Palantir Coding Challenge Question

PalantirCodingChallenge.java

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

/**
 * Credit for Algorithm from top answer in: 
 * http://stackoverflow.com/questions/7116438/algorithms-question-flipping-columns
 * 
 * This problem is given to interview candidates in the screening process
 * for Palantir Technologies.  I've seen it through several different 
 * candidates.  
 * 
 * Given an n*m Matrix of 'P' and 'T' characters, find the maximum number
 * of completely 'P' rows, after k column flips.
 * 
 * A column flip consists of taking one column in the matrix, and inverting
 * it's values (i.e. all 'P' to 'T' and all 'T' to 'P')
 * 
 * Here is an example matrix:
 * PPT
 * TPT
 * TTT
 * 
 * if k = 1, the optimal solution is:
 * PPP
 * TPP
 * TTP
 * and it should output 1. 
 * 
 * The number of all 'P' rows is 1, once the algorithm completes.
 * 
 * The input for the above test case would be:
 * 3 3
 * PPT
 * TPT
 * TTT
 * 1
 * 
 * The output would be:
 * 1
 * 
 * @author Joseph Liccini
 *
 */
public class PalantirCodingChallenge {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		char[][] matrix = new char[n][m];
		
		for (int i = 0; i < n; ++i) 
			matrix[i] = sc.next().toCharArray();
		
		int k = sc.nextInt();
		
		Map<String, Integer> map = new HashMap<String, Integer>();
		for (int i = 0; i < n; ++i) {
			char[] row = matrix[i];
			int count = 0;

			for (char c: row) 
				if (c == 'T') ++count;

			if (count == k || (k - count >= 0 && (k - count) % 2 != 0)) {
				String key = new String(row);
				if (map.containsKey(key))
					map.put(key, map.get(key)+1);
				else
					map.put(key, 1);
			}
		}
	
		int max = 0;
		for (int value: map.values())
			max = Math.max(value, max);

		System.out.println(max);

		sc.close();
	}

}

I think;
(k - count) % 2 != 0) should be
(k - count) % 2 == 0)