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Home Code Solutions

Ones and Zeroes – LeetCode Solution

Ones and Zeroes - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

Leetcode All Problems Solutions

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Table of Contents

  • Ones and Zeroes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Ones and Zeroes LeetCode Solution
  • Java Ones and Zeroes LeetCode Solution
  • Python 3 Ones and Zeroes LeetCode Solution
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Ones and Zeroes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

You are given an array of binary strings strs and two integers m and n.

Return the size of the largest subset of strs such that there are at most m 0‘s and n 1‘s in the subset.

A set x is a subset of a set y if all elements of x are also elements of y.

 

Example 1:

Input: strs = ["10","0001","111001","1","0"], m = 5, n = 3
Output: 4
Explanation: The largest subset with at most 5 0's and 3 1's is {"10", "0001", "1", "0"}, so the answer is 4.
Other valid but smaller subsets include {"0001", "1"} and {"10", "1", "0"}.
{"111001"} is an invalid subset because it contains 4 1's, greater than the maximum of 3.

Example 2:

Input: strs = ["10","0","1"], m = 1, n = 1
Output: 2
Explanation: The largest subset is {"0", "1"}, so the answer is 2.

 

Constraints:

  • 1 <= strs.length <= 600
  • 1 <= strs[i].length <= 100
  • strs[i] consists only of digits '0' and '1'.
  • 1 <= m, n <= 100

C++ Ones and Zeroes LeetCode Solution


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int findMaxForm(vector<string>& strs, int m, int n) {
  vector<vector<int>> memo(m+1, vector<int>(n+1, 0));
  int numZeroes, numOnes;

  for (auto &s : strs) {
    numZeroes = numOnes = 0;
    // count number of zeroes and ones in current string
    for (auto c : s) {
      if (c == '0')
	numZeroes++;
      else if (c == '1')
	numOnes++;
    }

    // memo[i][j] = the max number of strings that can be formed with i 0's and j 1's
    // from the first few strings up to the current string s
    // Catch: have to go from bottom right to top left
    // Why? If a cell in the memo is updated(because s is selected),
    // we should be adding 1 to memo[i][j] from the previous iteration (when we were not considering s)
    // If we go from top left to bottom right, we would be using results from this iteration => overcounting
    for (int i = m; i >= numZeroes; i--) {
	for (int j = n; j >= numOnes; j--) {
          memo[i][j] = max(memo[i][j], memo[i - numZeroes][j - numOnes] + 1);
	}
    }
  }
  return memo[m][n];
}

Java Ones and Zeroes LeetCode Solution


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public int findMaxForm(String[] strs, int m, int n) {
    int[][] dp = new int[m+1][n+1];
    for (String s : strs) {
        int[] count = count(s);
        for (int i=m;i>=count[0];i--) 
            for (int j=n;j>=count[1];j--) 
                dp[i][j] = Math.max(1 + dp[i-count[0]][j-count[1]], dp[i][j]);
    }
    return dp[m][n];
}
    
public int[] count(String str) {
    int[] res = new int[2];
    for (int i=0;i<str.length();i++)
        res[str.charAt(i) - '0']++;
    return res;
 }

 



Python 3 Ones and Zeroes LeetCode Solution


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class Solution:
    def findMaxForm(self, S: List[str], M: int, N: int) -> int:
        dp = [[0 for _ in range(N+1)] for _ in range(M+1)]
        for str in S:
            zeros = str.count("0")
            ones = len(str) - zeros
            for i in range(M, zeros - 1, -1):
                for j in range(N, ones - 1, -1):
                    dp[i][j] = max(dp[i][j], dp[i-zeros][j-ones] + 1)
        return dp[M][N]



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264


Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195


Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140


Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96


Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66


Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40


Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31


Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21


Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16


Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9


Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7


Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4


Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3


Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionOnes and Zeroespypy 3Python 2python 3rubyrustSolution
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