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

Python Implementation:

import itertools
list = [1, 1, 2, 3, 4, 5,]
uniquelist = set(list)
targetsum = 5
for n in itertools.combinations(uniquelist, 2):
    if n[0] + n[1] == targetsum:
        print str(n[0]) + " + " + str(n[1])

Output:

1 + 4
2 + 3

Answer 2

https://github.com/clockzhong/findSumPairNumber

I did it under O(m+n) complexity cost for both time and memory space. I suspect that's the best algorithm so far.

Answer 3

O(n)

def find_pairs(L,sum):
    s = set(L)
    edgeCase = sum/2
    if L.count(edgeCase) ==2:
        print edgeCase, edgeCase
    s.remove(edgeCase)      
    for i in s:
        diff = sum-i
        if diff in s: 
            print i, diff


L = [2,45,7,3,5,1,8,9]
sum = 10          
find_pairs(L,sum)

Methodology: a + b = c, so instead of looking for (a,b) we look for a = c - b

Answer 4

One Solution can be this, but not optimul (The complexity of this code is O(n^2)):

public class FindPairsEqualToSum {

private static int inputSum = 0;

public static List<String> findPairsForSum(int[] inputArray, int sum) {
    List<String> list = new ArrayList<String>();
    List<Integer> inputList = new ArrayList<Integer>();
    for (int i : inputArray) {
        inputList.add(i);
    }
    for (int i : inputArray) {
        int tempInt = sum - i;
        if (inputList.contains(tempInt)) {
            String pair = String.valueOf(i + ", " + tempInt);
            list.add(pair);
        }
    }
    return list;
   }
}

Answer 5

A Simple program in java for arrays having unique elements:

import java.util.*;
public class ArrayPairSum {
    public static void main(String[] args) { 
        int []a = {2,4,7,3,5,1,8,9,5};
        sumPairs(a,10);  
    }

    public static void sumPairs(int []input, int k){
      Set<Integer> set = new HashSet<Integer>();    
      for(int i=0;i<input.length;i++){

        if(set.contains(input[i]))
            System.out.println(input[i] +", "+(k-input[i]));
        else
            set.add(k-input[i]);
       }
    }
}

Answer 6

less than o(n) solution will be=>

function(array,k)
          var map = {};
          for element in array
             map(element) = true;
             if(map(k-element)) 
                 return {k,element}