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);