M-ZubairAhmed

while True:
    print("This program splits the seconds you enter into Days:Hours:Minutes:Seconds")
    n = int(input("enter seconds"))
    days = n // (24*60*60)
    days_rem = n %(24*60*60)
    hours = days_rem // (60*60)
    hours_rem = days_rem % (60*60)
    minutes = hours_rem//60
    seconds = hours_rem%60
    

M-ZubairAhmed

n = int(input("enter triangle number"))
    ini = 10
    for x in range(1,n+1):
        if (x==1):
            print (1)
        else:
            ini = ini * 10
            print (ini)

M-ZubairAhmed

import random
while True:
    if (random.randint(1,2) == 1):
        print("Right")
        #print(random_number)
        input("hit enter for next")
    if (random.randint(1,2) == 2):
        print("Left")
        #print(random_number)
        input("hit enter for next")        

M-ZubairAhmed

public static int binaryToDecimal (int binaryNumber){
  int digitIndex;
  int decimalNum = 0;
  int powerIndex = 0;
  while(( binaryNumber != 0) && ( binaryNumber >0)){
      digitIndex = binaryNumber % 10;
      decimalNum = decimalNum + (int)(Math.pow(2,powerIndex) * digitIndex);
      binaryNumber = binaryNumber / 10;
      powerIndex++;
    }

M-ZubairAhmed

int decimalNum;

//Converts decimal number to binary, where regex is 2
Integer.toString(decimalNum,2);

//Converts decimal number to hexadecimal, where regex is 16
Integer.toString(decimalNum,16);

M-ZubairAhmed

public class Main {
    public static void main(String[] args) {

        String originalString = "azzbyyyx";
        StringBuilder str = new StringBuilder(originalString);

        int loopCounter = 0;
        int repetitionCounter = 0;
        while (true) {
            if (loopCounter < str.length() - 1) {

M-ZubairAhmed

public static int convertBits(int initialNumber, int newNumber) {
  return Integer.bitCount(initialNumber ^ newNumber);    
  }
}

M-ZubairAhmed

            String[] common = {"the","of","and","a","to","in","is","you","that","it","he","was","for","on","are","as","with","his","they","I","at","be","this","have","from","or","one","had","by","word","but","not","what","all","were","we","when","your","can","said","there","use","an","each","which","she","do","how","their","if","will","up","other","about","out","many","then","them","these","so","some","her","would","make","like","him","into","time","has","look","two","more","write","go","see","number","no","way","could","people","my","than","first","water","been","call","who","oil","its","now","find","long","down","day","did","get","come","made","may","part"};

M-ZubairAhmed

//bNum = Initializing big interger to value.
BigInteger bNum = BigInteger.ONE;

// n = The input number whose factorial is to be calculated.
for (long i = 1; i <= n; i++) {
	bNum = bNum.multiply(bNum.valueOf(i));
}

//Storing the value of factorial in a string.
String factorialValue = String.valueOf(bNum);

M-ZubairAhmed

/*  When the given problem is to find number number of trailing zeros of the factorial, one is not required to 
find out the factorial.
Statement : factorial(5) = 120 	= 1*2*3*4*5
factorial(4) = 24	= 1*2*3*4
if it is observed the occurance of 5 or multiples of five is in direct proportion to number of trailing zeros
If however, expression contains powers of 5, then the trailing zeros increasing with power of 5.
factorial(25) = 1*...5*..10..*15*..20*..25
As expected number of trailing zero in above is expected to be 5, but since 25 is 5 power 2, it adds another zero 
hence final count is 6.
*/