In this problem, we have several pairs of words. We want to count the maximum number of times a letter will appear when we write out a word from each pair, no matter which word we pick from each pair.

We can start to tackle this problem by handling the case where there is exactly one pair of words to consider. When there is exactly one pair of words to consider, the maximum number of times the letter 'a' will appear is the maximum of the number of times the letter 'a' appears in the first word and the number of times the letter 'a' appears in the second word. We can replace 'a' with any letter and that will get us the maximum number of times any given letter will appear.

Now that we've handled the case where there is exactly one pair of words to consider, we need to figure out how to extend this to multiple pairs. Intuitively, the situation where we'll need to write 'a' the maximum possible number of times in total is when we need to write 'a' the maximum possible number of times for each individual pair. What we can do therefore is compute the maximum frequency count for a given letter in a pair and then add up the maximum frequency counts over all pairs.

Here is my Java code.

import java.io.*; import java.util.*; public class blocks { public static void main(String[] args) throws IOException { // initialize file I/O BufferedReader br = new BufferedReader(new FileReader("blocks.in")); PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("blocks.out"))); // allocate array to track total blocks needed // letter a maps to index 0, letter b maps to index 1, ... // letter z maps to index 25 int[] blocksNeeded = new int[26]; // read in the number of blocks int n = Integer.parseInt(br.readLine()); for(int i = 0; i < n; i++) { // read in each pair of words StringTokenizer st = new StringTokenizer(br.readLine()); String firstWord = st.nextToken(); String secondWord = st.nextToken(); // get the frequency counts int[] firstFrequency = getFrequency(firstWord); int[] secondFrequency = getFrequency(secondWord); // update the number of blocks needed for(int j = 0; j < 26; j++) { if(firstFrequency[j] > secondFrequency[j]) { blocksNeeded[j] += firstFrequency[j]; } else { blocksNeeded[j] += secondFrequency[j]; } } } // print the answer for(int i = 0; i < 26; i++) { pw.println(blocksNeeded[i]); } pw.close(); } /** * This function takes in a string and returns an integer array * of the number of blocks needed to spell out the given word. */ public static int[] getFrequency(String s) { int[] blocksNeeded = new int[26]; for(int i = 0; i < s.length(); i++) { int index = s.charAt(i) - 'a'; blocksNeeded[index]++; } return blocksNeeded; } }