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 :
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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