In this post, we will see algorithm problems with their solutions which are asked during Java interviews.

*How to check if a number is Palindrome?*

void checkPalindrome(int n){
int temp, sum = 0;
int input=n;
while(n>0) {
temp = n%10;
sum = (sum*10) + temp;
n = n/10;
}
if(input == sum){
System.out.println("Palindrome");
} else {
System.out.println("Not Palindrome");
}
}

*How to check if a number is Prime in Java8?*

void checkPrime(int n) {
if(n > 1 && IntStream.range(2, n).noneMatch(i -> i%n==0)) {
System.out.println("Prime");
} else {
System.out.println("Non-Prime");
}
}

*How to sort objects in reverse order in Java8?*

Student student1 = new Student(372,"Venkat",1);
Student student2 = new Student(2,"Sachin",4);
Student student3 = new Student(2345,"Ganguly",6);
Student student4 = new Student(72,"Karthik",2);
List studlist = new CopyOnWriteArrayList();
studlist.add(student1);
studlist.add(student2);
studlist.add(student3);
studlist.add(student4);
// Iterate in Java8
studlist.forEach(s -> System.out.println(s.name));
// Sort by Ids
studlist.sort((Student s1,Student s2) -> s1.getId() - s2.getId());
// Sort by Rank in reverse Order
studlist.sort((Student s1,Student s2) -> s2.getRank() - s1.getRank());

*Find second highest number in an Array?*

int arr[] = {45,89, 29,1, 9, 100};
int highest = 0, secondHighest = 0;
for(int i=0; i<arr.length;i++) { if(arr[i] > highest) {
highest = arr[i];
} else if(arr[i] > secondHighest) {
secondHighest = arr[i];
}
}

*Find Nth highest Salary from a SQL Table?*

SELECT MIN(SALARY) FROM EMPLOYEE
WHERE SALARY IN (SELECT DISTINCT TOP N
FROM EMPLOYEE ORDER BY SALARY desc);

*Print Only Numerics from a String?*

String sampleStr = "fdsha3430d3kdjafl0737434833";
String numericsOnlyStr = sampleStr.replaceAll("[^0-9]", "");

*Print Duplicates in an Array?*

for(int i=0;i<arr.length;i++) {
for(int j=i+1; j< arr.length; j++) {
if(arr[i] == arr[j]) {
System.out.println(arr[j]);
}
}
}

*Fetch Frequency of Elements repeated in an Array?*

Map<Integer, Integer> mp = new HashMap<>();
// Iterating through array elements
for (int i = 0; i < n; i++)
{
if (mp.containsKey(arr[i])) {
mp.put(arr[i], mp.get(arr[i]) + 1);
} else {
mp.put(arr[i], 1);
}
}
// Iterating through Map and Printing frequencies
for (Map.Entry<Integer, Integer> entry : mp.entrySet()) {
System.out.println(entry.getKey() + " " + entry.getValue());
}

*Find Triplets in an array whose sum is equal to n?*

public class Triplets {
public static List<List> findTriplets(int[] numbers, int sum) {
List<List> tripletsCombo = new ArrayList<List>();
HashSet set = new HashSet();
List triplets = new ArrayList();
if (numbers.length == 0 || sum <= 0) {
return tripletsCombo;
}
Arrays.sort(numbers);
for (int i = 0; i < numbers.length - 2; i++) {
int j = i + 1;
int k = numbers.length - 1;
while (j < k) {
if (numbers[i] + numbers[j] + numbers[k] == sum) {
String str = numbers[i] + "," + numbers[j] + "," + numbers[k];
// Check for the unique Triplet
if (!set.contains(str)) {
triplets.add(numbers[i]);
triplets.add(numbers[j]);
triplets.add(numbers[k]);
tripletsCombo.add(triplets);
triplets = new ArrayList();
set.add(str);
}
j++;
k--;
} else if (numbers[i] + numbers[j] + numbers[k] < sum) { j++; } else { // numbers[i] + numbers[j] + numbers[k] > sum
k--;
}
}
}
return tripletsCombo;
}
public static void main(String[] args) {
int[] numbers = { 2, 3, 1, 5, 4 };
int sum = 9;
List<List> triplets = findTriplets(numbers, sum);
if (triplets.isEmpty()) {
System.out.println("No triplets are found");
} else {
System.out.println(triplets);
}
}
}

*How to check if two strings are Anagrams?*

Two strings are called Anagrams if they contain same set of characters but in different order. Examples: **“Astronomer – Moon starer”, “A gentleman – Elegant man”, “Dormitory – Dirty Room”, “keep – peek”.**

void isAnagram(String input1, String input2) {
//Removing all white spaces from s1 and s2
String s1_nonSpaces = input1.replaceAll("\\s", "");
String s2_nonSpaces = input2.replaceAll("\\s", "");
boolean status = true;
if(s1_nonSpaces.length() != s2_nonSpaces.length()) {
status = false;
} else {
char[] s1Array = s1_nonSpaces.toLowerCase().toCharArray();
char[] s2Array = s2_nonSpaces.toLowerCase().toCharArray();
Arrays.sort(s1Array);
Arrays.sort(s2Array);
status = Arrays.equals(s1Array, s2Array);
}
System.out.print(status?"Anagrams":"Non-Anagrams");
}

*Swap numbers without using temp/third variable?*

void swapWithoutTemp(int a, int b) {
a = a+b;
b = a-b;
a = a-b;
}

*Find number of combinations for Sum of Two Elements from two arrays is equal to N?*

We have two arrays of numbers, suppose we take one element from first array and another element from second array. Their sum should be equal to N(given number).

sumOfTwoElementsInTwoArrays() {
int arr1[] = {4,8,10,12,7};
int arr2[] = {6,90,34,45};
int sumValue = 44;
HashSet complements = new HashSet();
int pairCount = 0;
for(int i=0;i<arr1.length;i++) {
complements.add(arr1[i] - sumValue);
}
for(int j=0;j<arr1.length;j++) {
if(complements.contains(arr2[j])) {
pairCount++;
}
}
System.out.print("Number of pairs is "+pairCount);
}

*First non repeated character in a String?*

String str = "BANANA";
char firsNonRepeatedCharacter;
HashMap<Character, Integer> hmp = new HashMap<Character, Integer>();
for(int z=0;z<s.length();z++) {
if(hmp.containsKey(str.charAt(z))) {
hmp.put(str.charAt(z), hmp.get(str.charAt(z))+1);
} else {
hmp.put(str.charAt(z), 1);
}
}
Set characterSet = hmp.keySet();
for(Character c:characterSet){
if(hmp.get(c).toString()equals("1")) {
firsNonRepeatedCharacter = c;
break;
}
}

*Find the number of occurrence of an element in an array using Java8?*

int b[] = {1,2,34,1};
List bList = Arrays.stream(b).boxed().collect(Collectors.toList());
System.out.println(bList.stream().filter(z -> z.toString().equalsIgnoreCase("1")).count());

*100 doors toggle open/close*

There are 100 doors in a row, all doors are initially closed. A person walks through all doors multiple times and toggle (if open then close, if close then open) them in following way:

In first walk, the person toggles every door, In second walk, the person toggles every second door, i.e., 2nd, 4th, 6th, 8th, …, In third walk, the person toggles every third door, i.e. 3rd, 6th, 9th, …

Find in nth walk, what will be the status of all doors

doorsOpenClosed(int no_of_walks) {
int door_id, walk_id;
int doors[] = new int[101];
for(int i=0;i<100;i++) {
doors[i] = 0;
}
for (walk_id = 1; walk_id <= no_of_walks; walk_id++) {
for (door_id = walk_id; door_id <= 100; door_id += walk_id) {
if(door_id%walk_id == 0) {
doors[door_id]=(doors[door_id] == 0)?1:0;
}
}
}
for (int j = 0; j <= 100; j++) {
if(doors[j] == 1) {
System.out.println("Open Door number::::"+j);
}
}
}