USACO 2022 January Contest Bronze Division - Herdle

Problem link: here

Solution Author: Stefan Dascalescu

Problem Solution

First, we deal with the letters placed correctly, then we will keep frequency arrays to determine the smallest quantities available for each of the letters given.

Source codes

The source codes in C++ and Python can be seen below.

#include <bits/stdc++.h>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    char grid[3][3], grid2[3][3];
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            cin >> grid[i][j];
    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            cin >> grid2[i][j];

    int green = 0, yellow = 0;

    int frq[26] = {0};
    int frq2[26] = {0};

    for(int i = 0; i < 3; i++)
        for(int j = 0; j < 3; j++)
            if(grid[i][j] == grid2[i][j])
                green++;
            else
            {
                ++frq[grid[i][j] - 'A'];
                ++frq2[grid2[i][j] - 'A'];
            }

    for(int i = 0; i < 26; i++)
        yellow += min(frq[i], frq2[i]);

    cout << green << '\n' << yellow << '\n';
    return 0;
}

grid = [list(input().strip()) for _ in range(3)]
grid2 = [list(input().strip()) for _ in range(3)]
green = 0
yellow = 0
frq = [0] * 26
frq2 = [0] * 26
for i in range(3):
    for j in range(3):
        if grid[i][j] == grid2[i][j]:
            green += 1
        else:
            frq[ord(grid[i][j]) - ord('A')] += 1
            frq2[ord(grid2[i][j]) - ord('A')] += 1
for i in range(26):
    yellow += min(frq[i], frq2[i])
print(green)
print(yellow)