g0401_0500.s0474_ones_and_zeroes.Solution Maven / Gradle / Ivy
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package g0401_0500.s0474_ones_and_zeroes;
// #Medium #Array #String #Dynamic_Programming
public class Solution {
/*
* credit: https://leetcode.com/problems/ones-and-zeroes/discuss/95811/Java-Iterative-DP-Solution-O(mn)-Space and
* https://leetcode.com/problems/ones-and-zeroes/discuss/95811/Java-Iterative-DP-Solution-O(mn)-Space/100352
*
* The problem can be interpreted as:
* What's the max number of str can we pick from strs with limitation of m "0"s and n "1"s.
*
* Thus we can define dp[i][j] as it stands for max number of str can we pick from strs with limitation
* of i "0"s and j "1"s.
*
* For each str, assume it has a "0"s and b "1"s, we update the dp array iteratively
* and set dp[i][j] = Math.max(dp[i][j], dp[i - a][j - b] + 1).
* So at the end, dp[m][n] is the answer.
*/
public int findMaxForm(String[] strs, int m, int n) {
int[][] dp = new int[m + 1][n + 1];
for (String str : strs) {
int[] count = count(str);
for (int i = m; i >= count[0]; i--) {
for (int j = n; j >= count[1]; j--) {
dp[i][j] = Math.max(dp[i][j], dp[i - count[0]][j - count[1]] + 1);
}
}
}
return dp[m][n];
}
private int[] count(String str) {
int[] res = new int[2];
for (int i = 0; i < str.length(); i++) {
res[str.charAt(i) - '0']++;
}
return res;
}
}