问题
This question was asked in the Google programming interview. I thought of two approaches for the same:
Find all the subsequences of length. While doing so compute the sum and of the two elements and check if it is equal to k. If ye, print Yes, else keep searching. This is a brute Force approach.
Sort the array in non-decreasing order. Then start traversing the array from its right end. Say we have the sorted array, {3,5,7,10} and we want the sum to be 17. We will start from element 10, index=3, let's denote the index with 'j'. Then include the current element and compute required_sum= sum - current_element. After that, we can perform a binary or ternary search in array[0- (j-1)] to find if there is an element whose value is equal to the required_sum. If we find such an element, we can break as we have found a subsequence of length 2 whose sum is the given sum. If we don't find any such element, then decrease the index of j and repeat the above-mentioned steps for resulting subarray of length= length-1 i.e. by excluding the element at index 3 in this case.
Here we have considered that array could have negative as well as positive integers.
Can you suggest a better solution than this? A DP solution maybe? A solution that can further reduce it's time complexity.
回答1:
This question can be easily solved with the help of set in O(N) time and space complexity.First add all the elements of array into set and then traverse each element of array and check whether K-ar[i] is present in set or not.
Here is the code in java with O(N) complexity :
boolean flag=false;
HashSet<Long> hashSet = new HashSet<>();
for(int i=0;i<n;i++){
if(hashSet.contains(k-ar[i]))flag=true;
hashSet.add(ar[i]);
}
if(flag)out.println("YES PRESENT");
else out.println("NOT PRESENT");
回答2:
Here is a Java implementation with the same time complexity as the algorithm used to sort the array. Note that this is faster than your second idea because we do not need to search the entire array for a matching partner each time we examine a number.
public static boolean containsPairWithSum(int[] a, int x) {
Arrays.sort(a);
for (int i = 0, j = a.length - 1; i < j;) {
int sum = a[i] + a[j];
if (sum < x)
i++;
else if (sum > x)
j--;
else
return true;
}
return false;
}
Proof by induction:
Let a[0,n] be an array of length n+1 and p = (p1, p2) where p1, p2 are integers and p1 <= p2 (w.l.o.g.). Assume a[0,n] contains p1 and p2. In the case that it does not, the algorithm is obviously correct.
Base case (i = 0, j = n):
a[0,-1] does not contain p1 and a[n,n+1] does not contain p2.
Hypothesis:
a[0,i-1] does not contain a[i] and a[j+1,n] does not contain p2.
Step case (i to i + 1 or j to j - 1):
- Assume
p1 = a[i]. Then, sincep1 + a[j] < p1 + p2, index j must be increased. But from the hypothesis we know thata[j+1,n-1]does not contain p2. Contradiction. It follows thatp1 != a[i]. - j to j - 1 analogously.
Because each iteration, a[0,i-1] and a[j+1,n], does not contain p1, and p2, a[i,j] does contain p1 and p2. Eventually, a[i] = p1 and a[j] = p2 and the algorithm returns true.
回答3:
if you want to find pair count,
pairs = [3,5,7,10]
k = 17
counter = 0
for i in pairs:
if k - i in pairs:
counter += 1
print(counter//2)
回答4:
This is java implementation with O(n) Time complexity and O(n) space. The idea is have a HashMap which will contain complements of every array element w.r.t target. If the complement is found, we have 2 array elements which sum to the target.
public boolean twoSum(int[] nums, int target) {
if(nums.length == 0 || nums == null) return false;
Map<Integer, Integer> complementMap = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int curr = nums[i];
if(complementMap.containsKey(target - curr)){
return true;
}
complementMap.put(curr, i);
}
return false;
}
回答5:
Python Solution:
def FindPairs(arr, k):
for i in range(0, len(arr)):
if k - arr[i] in arr:
return True
return False
A = [1, 4, 45, 6, 10, 8]
n = 100
print(FindPairs(A, n))
Or
def findpair(list1, k):
for i in range(0, len(list1)):
for j in range(0, len(list1)):
if k == list1[i] + list1[j]:
return True
return False
nums = [10, 5, 6, 7, 3]
k = 100
print(findpair(nums, k))
回答6:
Javascript solution:
function hasSumK(arr, k) {
hashMap = {};
for (let value of arr) {
if (hashMap[value]) { return true;} else { hashMap[k - value] = true };
}
return false;
}
回答7:
Using Scala, in a single pass with O(n) time and space complexity.
import collection.mutable.HashMap
def addUpToK(arr: Array[Int], k: Int): Option[Int] = {
val arrayHelper = new HashMap[Int,Int]()
def addUpToKHelper( i: Int): Option[Int] = {
if(i < arr.length){
if(arrayHelper contains k-arr(i) ){
Some(arr(i))
}else{
arrayHelper += (arr(i) -> (k-arr(i)) )
addUpToKHelper( i+1)
}
}else{
None
}
}
addUpToKHelper(0)
}
addUpToK(Array(10, 15, 3, 7), 17)
回答8:
Here is a C implementation
For Sorting O(n2) time and space complexity.
For Solving Problem We use
single pass with O(n) time and space complexity via Recursion.
/* Given a list of numbers and a number k , return weather any two numbers from the list add up to k.
For example, given [10,15,3,7] and k of 17 , return 10 + 7 is 17
Bonus: Can You Do in one pass ? */
#include<stdio.h>
int rec(int i , int j ,int k , int n,int array[])
{
int sum;
for( i = 0 ; i<j ;)
{
sum = array[i] + array[j];
if( sum > k)
{
j--;
}else if( sum < k)
{
i++;
}else if( sum == k )
{
printf("Value equal to sum of array[%d] = %d and array[%d] = %d",i,array[i],j,array[j]);
return 1;//True
}
}
return 0;//False
}
int main()
{
int n ;
printf("Enter The Value of Number of Arrays = ");
scanf("%d",&n);
int array[n],i,j,k,x;
printf("Enter the Number Which you Want to search in addition of Two Number = ");
scanf("%d",&x);
printf("Enter The Value of Array \n");
for( i = 0 ; i <=n-1;i++)
{
printf("Array[%d] = ",i);
scanf("%d",&array[i]);
}
//Sorting of Array
for( i = 0 ; i <=n-1;i++)
{
for( j = 0 ; j <=n-i-1;j++)
{
if( array[j]>array[j+1])
{
//swapping of two using bitwise operator
array[j] = array[j]^array[j+1];
array[j+1] = array[j]^array[j+1];
array[j] = array[j]^array[j+1];
}
}
}
k = x ;
j = n-1;
rec(i,j,k,n,array);
return 0 ;
}
OUTPUT
Enter The Value of Number of Arrays = 4
Enter the Number Which you Want to search in addition of Two Number = 17
Enter The Value of Array
Array[0] = 10
Array[1] = 15
Array[2] = 3
Array[3] = 7
Value equal to sum of array[1] = 7 and array[2] = 10
Process returned 0 (0x0) execution time : 54.206 s
Press any key to continue.
回答9:
Here is python's implementation
arr=[3,5,7,10]
k=17
flag=False
for i in range(0,len(arr)):
if k-arr[i] in arr:
flag=True
print( flag )
回答10:
The solution can be found out in just one pass of the array. Initialise a hash Set and start iterating the array. If the current element in the array is found in the set then return true, else add the complement of this element (x - arr[i]) to the set. If the iteration of array ended without returning it means that there is no such pair whose sum is equal to x so return false.
public boolean containsPairWithSum(int[] a, int x) {
Set<Integer> set = new HashSet<>();
for (int i = 0; i< a.length; i++) {
if(set.contains(a[i]))
return true;
set.add(x - a[i]);
}
return false;
}
回答11:
C++ solution:
int main(){
int n;
cin>>n;
int arr[n];
for(int i = 0; i < n; i++)
{
cin>>arr[i];
}
int k;
cin>>k;
int t = false;
for(int i = 0; i < n-1; i++)
{
int s = k-arr[i];
for(int j = i+1; j < n; j++)
{
if(s==arr[j])
t=true;
}
}
if (t){
cout<<"Thank you C++, very cool";
}
else{
cout<<"Damn it!";
}
return 0;
}
回答12:
Python
def add(num, k):
for i in range(len(num)):
for j in range(len(num)):
if num[i] + num[j] == k:
return True
return False
回答13:
Here's Python. O(n). Need to remove the current element whilst looping because the list might not have duplicate numbers.
def if_sum_is_k(list, k):
i = 0
list_temp = list.copy()
match = False
for e in list:
list_temp.pop(i)
if k - e in list_temp:
match = True
i += 1
list_temp = list.copy()
return match
回答14:
I came up with two solutions in C++. One was a naive brute force type which was in O(n^2) time.
int main() {
int N,K;
vector<int> list;
cin >> N >> K;
clock_t tStart = clock();
for(int i = 0;i<N;i++) {
list.push_back(i+1);
}
for(int i = 0;i<N;i++) {
for(int j = 0;j<N;j++) {
if(list[i] + list[j] == K) {
cout << list[i] << " " << list[j] << endl;
cout << "YES" << endl;
printf("Time taken: %.2fs\n", (double)(clock() - tStart)/CLOCKS_PER_SEC);
return 0;
}
}
}
cout << "NO" << endl;
printf("Time taken: %f\n", (double)(clock() - tStart)/CLOCKS_PER_SEC);
return 0;}
This solution as you could imagine will take a large amount of time on higher values of input.
My second solution I was able to implement in O(N) time. Using an unordered_set, much like the above solution.
#include <iostream>
#include <unordered_set>
#include <time.h>
using namespace std;
int main() {
int N,K;
int trig = 0;
int a,b;
time_t tStart = clock();
unordered_set<int> u;
cin >> N >> K;
for(int i = 1;i<=N;i++) {
if(u.find(abs(K - i)) != u.end()) {
trig = 1;
a = i;
b = abs(K - i);
}
u.insert(i);
}
trig ? cout << "YES" : cout << "NO";
cout << endl;
cout << a << " " << b << endl;
printf("Time taken %fs\n",(double) (clock() - tStart)/CLOCKS_PER_SEC);
return 0;
}
回答15:
Python code:
L = list(map(int,input("Enter List: ").split()))
k = int(input("Enter value: "))
for i in L:
if (k - i) in L:
print("True",k-i,i)
回答16:
function check(arr,k){
var result = false;
for (var i=0; i < arr.length; i++){
for (var j=i+1; j < arr.length; j++){
result = arr[i] + arr[j] == k;
console.log(`${arr[i]} + ${arr[j]} = ${arr[i] + arr[j]}`);
if (result){
break;
}
}
return result;
}
Javascript.
回答17:
C# solution:
bool flag = false;
var list = new List<int> { 10, 15, 3, 4 };
Console.WriteLine("Enter K");
int k = int.Parse(Console.ReadLine());
foreach (var item in list)
{
flag = list.Contains(k - item);
if (flag)
{
Console.WriteLine("Result: " + flag);
return;
}
}
Console.WriteLine(flag);
回答18:
My C# Implementation:
bool isPairPresent(int[] numbers,int value)
{
for (int i = 0; i < numbers.Length; i++)
{
for (int j = 0; j < numbers.Length; j++)
{
if (value - numbers[i] == numbers[j])
return true;
}
}
return false;
}
回答19:
Python Implementation: The code would execute in O(n) complexity with the use of dictionary. We would be storing the (desired_output - current_input) as the key in the dictionary. And then we would check if the number exists in the dictionary or not. Search in a dictionary has an average complexity as O(1).
def PairToSumK(numList,requiredSum):
dictionary={}
for num in numList:
if requiredSum-num not in dictionary:
dictionary[requiredSum-num]=0
if num in dictionary:
print(num,requiredSum-num)
return True
return False
arr=[10, 5, 3, 7, 3]
print(PairToSumK(arr,6))
回答20:
Javascript
const findPair = (array, k) => {
array.sort((a, b) => a - b);
let left = 0;
let right = array.length - 1;
while (left < right) {
const sum = array[left] + array[right];
if (sum === k) {
return true;
} else if (sum < k) {
left += 1;
} else {
right -= 1;
}
}
return false;
}
回答21:
Here's a javascript solution:
function ProblemOne_Solve()
{
const k = 17;
const values = [10, 15, 3, 8, 2];
for (i=0; i<values.length; i++) {
if (values.find((sum) => { return k-values[i] === sum} )) return true;
}
return false;
}
回答22:
I implemented with Scala
def hasSome(xs: List[Int], k: Int): Boolean = {
def check(xs: List[Int], k: Int, expectedSet: Set[Int]): Boolean = {
xs match {
case List() => false
case head :: _ if expectedSet contains head => true
case head :: tail => check(tail, k, expectedSet + (k - head))
}
}
check(xs, k, Set())
}
回答23:
I have tried the solution in Go Lang. However, it consumes O(n^2) time.
package main
import "fmt"
func twoNosAddUptoK(arr []int, k int) bool{
// O(N^2)
for i:=0; i<len(arr); i++{
for j:=1; j<len(arr);j++ {
if arr[i]+arr[j] ==k{
return true
}
}
}
return false
}
func main(){
xs := []int{10, 15, 3, 7}
fmt.Println(twoNosAddUptoK(xs, 17))
}
回答24:
Here's two very quick Python implementations (which account for the case that inputs of [1,2] and 2 should return false; in other words, you can't just double a number, since it specifies "any two").
This first one loops through the list of terms and adds each term to all of the previously seen terms until it hits the desired sum.
def do_they_add(terms, result):
first_terms = []
for second_term in terms:
for first_term in first_terms:
if second_term + first_term == result:
return True
first_terms.append(second_term)
return False
This one subtracts each term from the result until it reaches a difference that is in the list of terms (using the rule that a+b=c -> c-a=b). The use of enumerate and the odd list indexing is to exclude the current value, per the first sentence in this answer.
def do_they_add_alt(terms, result):
for i, term in enumerate(terms):
diff = result - term
if diff in [*terms[:i - 1], *terms[i + 1:]]:
return True
return False
If you do allow adding a number to itself, then the second implementation could be simplified to:
def do_they_add_alt(terms, result):
for term in terms:
diff = result - term
if diff in terms:
return True
return False
回答25:
solution in javascript
this function takes 2 parameters and loop through the length of list and inside the loop there is another loop which adds one number to other numbers in the list and check there sum if its equal to k or not
const list = [10, 15, 3, 7];
const k = 17;
function matchSum(list, k){
for (var i = 0; i < list.length; i++) {
list.forEach(num => {
if (num != list[i]) {
if (list[i] + num == k) {
console.log(`${num} + ${list[i]} = ${k} (true)`);
}
}
})
}
}
matchSum(list, k);
回答26:
My answer to Daily Coding Problem
# Python 2.7
def pairSumK (arr, goal):
return any(map(lambda x: (goal - x) in arr, arr))
arr = [10, 15, 3, 7]
print pairSumK(arr, 17)
回答27:
Here is the code in Python 3.7 with O(N) complexity :
def findsome(arr,k):
if len(arr)<2:
return False;
for e in arr:
if k>e and (k-e) in arr:
return True
return False
and also best case code in Python 3.7 with O(N^2) complexity :
def findsomen2 (arr,k):
if len(arr)>1:
j=0
if arr[j] <k:
while j<len(arr):
i =0
while i < len(arr):
if arr[j]+arr[i]==k:
return True
i +=1
j +=1
return False
回答28:
Javascript Solution
function matchSum(arr, k){
for( var i=0; i < arr.length; i++ ){
for(var j= i+1; j < arr.length; j++){
if (arr[i] + arr[j] === k){
return true;
}
}
}
return false;
}
回答29:
Using vavr library it can be done in pretty concise way:
List<Integer> numbers = List(10, 15, 3, 7);
int k = 17;
boolean twoElementsFromTheListAddUpToK = numbers
.filter(number -> number < k)
.crossProduct()
.map(tuple -> tuple._1 + tuple._2)
.exists(number -> number == k);
回答30:
Here is Swift solution:
func checkTwoSum(array: [Int], k: Int) -> Bool {
var foundPair = false
for n in array {
if array.contains(k - n) {
foundPair = true
break
} else {
foundPair = false
}
}
return foundPair
}
来源：https://stackoverflow.com/questions/51300360/given-a-list-of-numbers-and-a-number-k-return-whether-any-two-numbers-from-the