Find a pair of elements from an array whose sum equals a given number

Question

Given array of n integers and given a number X, find all the unique pairs of elements (a,b), whose summation is equal to X.

The following is my solution, it is O(nLo

Answer 1

There are 3 approaches to this solution:

Let the sum be T and n be the size of array

Approach 1:
The naive way to do this would be to check all combinations (n choose 2). This exhaustive search is O(n²).

Approach 2: 
 A better way would be to sort the array. This takes O(n log n)
Then for each x in array A, use binary search to look for T-x. This will take O(nlogn).
So, overall search is  O(n log n)

Approach 3 :
The best way would be to insert every element into a hash table (without sorting). This takes O(n) as constant time insertion.
Then for every x, we can just look up its complement, T-x, which is O(1).
Overall the run time of this approach is O(n).

You can refer more here.Thanks.

Answer 2

this is the implementation of O(n*lg n) using binary search implementation inside a loop.

#include <iostream>

using namespace std;

bool *inMemory;


int pairSum(int arr[], int n, int k)
{
    int count = 0;

    if(n==0)
        return count;
    for (int i = 0; i < n; ++i)
    {
        int start = 0;
        int end = n-1;      
        while(start <= end)
        {
            int mid = start + (end-start)/2;
            if(i == mid)
                break;
            else if((arr[i] + arr[mid]) == k && !inMemory[i] && !inMemory[mid])
            {
                count++;
                inMemory[i] = true;
                inMemory[mid] = true;
            }
            else if(arr[i] + arr[mid] >= k)
            {
                end = mid-1;
            }
            else
                start = mid+1;
        }
    }
    return count;
}


int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    inMemory = new bool[10];
    for (int i = 0; i < 10; ++i)
    {
        inMemory[i] = false;
    }
    cout << pairSum(arr, 10, 11) << endl;
    return 0;
}

Answer 3

My Solution - Java - Without duplicates

    public static void printAllPairSum(int[] a, int x){
    System.out.printf("printAllPairSum(%s,%d)\n", Arrays.toString(a),x);
    if(a==null||a.length==0){
        return;
    }
    int length = a.length;
    Map<Integer,Integer> reverseMapOfArray = new HashMap<>(length,1.0f);
    for (int i = 0; i < length; i++) {
        reverseMapOfArray.put(a[i], i);
    }

    for (int i = 0; i < length; i++) {
        Integer j = reverseMapOfArray.get(x - a[i]);
        if(j!=null && i<j){
            System.out.printf("a[%d] + a[%d] = %d + %d = %d\n",i,j,a[i],a[j],x);
        }
    }
    System.out.println("------------------------------");
}

Answer 4

in C#:

        int[] array = new int[] { 1, 5, 7, 2, 9, 8, 4, 3, 6 }; // given array
        int sum = 10; // given sum
        for (int i = 0; i <= array.Count() - 1; i++)
            if (array.Contains(sum - array[i]))
                Console.WriteLine("{0}, {1}", array[i], sum - array[i]);

Answer 5

Just attended this question on HackerRank and here's my 'Objective C' Solution:

-(NSNumber*)sum:(NSArray*) a andK:(NSNumber*)k {
    NSMutableDictionary *dict = [NSMutableDictionary dictionary];
    long long count = 0;
    for(long i=0;i<a.count;i++){

        if(dict[a[i]]) {
            count++;
            NSLog(@"a[i]: %@, dict[array[i]]: %@", a[i], dict[a[i]]);
        }
        else{
            NSNumber *calcNum = @(k.longLongValue-((NSNumber*)a[i]).longLongValue);
            dict[calcNum] = a[i];
        }

    }
    return @(count);
}

Hope it helps someone.

Answer 6

# Let arr be the given array.
# And K be the give sum


for i=0 to arr.length - 1 do
  # key is the element and value is its index.
  hash(arr[i]) = i  
end-for

for i=0 to arr.length - 1 do
  # if K-th element exists and it's different then we found a pair
  if hash(K - arr[i]) != i  
    print "pair i , hash(K - arr[i]) has sum K"
  end-if
end-for