Dear Readers, these are the commonly asked java programs to check your ability on writing the logic.
Please find the list of commonly asked programs.
- Print if the strings are Anagram or not?
An anagram is a word or phrase that’s formed by rearranging the letters of another word or phrase.
import java.util.Arrays;
public class Anagram {
public static void main(String[] args) {
String str1 = "own";
String str2 = "now";
char[] chArray1 = str1.toCharArray();
char[] chArray2 = str2.toCharArray();
Arrays.sort(chArray1);
Arrays.sort(chArray2);
System.out.println(Arrays.equals(chArray1, chArray2));
}
}
output:
true2. Array Rotation
Rotate the array clockwise and anticlockwise by no of positions.
package com.commonlyasked;
import java.util.Arrays;
public class ArrayRotation {
public static void main(String[] args) {
int[] arr = {1, 2, 3, 4, 5, 6, 7};
int num = 2;
leftRotation(arr,num);
rightRotation(arr, num);
}
private static void leftRotation(int[] arr, int num) {
for (int i = 0; i < num; i++) {
int temp = arr[0];
for (int j = 0; j < arr.length - 1; j++) {
arr[j] = arr[j + 1];
}
arr[arr.length - 1] = temp;
}
System.out.println("Input Array After Left Rotation By " + num +
" Positions :");
System.out.println(Arrays.toString(arr));
}
private static void rightRotation(int[] arr, int num) {
for (int i = 0; i < num; i++) {
int temp = arr[arr.length - 1];
for (int j = arr.length - 1; j > 0; j--) {
arr[j] = arr[j - 1];
}
arr[0] = temp;
}
System.out.println("Input Array After Right Rotation By " + num +
" Positions :");
System.out.println(Arrays.toString(arr));
}
}
output:
Input Array After Left Rotation By 2 Positions :
[3, 4, 5, 6, 7, 1, 2]
Input Array After Right Rotation By 2 Positions :
[1, 2, 3, 4, 5, 6, 7]3. Count the no of 1’s without using any inbuilt function.
package com.commonlyasked;
public class CountOfos1sWithoutInBuilt {
public static void main(String[] args) {
int[] arr = new int[]{1, 1, 0, 1, 0, 1, 1, 0};
int sum = 0;
for (int i : arr) {
sum += i;
}
System.out.println("count of 1's::" + sum);
System.out.println("count of 0's::" + (arr.length - sum));
}
}
output:
count of 1's::5
count of 0's::34. Count the no of ways to reach the nth stair.
There are n stairs, a person standing at the bottom wants to climb stairs to reach the nth stair. The person can climb either 1 stair or 2 stairs at a time, the task is to count the number of ways that a person can reach at the top.
package com.commonlyasked;
/*
In the above approach it can be seen that dp[i] only depends on previous states.
We can optimize the space complexity of the dynamic programming solution to O(1) by using only two variables prev1 and prev2
to keep track of the previous two values of dp[i-1] and dp[i-2]. Since we only need these two values to calculate the next value,
we don’t need to store the entire array.
Time Complexity: O(N)
Auxiliary Space: O(1)
*/
public class CountWayToReachNStair {
static int countWays(int n) {
// declaring two variables to store the count
int prev = 1;
int prev2 = 1;
// Running for loop to count all possible ways
for (int i = 2; i <= n; i++) {
int curr = prev + prev2;
prev2 = prev;
prev = curr;
}
return prev;
}
public static void main(String[] args) {
int n = 5;
System.out.println("Number of Ways : "
+ countWays(n));
}
}
output:
Number of Ways : 85. Factorial of a number recursively/ iterative way
The factorial of a whole number is the function that multiplies the number by every natural number below it.
n! or “n factorial” means:
- n! = 1 · 2 · 3 · … · n = Product of the first n positive integers = n(n-1)(n-2)…………………….(3)(2)(1)
Example: 5 factorial, that is, 5! can be written as: 5! = 5 × 4 × 3 × 2 × 1 = 120.
package com.commonlyasked;
public class Factorial {
// Recursion way
static int fact(int num) {
if (num < 0) return -1;
if (num == 0) return 1;
else
return (num * fact(num - 1));
}
public static void main(String[] args) {
//Iterative way
int fact = 1;
int num = 5;
if (num < 0) System.out.println("Factorial cant be calculated for
negative no");
if (num == 0) System.out.println("Factorial of 0 is::" + 1);
for (int i = 1; i <= num; i++) {
fact = fact * i;
}
System.out.println("Factorial of " + num + " using iterative way "
+ fact);
System.out.println("Factorial of " + num + " using recursive way "
+ fact(num));
}
}
output :
Factorial of 5 using iterative way 120
Factorial of 5 using recursive way 120 6. Fibonacci of a number
The Fibonacci sequence is the series of numbers where each number is the sum of the two preceding numbers
Using this numbering, the Fibonacci sequence can be defined by the following three equations:
- F0 = 0 (applies only to the first integer)
- F1 = 1 (applies only to the second integer)
- Fn = Fn-1 + Fn-2 (applies to all other integers)
package com.commonlyasked;
public class Fibonacci {
//recursion
public static int fibonaci(int num) {
if (num < 0) return 0;
if (num == 1 || num == 2) return 1;
else
return fibonaci(num - 2) + fibonaci(num - 1);
}
public static void main(String[] args) {
//iterative way
int num = 5;
int n1;
int n2 = 0, n3 = 1;
System.out.println("Fibonaci series using iteration::");
for (int i = 0; i < num; i++) {
n1 = n2;
n2 = n3;
n3 = n1 + n2;
System.out.print(n1 + " ");
}
System.out.println("\nFibonaci series using recursion::");
for (int i = 0; i < num; i++) {
System.out.print(fibonaci(i) + " ");
}
}
}
ouput:
Fibonaci series using iteration::
0 1 1 2 3
Fibonaci series using recursion::
0 1 1 2 3 7. Palindrome
A palindrome is a word, number, phrase, or other sequence of symbols that reads the same backwards as forwards.
For example Madam , Mom etc.
package com.commonlyasked;
public class Palindrome {
public static void main(String[] args) {
String original = "mom";
String reverse = "";
for (int i = original.length() - 1; i >= 0; i--) {
reverse += original.charAt(i);
}
System.out.println("palindrome or not::" + reverse.equals(original));
}
}
output:
palindrome or not::true8. Prime no generation between 2 numbers.
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In other words, a prime number is a number that is divisible only by 1 and itself, and has no other factors.
For example, 2, 3, 5, 7, 11, 13, 17, 19, 23, and so on, are prime numbers because they cannot be divided evenly by any other numbers except 1 and themselves.
package com.commonlyasked;
public class PrimeNo {
public static void main(String[] args) {
int first = 10;
int second = 20;
for (int i = first; i < second; i++) {
boolean value = isPrime(i);
if (value) System.out.println("Prime no::" + i);
}
}
private static boolean isPrime(int num) {
if (num <= 1) return false;
for (int i = 2; i < num; i++) {
if (num % i == 0) return false;
}
return true;
}
}
output :
Prime no::11
Prime no::13
Prime no::17
Prime no::199. Reverse a number
package com.commonlyasked;
public class ReverseANumber {
public static void main(String[] args) {
int num=123;
int reverse = 0;
System.out.println("Before Reversal of a number::"+ num);
while (num !=0) {
int remainder = num % 10;
reverse = reverse * 10 + remainder;
num = num / 10;
}
System.out.println("After Reversal of a number::"+ reverse);
}
}
Output:
Before Reversal of a number::123
After Reversal of a number::32110. Find Second Highest number in an unsorted array
package com.commonlyasked;
public class SecondHighest {
public static void main(String[] args) {
int[] arr = new int[]{4, 5, 2, 1, 9, 7};
int highest = Integer.MIN_VALUE;
int secondHighest = Integer.MIN_VALUE;
for (int i = 0; i <= arr.length - 1; i++) {
if (arr[i] > highest) {
secondHighest = highest;
highest = arr[i];
} else if (arr[i] > secondHighest) {
secondHighest = arr[i];
}
}
System.out.println("Highest:: " + highest + "\nsecond highest::" +
secondHighest);
}
}
output:
Highest:: 9
second highest::711. Reverse a string ( word or a sentence) iterative/recursive way.
package com.commonlyasked;
public class StringReverse {
public static void main(String[] args) {
//Reverse a word
String word = "Hello";
for (int i = word.toCharArray().length - 1; i >= 0; i--) {
System.out.print(word.toCharArray()[i]);
}
//reverse a sentence
String str = "Hello today is expiry";
String[] arr = str.split(" ");
String result = "";
for (int i = arr.length - 1; i >= 0; i--) {
result += arr[i] + " ";
}
System.out.println("\n" + result);
System.out.println("\nUsing recursion");
reverse(word);
}
// recursion to reverse a sentence
static void reverse(String str) {
if ((str == null) || (str.length() <= 1))
System.out.println(str);
else {
System.out.print(str.charAt(str.length() - 1));
reverse(str.substring(0, str.length() - 1));
}
}
}
output :
olleH
expiry is today Hello
Using recursion
olleH12 . Find all pairs equal to a given sum
package com.commonlyasked;
import java.util.HashMap;
import java.util.Map;
public class FindPairsOfGivenSum {
public static void main(String[] args) {
int[] nums = new int[]{15, 12, 4, 16, 9, 8, 24, 0};
int sum = 24;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length - 1; i++) {
if (map.containsKey(sum - nums[i])) {
System.out.println("pair found::" + (sum - nums[i]) + " " + nums[i]);
}
map.put(nums[i], i);
}
}
}
output:
pair found::15 9
pair found::16 813. Swap 2 numbers without using any inbuilt function
package com.commonlyasked;
public class SwapNumbers {
public static void main(String[] a) {
int x = 10;
int y = 5;
System.out.println("Before swapping:"
+ " x = " + x + ", y = " + y);
x = x + y;
y = x - y;
x = x - y;
System.out.println("After swapping:"
+ " x = " + x + ", y = " + y);
}
}
output:
Before swapping: x = 10, y = 5
After swapping: x = 5, y = 10- Thank you for reading this article. Please provide your valuable suggestions/ feedback.
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