CareerCup

1)start from top right corner
2) until curr_elem==elem_to_search || you go out of matrix, go down if curr_elem<elem_to_search else go left

complexity= O(m+n)

Do u mean that rows and columns are sorted in themselves independently ? ie.
3 6 7 12
7 9 10 13
9 12 13 14

number of solutions are possible :
1) if min(m,n)==1 use binary search.

2) if min(m,n)<<small linear search in zig-zag manner starting at top right corner. O(max(m,n))

3) if both m and n large use 2D binary-search O(log(n)*log(m))

public static void findInArray(int[][] inpArr, int num) {
		int count = 1;
		boolean flag = true;
		int i = 0;
		int j = (inpArr[0].length)-1;
		int currentNum = inpArr[i][j];
		while(num!=currentNum) {
			count++;
			if(num>currentNum) {
				i++;
			} else {
				j--;
			}
			try {
			currentNum = inpArr[i][j];
			} catch(ArrayIndexOutOfBoundsException e) {
				System.out.println(num + " not found in input array");
				flag = false;
				break;
			}
		}
		if(flag) System.out.println(num + "fount at (" + i +","+ j + ") position in " + count + " comparision(s).");
	}