How to print maximum number of A’s using given four keys

int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // Initialize result
    int max = 0;
 
    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    int b;
    for (b=N-3; b>=1; b--)
    {
            // If the breakpoint is at b'th keystroke
            int curr = (N-b-1)*findoptimal(b);
            if (curr > max)
                max = curr;
     }
     return max;
}

Explanation
For N = 7
Case1
B = n-3 = 4
AAAA Ctrl-A Ctrl-C Ctrl-V = 8;    n-b-1 = 7-4-1 = 2 * findoptimal(4) = 2*4 = 8

Case2
B = 3
AAA Ctrl-A Ctrl-C Ctrl-V Ctrl-V= 9;  n-b-1 = 7-3-1 = 3* findoptimal(3) = 3*3 = 9

Case3
B = 2
AA Ctrl-A Ctrl-C Ctrl-V Ctrl-V Ctrl-V =  8; n-b-1 = 7-2-1 = 4* findoptimal(2) = 4*2 = 8

Case4
B = 1
A Ctrl-A Ctrl-C Ctrl-V Ctrl-V Ctrl-V Ctrl-V = 5; n-b-1 = 7-1-1 = 5* findoptimal(1) = 5*1 = 5

For N = 10
B = n-3 = 7

n-b-1 = 10-7-1 = 2*findoptimal(7)

Minimum number of jumps to reach end

/*
 * We use "last" to keep track of the maximum distance that has been reached
 * by using the minimum steps "ret", whereas "curr" is the maximum distance
 * that can be reached by using "ret+1" steps. Thus,
 * curr = max(i+A[i]) where 0 <= i <= last.
 */
    
   int jump(int A[], int n) {
        int ret = 0;
        int last = 0;
        int curr = 0;
        for (int i = 0; i < n; ++i) {
            if (i > last) {
                last = curr;
                ++ret;
            }
            curr = max(curr, i+A[i]);
        }

        return ret;
    }

Check for Uppercase, Lowercase, Digits & Whitespace

for (char c : password.toCharArray()) {

if (Character.isUpperCase(password.charAt(i))){ upperCase++; }

if (Character.isLowerCase(password.charAt(i))){ lowerCase++; }

if (Character.isDigit(password.charAt(i)))   { numberCount++;}

if (Character.isWhitespace(password.charAt(i)))   { blankCount++;}

}

Counting integers in a string

String s = "abcd123";
    int counter = 0;
    for(char c : s.toCharArray()) 
    {
        if( c >= '0' && c<= '9') 
        {
            ++counter;
        }
    }

OR

String s = "abcd123";
int count = 0;
for (int i = 0, len = s.length(); i < len; i++) 
{
    if (Character.isDigit(s.charAt(i))) 
    {
        count++;
    }
}

Square of a number without using *, / ,+and pow()

If n is even, it can be written as

  n = 2*x 
  n2 = (2*x)2 = 4*x2

If n is odd, it can be written as

  n = 2*x + 1
  n2 = (2*x + 1)2 = 4*x2 + 4*x + 1

square(n) = 0 if n == 0
  if n is even 
     square(n) = 4*square(n/2) 
  if n is odd
     square(n) = 4*square(floor(n/2)) + 4*floor(n/2) + 1

Examples
square(6) = 4*square(3)
square(3) = 4*(square(1)) + 4*1 + 1 = 9
square(7) = 4*square(3) + 4*3 + 1 = 4*9 + 4*3 + 1 = 49

CODE

int square(int n)
{
    // Base case
    if (n==0) return 0;
 
    // Handle negative number
    if (n < 0) n = -n;
 
    // Get floor(n/2) using right shift
    int x = n>>1;
 
    // If n is odd
    if (n&1)
        return ((square(x)<<2) + (x<<2) + 1);
    else // If n is even
        return (square(x)<<2);

generate PDF in Java

public String generateLeaversReport(ArrayList<UserReportBean> leaversAList,String leaversFromDate,String leaversToDate,String filteredLocation) throws Exception
{
    Document document = null;
    File f= null;
    try
    {
        document = new Document(PageSize.A4,10,10,10,10);
        f = new File("Leavers.pdf");
        FileOutputStream fos = new FileOutputStream(f);
        PdfWriter.getInstance(document, fos );
        document.open();
        Table table=new Table(29);
        table.setBorder(0);
        table.setWidth(100f);
        table.setWidths(new float[]{0.5f,1.2f,1.0f,0.6f,1.25f,1.25f,1.5f,1.25f,1.20f,0.75f,1f,1.20f,0.75f,0.75f,0.75f,1.5f,0.8f,1.10f,0.8f,0.8f,0.8f,0.8f,0.9f,0.6f,0.6f,0.6f,0.6f,0.6f,0.6f});
        table.setCellsFitPage(true);
        table.setPadding(2);
        table.setSpacing(0);

        Cell hdrCell = new Cell(new Chunk(StringUtils.trimToEmpty("Leavers Report For the Period :  "+leaversFromDate+" To "+leaversToDate+" For "+filteredLocation),
                FontFactory.getFont(FontFactory.HELVETICA_BOLD, 8, Font.BOLD, new Color(0, 0, 0))));
     
        hdrCell.setColspan(29);
        hdrCell.setHeader(true);
        hdrCell.setBorder(Rectangle.NO_BORDER);
        hdrCell.setHorizontalAlignment(Cell.ALIGN_CENTER);
        table.addCell(hdrCell);
        
        table.addCell(getHeaderCell_New("S.No.",1));
        table.addCell(getHeaderCell_New("AssociateID",1));
        table.addCell(getHeaderCell_New("First Name",1));
        table.addCell(getHeaderCell_New("Middle Name",1));
        table.addCell(getHeaderCell_New("Last Name",1));
        
        for(int i=0;i<leaversAList.size();i++)
        {
            UserReportBean bean=leaversAList.get(i);
            table.addCell(getCell_New(String.valueOf(i+1),1));
            table.addCell(getCell_New(bean.getAssociateID(),1));
            table.addCell(getCell_New(bean.getFirstName(),1));
            table.addCell(getCell_New(bean.getMiddleName(),1));
            table.addCell(getCell_New(bean.getLastName(),1));
        }
        document.add(table);
    }
    catch(Exception ex)
    {
        ex.printStackTrace();
    }
    finally
    {
        if(document!=null)
        {
            document.close();
        }
    }
    return f!=null?f.getAbsolutePath():null;
}

Find word in dictionary

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,
Given:
start = “hit”
end = “cog”
dict = [“hot”,”dot”,”dog”,”lot”,”log”]
Answer

public int ladderLength(String start, String end, HashSet<String> dict) {
    if (dict.size() == 0)
        return 0;
 
    dict.add(end);
 
    LinkedList<String> wordQueue = new LinkedList<String>();
    LinkedList<Integer> distanceQueue = new LinkedList<Integer>();
 
    wordQueue.add(start);
    distanceQueue.add(1);
 
    //track the shortest path
    int result = Integer.MAX_VALUE;
    while (!wordQueue.isEmpty()) {
        String currWord = wordQueue.pop();
        Integer currDistance = distanceQueue.pop();
 
        if (currWord.equals(end)) {
            result = Math.min(result, currDistance);
        }
 
        for (int i = 0; i < currWord.length(); i++) {
            char[] currCharArr = currWord.toCharArray();
            for (char c = 'a'; c <= 'z'; c++) {
                currCharArr[i] = c;
 
                String newWord = new String(currCharArr);
                if (dict.contains(newWord)) {
                    wordQueue.add(newWord);
                    distanceQueue.add(currDistance + 1);
                    dict.remove(newWord);
                }
            }
        }
    }
 
    if (result < Integer.MAX_VALUE)
        return result;
    else
        return 0;
}

Removing element from hashmap

Iterator<Map.Entry<String,String>> iter = TestMap.entrySet().iterator();
while (iter.hasNext()) {
    Map.Entry<String,String> entry = iter.next();
    if("Sample".equalsIgnoreCase(entry.getValue())){
        iter.remove();
    }
}

FINDING NTH HIGHEST NUMBER

EmpID   Name    Salary
1   A   100
2   B   800
3   C   300
4   D   400
5   E   500
6   F   200
7   G   600

SELECT * FROM Employee E1
WHERE (N-1) = (
                SELECT COUNT(DISTINCT(E2.Salary))
                FROM Employee E2
                WHERE E2.Salary > E1.Salary
              )

Explanation

Suppose you want to find 5th highest salary, which means there are total 4 employees who have salary greater than 5th highest employee. So for each row from the outer query check the total number of salaries which are greater than current salary. Outer query will work for 100 first and check for number of salaries greater than 100. It will be 6, do not match (5-1) = 6 where clause of outerquery. Then for 800, and check for number of salaries greater than 800, 4=0 false then work for 300 and finally there are totally 4 records in the table which are greater than 300. Therefore 4=4 will meet the where clause and will return 3 C 300.

Another way

SELECT *
FROM(
SELECT BONUSMONTHID, dense_rank() over (ORDER BY BONUSMONTHID DESC) AS DenseRank
FROM REFERRALBONUS )
where DenseRank = 2

8 PEOPLE PUZZLE

There are 8 people, who need to cross a river in  single boat. People are 1 Cop, 1 Thief, 1 Man and his 2 Daughters, 1 Women and her 2 sons. As usual the conditions that has to be satisfied while crossing the River are given below:
•    When Police man is not on that river bank, Thief will beat all the people available on that river bank.
•    When Women is Not on that river bank, Man will beat her 2 sons.
•    When Man is Not on that river bank, Woman will beat his 2 daughters.
•    Only Two people can travel in the boat at any given time.
•    2 Sons and 2 daughters doesn’t know to drive the boat.
ANSWER
Please follow Step by Step process as defined below:
Initial stage Cop, Thief, Men, Daughters, Women, Sons are at River Bank – 1
1.    Pick up Cop-Thief pair and travel to River bank – 2
2.    Drop the Thief at River bank – 2 and cop only will reach back River bank – 1
3.    Cop will Pick up Daughter – 1 at river bank -1 and travels to River Bank -2
4.    Cop will drop the daughter – 1 at river bank – 2 and picks thief along with him and reaches back to River bank – 1
5.    Both Cop and thief pair will drop at River bank -1
6.    Now Man along with his Daughter – 2 travels from River bank -1 to River Bank -2
7.    Man will drop his Daughter -2 at river bank – 2 and reaches back River Bank – 1 single
8.    Now Man will pick up Women at River bank -1 and reaches to River bank – 2
9.    Man will get down at river bank-2 and Women alone will return back to River bank – 1
10.    Women will get down at River bank -1 . Cop and Thief pair will travel from River bank – 1 to River bank – 2
11.    Cop and Thief will get down at River bank – 2. Man alone will travel back to River bank -1
12.    Man picks up woman at River bank -1 and reaches back to River bank -2
13.    Man will get down at River bank – 2 and women alone will sail back to River bank -1
14.    Women will pick up her Son – 1 at River bank – 1 and travels back to River bank – 2
15.    Women along with her son – 1 will get down at River bank -2.
16.    Cop and Thief  pair will travel from River bank – 2 back to River bank – 1
17.    Cop will drop Thief at River bank -1 and picks Son -2 and reaches River bank -2
18.    Cop will drop son-2 at river bank -2 and sails back to River bank -1 to pick up thief
19.    Both Cop an Thief pair will reach back River bank – 2 safely.