Find 2 numbers in an unsorted array equal to a given sum

Question

We need to find pair of numbers in an array whose sum is equal to a given value.

A = {6,4,5,7,9,1,2}

Sum = 10 Then the pairs are - {6,4} ,

Answer 1

Python 2.7 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

Shouldn't iterating from both ends just solve the problem?

Sort the array. And start comparing from both ends.

if((arr[start] + arr[end]) < sum) start++;
if((arr[start] + arr[end]) > sum) end--;
if((arr[start] + arr[end]) = sum) {print arr[start] "," arr[end] ; start++}
if(start > end)  break;

Time Complexity O(nlogn)

Answer 3

https://github.com/clockzhong/findSumPairNumber

#! /usr/bin/env python
import sys
import os
import re


#get the number list
numberListStr=raw_input("Please input your number list (seperated by spaces)...\n")
numberList=[int(i) for i in numberListStr.split()]
print 'you have input the following number list:'
print numberList

#get the sum target value
sumTargetStr=raw_input("Please input your target number:\n")
sumTarget=int(sumTargetStr)
print 'your target is: '
print sumTarget


def generatePairsWith2IndexLists(list1, list2):
    result=[]
    for item1 in list1:
        for item2 in list2:
            #result.append([item1, item2])
            result.append([item1+1, item2+1])
    #print result
    return result

def generatePairsWithOneIndexLists(list1):
    result=[]
    index = 0
    while index< (len(list1)-1):
        index2=index+1
        while index2 < len(list1):
            #result.append([list1[index],list1[index2]])
            result.append([list1[index]+1,list1[index2]+1])
            index2+=1
        index+=1
    return result


def getPairs(numList, target):
    pairList=[]
    candidateSlots=[] ##we have (target-1) slots 

    #init the candidateSlots list
    index=0
    while index < target+1:
        candidateSlots.append(None)
        index+=1

    #generate the candidateSlots, contribute O(n) complexity
    index=0
    while index<len(numList):
        if numList[index]<=target and numList[index]>=0:
            #print 'index:',index
            #print 'numList[index]:',numList[index]     
            #print 'len(candidateSlots):',len(candidateSlots)
            if candidateSlots[numList[index]]==None:
                candidateSlots[numList[index]]=[index]
            else:
                candidateSlots[numList[index]].append(index)
        index+=1

    #print candidateSlots

    #generate the pairs list based on the candidateSlots[] we just created
    #contribute O(target) complexity
    index=0
    while index<=(target/2):
        if candidateSlots[index]!=None and candidateSlots[target-index]!=None:
            if index!=(target-index):
                newPairList=generatePairsWith2IndexLists(candidateSlots[index], candidateSlots[target-index])
            else:
                newPairList=generatePairsWithOneIndexLists(candidateSlots[index])
            pairList+=newPairList
        index+=1

    return pairList

print getPairs(numberList, sumTarget)

I've successfully implemented one solution with Python under O(n+m) time and space cost. The "m" means the target value which those two numbers' sum need equal to. I believe this is the lowest cost could get. Erict2k used itertools.combinations, it'll also cost similar or higher time&space cost comparing my algorithm.

Answer 4

    public void printPairsOfNumbers(int[] a, int sum){
    //O(n2)
    for (int i = 0; i < a.length; i++) {
        for (int j = i+1; j < a.length; j++) {
            if(sum - a[i] == a[j]){
                //match..
                System.out.println(a[i]+","+a[j]);
            }
        }
    }

    //O(n) time and O(n) space
    Set<Integer> cache = new HashSet<Integer>();
    cache.add(a[0]);
    for (int i = 1; i < a.length; i++) {
        if(cache.contains(sum - a[i])){
            //match//
            System.out.println(a[i]+","+(sum-a[i]));
        }else{
            cache.add(a[i]);
        }
    }

}

Answer 5

If numbers aren't very big, you can use fast fourier transform to multiply two polynomials and then in O(1) check if coefficient before x^(needed sum) sum is more than zero. O(n log n) total!

Answer 6

if its a sorted array and we need only pair of numbers and not all the pairs we can do it like this:

public void sums(int a[] , int x){ // A = 1,2,3,9,11,20 x=11
    int i=0 , j=a.length-1;
    while(i < j){
      if(a[i] + a[j] == x) system.out.println("the numbers : "a[x] + " " + a[y]);
      else if(a[i] + a[j] < x) i++;
      else j--;
    }
}

1 2 3 9 11 20 || i=0 , j=5 sum=21 x=11
1 2 3 9 11 20 || i=0 , j=4 sum=13 x=11
1 2 3 9 11 20 || i=0 , j=4 sum=11 x=11
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