CareerCup

int col,row, prevcol=0, prevrow=0;
for(int i=0; i<s.length(); i++)
{
	col  = ( s[i]-'a' ) % ROW_LEN;
	row = ( s[i]-'a' )   / ROW_LEN;

	while(col>prevcol) { putchar('r'); prevcol++; }
	while(col<prevcol) { putchar('l'); prevcol--; }

	while(row>prevrow) { putchar('d'); prevrow++; }
	while(row<prevrow) { putchar('u'); prevrow--; }
	
	putchar('!');
}

I don't know why you posted it multiple times :) but it's a good solution. I wasn't able to code it in allotted time although I had similar approach. Thanks for sharing your code.

This is a highschool question.

Scan the array to find next position (x,y).

Compare with previous one (xp, yp).

Keep going like that.

O(N^2 * stringlength).

public class OnScreenKeyboard {
	public static void main(String[] a) {
		OnScreenKeyboard onScreenKeyboard = new OnScreenKeyboard();
		onScreenKeyboard.findLocation("dacixdsadsrascv", 5);
	}

	private void findLocation(String string, int length) {

		int currentRow = 1, currentCol = 1, row = 0, col = 0, ascii = 0;
		char array[] = string.toCharArray();
		StringBuilder sb = new StringBuilder();

		for (char element : array) {
			ascii = ((int) element) - 96;
			row = (ascii / length) + 1;
			col = ascii % length;

			for (int i = Math.abs(row - currentRow); i > 0; i--) {
				sb.append((currentRow < row) ? "d" : "u");
			}

			for (int i = Math.abs(col - currentCol); i > 0; i--) {
				sb.append((currentCol < col) ? "r" : "l");

			}
			sb.append("!");
			currentRow = row;
			currentCol = col;
		}
		System.out.print(sb.toString());
	}
}

for(int i=0; i<s.length(); i++)
{
	col  = ( s[i]-'a' ) % ROW_LEN;
	row = ( s[i]-'a' )   / ROW_LEN;

	while(col>prevcol) { putchar('r'); col--; }
	while(col<prevcol) { putchar('l'); col++; }

	while(row>prevrow) { putchar('d'); col--; }
	while(row<prevrow) { putchar('u'); col++; }
	
	putchar('!');
}