I am attempting a Java mock interview on LeetCode. I have the following problem:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
I was attempting to implement a recursive solution. However, I am receiving errors upon trying to run my code.
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] sumnums = nums.clone();
//int[2] sol = {0,1};
int[] sol = new int[]{0,1};
sol = new int[2];
int j=sumnums.length;
int t=target;
for(int i=0;i<sumnums.length;i++){
if ((sumnums[i]+sumnums[j])==t){
sol[0]=i;
sol[1]=j;
//return sol;
}
}
j=j-1;
twoSum(sumnums,t);
return sol;
}
}
Error(s):
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 4
at Solution.twoSum(Solution.java:12)
at __DriverSolution__.__helper__(__Driver__.java:8)
at __Driver__.main(__Driver__.java:54)
It appears to me the error may have to do with the following line of code:
if ((sumnums[i]+sumnums[j])==t){
Therefore, I am wondering if this is a syntax related error. I am attempting to check to see if two numbers add up to a different number.
Since this is a naive attempt at a recursive solution, I am happy to take any other criticism. But I am mostly concerned with getting my attempt at this problem to work and run with all testcases.
Thanks.
METHOD 1. Naive approach: Use two for loops
The naive approach is to just use two nested for loops and check if the sum of any two elements in the array is equal to the given target.
Time complexity: O(n^2)
// Time complexity: O(n^2)
private static int[] findTwoSum_BruteForce(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] == target) {
return new int[] { i, j };
}
}
}
return new int[] {};
}
METHOD 2. Use a HashMap (Most efficient)
You can use a HashMap to solve the problem in O(n) time complexity. Here are the steps:
Initialize an empty HashMap.
Iterate over the elements of the array.
For every element in the array -
If the element exists in the Map, then check if it’s the complement (target - element) also exists in the Map or not. If the complement exists then return the indices of the current element and the complement.
Otherwise, put the element in the Map, and move to the next iteration.
Time complexity: O(n)
// Time complexity: O(n)
private static int[] findTwoSum(int[] nums, int target) {
Map<Integer, Integer> numMap = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (numMap.containsKey(complement)) {
return new int[] { numMap.get(complement), i };
} else {
numMap.put(nums[i], i);
}
}
return new int[] {};
}
METHOD 3. Use Sorting along with the two-pointer sliding window approach
There is another approach which works when you need to return the numbers instead of their indexes. Here is how it works:
Sort the array.
Initialize two variables, one pointing to the beginning of the array (left) and another pointing to the end of the array (right).
Loop until left < right, and for each iteration
if arr[left] + arr[right] == target, then return the indices.
if arr[left] + arr[right] < target, increment the left index.
else, decrement the right index.
This approach is called the two-pointer sliding window approach. It is a very common pattern for solving array related problems.
Time complexity: O(n*log(n))
// Time complexity: O(n*log(n))
private static int[] findTwoSum_Sorting(int[] nums, int target) {
Arrays.sort(nums);
int left = 0;
int right = nums.length - 1;
while(left < right) {
if(nums[left] + nums[right] == target) {
return new int[] {nums[left], nums[right]};
} else if (nums[left] + nums[right] < target) {
left++;
} else {
right--;
}
}
return new int[] {};
}
Why not use a HashMap? Here is how it will work. We will iterate through the array and check if the target - nums[i] exists in the map. If not, we will store Key, Value as the number and its index respectively i.e K = nums[i], V = i
Here is how it will work:
Consider nums = [2, 7, 11, 15], target = 9
We will start iterating the array
First comes 2. Here we will check ( 9 - 2 ) i.e 7 does not exist in the hashmap, so we will store 2 as key and its index 0 as its value
Then, comes 7. Here we will check ( 9 - 7 ) i.e 2 which exists in the map and so we will return the index of 2 and index of 7 i.e returning [ 0, 1 ]
public int[] twoSum(int[] nums, int target) {
HashMap<Integer,Integer> map = new HashMap<>();
for(int i = 0; i < nums.length; i++)
{
if(map.containsKey(target - nums[i]))
{
return new int[] {map.get(target-nums[i]),i};
}
else
map.put(nums[i],i);
}
return new int[] {-1,-1};
}
As you mentioned
First comes 2. Here we will check ( 9 - 2 ) i.e 7 does not exist in the hashmap
actually 7 does exist in the HashMap right
nums = [2, 7, 11, 15], target = 9
public static int[] twoSum(int[] nums, int target) {
int[] sumnums = nums.clone();
//int[2] sol = {0,1};
int[] sol = new int[]{0,1};
sol = new int[2];
int j=sumnums.length; // Every recursion j will be initialized as sumnums.length instead of J-1
int t=target;
for(int i=0;i<sumnums.length;i++){
// if ((sumnums[i]+sumnums[j])==t){
// Remember that Java arrays start at index 0, so this value(sumnums[j]) is not exist in the array.
if ((sumnums[i]+sumnums[j-1])==t){
sol[0]=i;
sol[1]=j;
return sol;
}
}
j=j-1;
twoSum(sumnums,t);
return sol;
}
One of the many possible solutions:
public class TwoSumIndex {
public static void main(String... args) {
int[] arr = {2, 7, 11, 15, 9, 0, 2, 7};
System.out.println(findTwoSums(arr, 9));
}
private static List<List<Integer>> findTwoSums(int[] arr, int target) {
Set<Integer> theSet = Arrays.stream(arr).mapToObj(Integer::valueOf).collect(Collectors.toSet());
List<Integer> arrList = Arrays.stream(arr).mapToObj(Integer::valueOf).collect(Collectors.toList());
List<Pair<Integer, Integer>> theList = new ArrayList<>();
List<Integer> added = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
int a = target - arr[i];
if (theSet.contains(a) && added.indexOf(i) < 0) { // avoid duplicates;
Integer theOther = arrList.indexOf(a);
theList.add(new Pair(i, theOther));
added.add(i);
added.add(theOther);
}
}
return theList.stream().map(pair -> new ArrayList<>(Arrays.asList(pair.getKey(), pair.getValue())))
.collect(Collectors.toList());
}
}
There are several things you need to know:
duplicates in the input array are allowed;
un-ordered array in the input is allowed;
no index pair duplicate in the output;
time complexity is theoretically O(N^2) but actually will be much lower since theSet.contains(a) O(logN) will filter out all failed indexes and then do the index duplicate O(N) checking, so the actual time complexity should be O(NlogN);
The output for the demo:
[[0, 1], [4, 5], [6, 1], [7, 0]]
I use the simpler approch
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] indices= new int[2];
int len=nums.length;
for(int i=0; i<len; i++){
for(int j=i+1; j<len;j++){
if((nums[i]+nums[j])==target){
indices[0] = i;
indices[1] = j;
return indices;
}
}
}
return null;
}
}
Complete code for beginners
//Predefined values
import java.util.Arrays;
public class TwoSum {
public static void main(String args[]) {
Solution solution = new Solution();
int target = 9;
int num[] = {2, 7, 11, 15};
num = solution.twoSum(num, target);
System.out.println(Arrays.toString(num));
}
}
class Solution {
public int[] twoSum(int[] nums, int target) {
int result[] = new int[2];
int sum;
for (int i = 0; i + 1 < nums.length; i++) {
// adding the alternate numbers
sum = nums[i] + nums[i + 1];
if (sum == target) {
result[0] = i;
result[1] = i + 1;
return result;
}
}
return null;
}
}
************************************************************************
//User-defined
import java.util.Arrays;
import java.util.Scanner;
/**
*
* #author shelc
*/
public class TwoSum {
public static void main(String[] args) {
Solution solution = new Solution();
Scanner scanner = new Scanner(System.in);
int n;
System.out.print("Enter the number of elements : ");
n = scanner.nextInt();
int array[] = new int[10];
System.out.println("Enter the array elemets : ");
for (int i = 0; i < n; i++) {
array[i] = scanner.nextInt();
}
int target;
System.out.println("Enter the target : ");
target = scanner.nextInt();
int nums[] = solution.twoSum(array, target);
System.out.println(Arrays.toString(nums));
}
}
class Solution {
public int[] twoSum(int nums[], int target) {
int result[] = new int[2];
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if ((nums[i] + nums[j]) == target) {
result[0] = i;
result[1] = j;
return result;
}
}
}
return null;
}
}
public static int[] twoSum(int[] nums, int target) {
int[] arr = new int[2];
int arrIndex =0;
for(int i =0; i<nums.length; i++){
for(int j = 0;j<nums.length-1;j++){
if( i!=j) {// don't check to itself
if (nums[i] + nums[j] == target) {
arr[arrIndex] = i;
arr[arrIndex + 1] = j;
break;
}
}
}
}
Arrays.sort(arr);
return arr;
}
I am trying to reverse an int array in Java.
This method does not reverse the array.
for(int i = 0; i < validData.length; i++)
{
int temp = validData[i];
validData[i] = validData[validData.length - i - 1];
validData[validData.length - i - 1] = temp;
}
What is wrong with it?
To reverse an int array, you swap items up until you reach the midpoint, like this:
for(int i = 0; i < validData.length / 2; i++)
{
int temp = validData[i];
validData[i] = validData[validData.length - i - 1];
validData[validData.length - i - 1] = temp;
}
The way you are doing it, you swap each element twice, so the result is the same as the initial list.
With Commons.Lang, you could simply use
ArrayUtils.reverse(int[] array)
Most of the time, it's quicker and more bug-safe to stick with easily available libraries already unit-tested and user-tested when they take care of your problem.
Collections.reverse(Arrays.asList(yourArray));
java.util.Collections.reverse() can reverse java.util.Lists and java.util.Arrays.asList() returns a list that wraps the the specific array you pass to it, therefore yourArray is reversed after the invocation of Collections.reverse().
The cost is just the creation of one List-object and no additional libraries are required.
A similar solution has been presented in the answer of Tarik and their commentors, but I think this answer would be more concise and more easily parsable.
public class ArrayHandle {
public static Object[] reverse(Object[] arr) {
List<Object> list = Arrays.asList(arr);
Collections.reverse(list);
return list.toArray();
}
}
I think it's a little bit easier to follow the logic of the algorithm if you declare explicit variables to keep track of the indices that you're swapping at each iteration of the loop.
public static void reverse(int[] data) {
for (int left = 0, right = data.length - 1; left < right; left++, right--) {
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
}
}
I also think it's more readable to do this in a while loop.
public static void reverse(int[] data) {
int left = 0;
int right = data.length - 1;
while( left < right ) {
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
// move the left and right index pointers in toward the center
left++;
right--;
}
}
Use a stream to reverse
There are already a lot of answers here, mostly focused on modifying the array in-place. But for the sake of completeness, here is another approach using Java streams to preserve the original array and create a new reversed array:
int[] a = {8, 6, 7, 5, 3, 0, 9};
int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length-i]).toArray();
Guava
Using the Google Guava library:
Collections.reverse(Ints.asList(array));
In case of Java 8 we can also use IntStream to reverse the array of integers as:
int[] sample = new int[]{1,2,3,4,5};
int size = sample.length;
int[] reverseSample = IntStream.range(0,size).map(i -> sample[size-i-1])
.toArray(); //Output: [5, 4, 3, 2, 1]
for(int i=validData.length-1; i>=0; i--){
System.out.println(validData[i]);
}
Simple for loop!
for (int start = 0, end = array.length - 1; start <= end; start++, end--) {
int aux = array[start];
array[start]=array[end];
array[end]=aux;
}
If working with data that is more primitive (i.e. char, byte, int, etc) then you can do some fun XOR operations.
public static void reverseArray4(int[] array) {
int len = array.length;
for (int i = 0; i < len/2; i++) {
array[i] = array[i] ^ array[len - i - 1];
array[len - i - 1] = array[i] ^ array[len - i - 1];
array[i] = array[i] ^ array[len - i - 1];
}
}
This will help you
int a[] = {1,2,3,4,5};
for (int k = 0; k < a.length/2; k++) {
int temp = a[k];
a[k] = a[a.length-(1+k)];
a[a.length-(1+k)] = temp;
}
This is how I would personally solve it. The reason behind creating the parametrized method is to allow any array to be sorted... not just your integers.
I hope you glean something from it.
#Test
public void reverseTest(){
Integer[] ints = { 1, 2, 3, 4 };
Integer[] reversedInts = reverse(ints);
assert ints[0].equals(reversedInts[3]);
assert ints[1].equals(reversedInts[2]);
assert ints[2].equals(reversedInts[1]);
assert ints[3].equals(reversedInts[0]);
reverseInPlace(reversedInts);
assert ints[0].equals(reversedInts[0]);
}
#SuppressWarnings("unchecked")
private static <T> T[] reverse(T[] array) {
if (array == null) {
return (T[]) new ArrayList<T>().toArray();
}
List<T> copyOfArray = Arrays.asList(Arrays.copyOf(array, array.length));
Collections.reverse(copyOfArray);
return copyOfArray.toArray(array);
}
private static <T> T[] reverseInPlace(T[] array) {
if(array == null) {
// didn't want two unchecked suppressions
return reverse(array);
}
Collections.reverse(Arrays.asList(array));
return array;
}
Your program will work for only length = 0, 1.
You can try :
int i = 0, j = validData.length-1 ;
while(i < j)
{
swap(validData, i++, j--); // code for swap not shown, but easy enough
}
There are two ways to have a solution for the problem:
1. Reverse an array in space.
Step 1. Swap the elements at the start and the end index.
Step 2. Increment the start index decrement the end index.
Step 3. Iterate Step 1 and Step 2 till start index < end index
For this, the time complexity will be O(n) and the space complexity will be O(1)
Sample code for reversing an array in space is like:
public static int[] reverseAnArrayInSpace(int[] array) {
int startIndex = 0;
int endIndex = array.length - 1;
while(startIndex < endIndex) {
int temp = array[endIndex];
array[endIndex] = array[startIndex];
array[startIndex] = temp;
startIndex++;
endIndex--;
}
return array;
}
2. Reverse an array using an auxiliary array.
Step 1. Create a new array of size equal to the given array.
Step 2. Insert elements to the new array starting from the start index, from the
given array starting from end index.
For this, the time complexity will be O(n) and the space complexity will be O(n)
Sample code for reversing an array with auxiliary array is like:
public static int[] reverseAnArrayWithAuxiliaryArray(int[] array) {
int[] reversedArray = new int[array.length];
for(int index = 0; index < array.length; index++) {
reversedArray[index] = array[array.length - index -1];
}
return reversedArray;
}
Also, we can use the Collections API from Java to do this.
The Collections API internally uses the same reverse in space approach.
Sample code for using the Collections API is like:
public static Integer[] reverseAnArrayWithCollections(Integer[] array) {
List<Integer> arrayList = Arrays.asList(array);
Collections.reverse(arrayList);
return arrayList.toArray(array);
}
There are some great answers above, but this is how I did it:
public static int[] test(int[] arr) {
int[] output = arr.clone();
for (int i = arr.length - 1; i > -1; i--) {
output[i] = arr[arr.length - i - 1];
}
return output;
}
It is most efficient to simply iterate the array backwards.
I'm not sure if Aaron's solution does this vi this call Collections.reverse(list); Does anyone know?
public void getDSCSort(int[] data){
for (int left = 0, right = data.length - 1; left < right; left++, right--){
// swap the values at the left and right indices
int temp = data[left];
data[left] = data[right];
data[right] = temp;
}
}
Solution with o(n) time complexity and o(1) space complexity.
void reverse(int[] array) {
int start = 0;
int end = array.length - 1;
while (start < end) {
int temp = array[start];
array[start] = array[end];
array[end] = temp;
start++;
end--;
}
}
public void display(){
String x[]=new String [5];
for(int i = 4 ; i > = 0 ; i-- ){//runs backwards
//i is the nums running backwards therefore its printing from
//highest element to the lowest(ie the back of the array to the front) as i decrements
System.out.println(x[i]);
}
}
Wouldn't doing it this way be much more unlikely for mistakes?
int[] intArray = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int[] temp = new int[intArray.length];
for(int i = intArray.length - 1; i > -1; i --){
temp[intArray.length - i -1] = intArray[i];
}
intArray = temp;
Using the XOR solution to avoid the temp variable your code should look like
for(int i = 0; i < validData.length; i++){
validData[i] = validData[i] ^ validData[validData.length - i - 1];
validData[validData.length - i - 1] = validData[i] ^ validData[validData.length - i - 1];
validData[i] = validData[i] ^ validData[validData.length - i - 1];
}
See this link for a better explanation:
http://betterexplained.com/articles/swap-two-variables-using-xor/
2 ways to reverse an Array .
Using For loop and swap the elements till the mid point with time complexity of O(n/2).
private static void reverseArray() {
int[] array = new int[] { 1, 2, 3, 4, 5, 6 };
for (int i = 0; i < array.length / 2; i++) {
int temp = array[i];
int index = array.length - i - 1;
array[i] = array[index];
array[index] = temp;
}
System.out.println(Arrays.toString(array));
}
Using built in function (Collections.reverse())
private static void reverseArrayUsingBuiltInFun() {
int[] array = new int[] { 1, 2, 3, 4, 5, 6 };
Collections.reverse(Ints.asList(array));
System.out.println(Arrays.toString(array));
}
Output : [6, 5, 4, 3, 2, 1]
public static void main(String args[]) {
int [] arr = {10, 20, 30, 40, 50};
reverse(arr, arr.length);
}
private static void reverse(int[] arr, int length) {
for(int i=length;i>0;i--) {
System.out.println(arr[i-1]);
}
}
below is the complete program to run in your machine.
public class ReverseArray {
public static void main(String[] args) {
int arr[] = new int[] { 10,20,30,50,70 };
System.out.println("reversing an array:");
for(int i = 0; i < arr.length / 2; i++){
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
for (int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}
}
}
For programs on matrix using arrays this will be the good source.Go through the link.
private static int[] reverse(int[] array){
int[] reversedArray = new int[array.length];
for(int i = 0; i < array.length; i++){
reversedArray[i] = array[array.length - i - 1];
}
return reversedArray;
}
Here is a simple implementation, to reverse array of any type, plus full/partial support.
import java.util.logging.Logger;
public final class ArrayReverser {
private static final Logger LOGGER = Logger.getLogger(ArrayReverser.class.getName());
private ArrayReverser () {
}
public static <T> void reverse(T[] seed) {
reverse(seed, 0, seed.length);
}
public static <T> void reverse(T[] seed, int startIndexInclusive, int endIndexExclusive) {
if (seed == null || seed.length == 0) {
LOGGER.warning("Nothing to rotate");
}
int start = startIndexInclusive < 0 ? 0 : startIndexInclusive;
int end = Math.min(seed.length, endIndexExclusive) - 1;
while (start < end) {
swap(seed, start, end);
start++;
end--;
}
}
private static <T> void swap(T[] seed, int start, int end) {
T temp = seed[start];
seed[start] = seed[end];
seed[end] = temp;
}
}
Here is the corresponding Unit Test
import static org.hamcrest.CoreMatchers.is;
import static org.junit.Assert.assertThat;
import org.junit.Before;
import org.junit.Test;
public class ArrayReverserTest {
private Integer[] seed;
#Before
public void doBeforeEachTestCase() {
this.seed = new Integer[]{1,2,3,4,5,6,7,8};
}
#Test
public void wholeArrayReverse() {
ArrayReverser.<Integer>reverse(seed);
assertThat(seed[0], is(8));
}
#Test
public void partialArrayReverse() {
ArrayReverser.<Integer>reverse(seed, 1, 5);
assertThat(seed[1], is(5));
}
}
Here is what I've come up with:
// solution 1 - boiler plated
Integer[] original = {100, 200, 300, 400};
Integer[] reverse = new Integer[original.length];
int lastIdx = original.length -1;
int startIdx = 0;
for (int endIdx = lastIdx; endIdx >= 0; endIdx--, startIdx++)
reverse[startIdx] = original[endIdx];
System.out.printf("reverse form: %s", Arrays.toString(reverse));
// solution 2 - abstracted
// convert to list then use Collections static reverse()
List<Integer> l = Arrays.asList(original);
Collections.reverse(l);
System.out.printf("reverse form: %s", l);
static int[] reverseArray(int[] a) {
int ret[] = new int[a.length];
for(int i=0, j=a.length-1; i<a.length && j>=0; i++, j--)
ret[i] = a[j];
return ret;
}
public static int[] reverse(int[] array) {
int j = array.length-1;
// swap the values at the left and right indices //////
for(int i=0; i<=j; i++)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
j--;
}
return array;
}
public static void main(String []args){
int[] data = {1,2,3,4,5,6,7,8,9};
reverse(data);
}
I got an 100 random elements array, each element is in range of 0-10, and i need to count each integer how many times it was typed (e.g. 1,2,2,3,8,8,4...)
OUTPUT:
1 - 1
2 - 2
3 - 1
8 - 2
4 - 1
My code so far is:
import java.util.Random;
public class Asses1 {
public static void main(String[] args) {
getNumbers();
}
private static int randInt() {
int max = 10;
int min = 0;
Random rand = new Random();
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
public static int[] getNumbers() {
int number = 100;
int[] array = new int[number];
for (int i = 0; i < array.length; i++) {
System.out.println(randInt());
}
System.out.println(number+" random numbers were displayed");
return array;
}
}
Add this method, which will do the counting:
public static void count(int[] x) {
int[] c=new int[11];
for(int i=0; i<x.length; i++)
c[x[i]]++;
for(int i=0; i<c.length; i++)
System.out.println(i+" - "+c[i]);
}
and change the main into this so that you call the previous method:
public static void main(String[] args) {
count(getNumbers());
}
Also, change the for loop in getNumbers into this in order to fill array with the generated numbers, not just printing them:
for (int i = 0; i < array.length; i++) {
array[i] = randInt();
System.out.println(array[i]);
}
Here is how it can be done in java 8
// Retrieve the random generated numbers
int[] numbers = getNumbers();
// Create an array of counters of size 11 as your values go from 0 to 10
// which means 11 different possible values.
int[] counters = new int[11];
// Iterate over the generated numbers and for each number increment
// the counter that matches with the number
Arrays.stream(numbers).forEach(value -> counters[value]++);
// Print the content of my array of counters
System.out.println(Arrays.toString(counters));
Output:
[12, 11, 7, 6, 9, 12, 8, 8, 10, 9, 8]
NB: Your method getNumbers is not correct you should fix it as next:
public static int[] getNumbers() {
int number = 100;
int[] array = new int[number];
for (int i = 0; i < array.length; i++) {
array[i] = randInt();
}
System.out.println(number+" random numbers were displayed");
return array;
}
int[] array2 = new int[11];
for (int i = 0; i < array.length; i++){
array2[randInt()]++
}
for (int i = 0; i < array.length; i++)
System.out.println(String.valueOf(i) + " - " + String.valueOf(array2[i]));
What I have done is:
Create an helping array array2 for storing number of occurences of each number.
When generating numbers increment number of occurences in helping array.
Map<Integer,Integer> map=new HashMap<Integer,Integer>();
int temp;
for (int i = 0; i < array.length; i++) {
temp=randInt();
if(map.containsKey(temp)){
map.put(temp, map.get(temp)+1);
}else{
map.put(temp, 1);
}
}