String Construction - Google Top Interview Questions


Problem Statement :


You are given a list of strings strings where each string contains "A"s and "B"s. 

You are also given integers a and b.

Return the maximum number of strings that can be constructed given that you can use at most a "A"s and at most b "B"s, without reuse.

Constraints

n ≤ 50 where n is the length of strings

a, b ≤ 50

Example 1

Input

strings = ["AABB", "AAAB", "A", "B"]

a = 4

b = 2

Output

3

Explanation

We can take these strings using 4 "A"s and 2 "B"s ["AAAB","A","B"]


Solution :



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                        Solution in C++ :

int dp[52][52][52];

int solve(vector<string>& strings, int a, int b) {
    int N = strings.size();
    vector<pair<int, int>> pairs;
    pairs.reserve(N);
    for (const string& s : strings) {
        int num_a = count(s.begin(), s.end(), 'A');
        int num_b = s.size() - num_a;
        pairs.emplace_back(num_a, num_b);
    }
    for (int k = 1; k <= N; k++) {
        auto p = pairs[k - 1];
        for (int i = 0; i <= a; i++) {
            for (int j = 0; j <= b; j++) {
                int t = dp[k - 1][i][j];  // don't use the current string
                if (i >= p.first && j >= p.second) {
                    // do use the current string
                    t = max(t, 1 + dp[k - 1][i - p.first][j - p.second]);
                }
                dp[k][i][j] = t;
            }
        }
    }
    return dp[N][a][b];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    private static final int len = 52;
    public int solve(String[] arr, int a, int b) {
        int[][][] dp = new int[len][len][len];

        for (int i = 1; i <= arr.length; i++) {
            String str = arr[i - 1];
            int countA = getCount(str, 'A');
            int countB = getCount(str, 'B');
            for (int j = 0; j <= a; j++) {
                for (int k = 0; k <= b; k++) {
                    dp[i][j][k] = dp[i - 1][j][k];
                    if (j - countA >= 0 && k - countB >= 0)
                        dp[i][j][k] =
                            Math.max(dp[i - 1][j][k], 1 + dp[i - 1][j - countA][k - countB]);
                }
            }
        }
        return dp[arr.length][a][b];
    }

    private int getCount(String str, char chr) {
        int count = 0;
        for (char ch : str.toCharArray()) {
            if (ch == chr)
                ++count;
        }
        return count;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, strings, a, b):
        dp = defaultdict(int, {(a, b): 0})
        ans = 0
        for s in strings:
            f = Counter(s)
            a_count = f["A"]
            b_count = f["B"]
            for (i, j), val in list(dp.items()):
                if i >= a_count and j >= b_count:
                    dp[(i - a_count, j - b_count)] = max(dp[(i - a_count, j - b_count)], val + 1)
                    ans = max(ans, dp[(i - a_count, j - b_count)])
        return ans


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