long[] a, long sum) {
        int n = a.length - 1;
        sort(a, MSB, 0, n);

        for (int i = 0, j = n; i < j; ) {
            long t = a[i] + a[j];
            if (t == sum) {
                return new SimpleImmutableEntry<>(i, j);
            } else if (t < sum) {
                i++;
            } else {
                j--;
            }
        }
        return null;
    }

    // Binary MSD radix sort: https://en.wikipedia.org/wiki/Radix_sort#In-place_MSD_radix_sort_implementations
    private static void sort(long[] a, int d, int lo, int hi) {
        if (hi < lo || d < 1) return;

        int left = lo - 1;
        int right = hi + 1;

        for (int i = left + 1; i < right; ) {
            if (isBitSet(a[i], d)) {
                swap(a, i, --right);
            } else {
                left++;
                i++;
            }
        }
        sort(a, d - 1, lo, left);
        sort(a, d - 1, right, hi);
    }

    private static boolean isBitSet(long x, int k) {
        boolean set = (x & 1L << (k - 1)) != 0;

        // invert signed bit so that all positive integers come after negative ones
        return (k == MSB) != set;
    }

    private static void swap(long[] a, int i, int j) {
        long tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }
}

Upvotes: 0

NiVeR
NiVeR

Reputation: 9806

I analyzed your observations and regarding the one that seems to meet the requirement:

Insert the elements in a hash table in one pass. Then make a 2nd pass, looking for T - a[i] in the hash table. Space complexity is O(n) for the hash table, and O(n) for the 2 passes. This solution meets the time requirement stated in the question.

I believe that this approach doesn't meet the requirements as the hashtable worst case insert complexity theoretically is O(n).

As you said that you studied the Radix Sort, I believe that's the way to go, you can sort the array within the time requirements and afterwards you can use the Two Pointers technque to check if there is the sum T:

int left = 0, right = array_size - 1;
boolean found = false;
while (left < right) {

    if (a[left] + a[right] == T) {              
        found = true;
    }
    else if (a[left] + a[right] < T) {
        left++;
    }
    else {
        right--;
    }
}

Upvotes: 2

pamcevoy
pamcevoy

Reputation: 1244

As you traverse the array, put the values into a hash mapping the value to index. Since we are only looking for the sum of two numbers, look for the sum of the current number and the remainder in the hash to get the target.

public static int[] twoSumInOnePass(int[] values, int target) throws Exception {
    // value => index
    Map<Integer, Integer> valueToIndexMap = new HashMap<Integer, Integer>();
    for (int i = 0; i < values.length; i++) {
        int remainder = target - values[i];
        if (valueToIndexMap.containsKey(remainder)) {
            return new int[] { valueToIndexMap.get(remainder), i };
        }
        valueToIndexMap.put(values[i], i);
    }

    throw new Exception("Could not find indexes that sum to " + target);
}

https://leetcode.com/problems/two-sum/description/

Upvotes: 0

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