最佳地将二进制字符串中的0和1容纳到K个存储桶中

给定二进制字符串S ，由0和1组成。您必须以满足以下条件的方式将0和1容纳到K个存储桶中：

1. 将0和1填充到存储桶中，以保持0和1的相对顺序。例如，您不能将S [1]放入存储区2中，而将S [0]放入存储区1中。您必须保留二进制字符串的原始顺序。
2. 不得将存储桶留空，并且字符串的任何元素都不应保留。
3. 在所有可能的住宿安排中，每个铲斗的所有乘积的总和(0的数量* 1的数量)应为最小值。

如果无法解决，请打印-1。

例子：

Input: S = "0001", K = 2
Output: 0
We have 3 choices {"0", "001"}, {"00", "01"}, {"000", 1}
First choice, we will get 1*0 + 2*1 = 2
Second choice, we will get 2*0 + 1*1 = 1
Third choice, we will get 3*0 + 0*1 = 0
Out of all the 3 choices, the third choice 
is giving the minimum answer.

Input: S = "0101", K = 1
Output: 1 

递归实现：您必须将二进制字符串容纳在K个存储桶中，而不会干扰上述条件。然后，可以通过先填充第i个存储桶(从0开始)，然后将元素从起始位置放置到N (N =二进制字符串的长度)，然后继续将计数零和一添加到起始索引，从而得出简单的递归解决方案。对于每次迭代，如果直到开始都存在x个零和y个，则重复进行f(start，K)= x * y + f(start + 1，K – 1)，因为下一次适应将从(start + 1)-索引和剩余存储桶将K – 1。

因此，递归公式将是–

F(start, current_bucket) =  |           |
                            |       min |  F(i + 1, next_bucket) + (ones * zeroes in current_bucket)  
                            |           |   
                            | i = start to N

自上而下的动态方法：
通过将start和bucket变量的不同组合的结果保存到2-D DP数组中，可以将递归关系更改为Dynamic Solution 。我们可以使用这样的事实，即应该保留字符串的顺序。您可以使用一个二维数组来保存大小为[[字符串大小*桶的大小]的状态，其中dp [i] [j]会告诉我们直到使用j + 1桶的字符串的第i个索引为止的最小容纳量。我们的最终答案将在dp [N-1] [K-1]中。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// 2-D dp array saving different states
// dp[i][j] = minimum value of accommodation
// till i'th index of the string
// using j+1 number of buckets.
vector > dp;
 
// Function to find the minimum required
// sum using dynamic programming
int solveUtil(int start, int bucket, string str, int K)
{
    int N = str.size();
 
    // If both start and bucket reached end then
    // return 0 else that arrangement is not possible
    // so return INT_MAX
    if (start == N) {
        if (bucket == K)
            return 0;
        return INT_MAX;
    }
 
    // Corner case
    if (bucket == K)
        return INT_MAX;
 
    // If state if already calculated
    // then return its answer
    if (dp[start][bucket] != -1)
        return dp[start][bucket];
 
    int zeroes = 0;
    int ones = 0;
    int ans = INT_MAX;
 
    // Start filling zeroes and ones which to be accommodated
    // in jth bucket then ans for current arrangement will be
    // ones*zeroes + recur(i+1, buket+1)
    for (int i = start; i < N; ++i) {
        if (str[i] == '1')
            ones++;
        else
            zeroes++;
 
        if (ones * zeroes > ans)
            break;
 
        int temp = solveUtil(i + 1, bucket + 1, str, K);
 
        // If this arrangement is not possible then
        // don't calculate further
        if (temp != INT_MAX) {
            ans = min(ans, temp + (ones * zeroes));
        }
    }
 
    return dp[start][bucket] = ans;
}
 
// Function to initialze the dp and call
// solveUtil() method to get the answer
int solve(string str, int K)
{
    int N = str.size();
    dp.clear();
    dp.resize(N, vector(K, -1));
 
    // Start with 0-th index and 1 bucket
    int ans = solveUtil(0, 0, str, K);
 
    return ans == INT_MAX ? -1 : ans;
}
 
// Driver code
int main()
{
    string S = "0101";
 
    // K buckets
    int K = 2;
 
    cout << solve(S, K) << endl;
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// 2-D dp array saving different states
// dp[i][j] = minimum value of accommodation
// till i'th index of the string
// using j+1 number of buckets.
static int[][] dp;
 
// Function to find the minimum required
// sum using dynamic programming
static int solveUtil(int start, int bucket,
                     String str, int K)
{
    int N = str.length();
     
    // If both start and bucket reached end
    // then return 0 else that arrangement
    // is not possible so return INT_MAX
    if (start == N)
    {
        if (bucket == K)
        {
            return 0;
        }
        return Integer.MAX_VALUE;
    }
     
    // Corner case
    if (bucket == K)
    {
        return Integer.MAX_VALUE;
    }
     
    // If state if already calculated
    // then return its answer
    if (dp[start][bucket] != -1)
    {
        return dp[start][bucket];
    }
    int zeroes = 0;
    int ones = 0;
    int ans = Integer.MAX_VALUE;
     
    // Start filling zeroes and ones which to be
    // accommodated in jth bucket then ans for
    // current arrangement will be
    // ones*zeroes + recur(i+1, buket+1)
    for(int i = start; i < N; ++i)
    {
        if (str.charAt(i) == '1')
        {
            ones++;
        }
        else
        {
            zeroes++;
        }
        if (ones * zeroes > ans)
        {
            break;
        }
        int temp = solveUtil(i + 1, bucket + 1, str, K);
         
        // If this arrangement is not possible then
        // don't calculate further
        if (temp != Integer.MAX_VALUE)
        {
            ans = Math.min(ans, temp + (ones * zeroes));
        }
    }
    return dp[start][bucket] = ans;
}
 
// Function to initialze the dp and call
// solveUtil() method to get the answer
static int solve(String str, int K)
{
    int N = str.length();
    dp = new int[N][K];
     
    for(int[] row : dp)
    {
        Arrays.fill(row, -1);
    }
     
    // Start with 0-th index and 1 bucket
    int ans = solveUtil(0, 0, str, K);
    return ans == Integer.MAX_VALUE ? -1 : ans;
}
 
// Driver code
public static void main(String[] args)
{
    String S = "0101";
     
    // K buckets
    int K = 2;
     
    System.out.println(solve(S, K));
}
}
 
// This code is contributed by rag2127

Python3

# Python3 implementation of the approach
 
# 2-D dp array saving different states
# dp[i][j] = minimum value of accommodation
# till i'th index of the str1ing
# using j+1 number of buckets.
 
# Function to find the minimum required
# sum using dynamic programming
def solveUtil(start, bucket, str1, K,dp):
 
    N = len(str1)
 
    # If both start and bucket reached end then
    # return 0 else that arrangement is not possible
    # so return INT_MAX
    if (start == N) :
        if (bucket == K):
            return 0
        return 10**9
 
 
    # Corner case
    if (bucket == K):
        return 10**9
 
    # If state if already calculated
    # then return its answer
    if (dp[start][bucket] != -1):
        return dp[start][bucket]
 
    zeroes = 0
    ones = 0
    ans = 10**9
 
    # Start filling zeroes and ones which to be accommodated
    # in jth bucket then ans for current arrangement will be
    # ones*zeroes + recur(i+1, buket+1)
    for i in range(start,N):
        if (str1[i] == '1'):
            ones += 1
        else:
            zeroes += 1
 
        if (ones * zeroes > ans):
            break
 
        temp = solveUtil(i + 1, bucket + 1, str1, K,dp)
 
        # If this arrangement is not possible then
        # don't calculate further
        if (temp != 10**9):
            ans = min(ans, temp + (ones * zeroes))
 
    dp[start][bucket] = ans
 
    return ans
 
 
# Function to initialze the dp and call
# solveUtil() method to get the answer
def solve(str1, K):
 
    N = len(str1)
 
    dp = [[-1 for i in range(K)] for i in range(N)]
 
    # Start with 0-th index and 1 bucket
    ans = solveUtil(0, 0, str1, K,dp)
 
    if ans == 10**9:
        return -1
    else:
        return ans
 
 
# Driver code
 
s = "0101"
S=[i for i in s]
 
# K buckets
K = 2
 
print(solve(S, K))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
class GFG
{
     
    // 2-D dp array saving different states
    // dp[i][j] = minimum value of accommodation
    // till i'th index of the string
    // using j+1 number of buckets.
    static int[,] dp;
      
    // Function to find the minimum required
    // sum using dynamic programming
    static int solveUtil(int start, int bucket,
                         string str, int K)
    {
        int N = str.Length;
          
        // If both start and bucket reached end
        // then return 0 else that arrangement
        // is not possible so return INT_MAX
        if (start == N)
        {
            if (bucket == K)
            {
                return 0;
            }
            return Int32.MaxValue;
        }
          
        // Corner case
        if (bucket == K)
        {
            return Int32.MaxValue;
        }
          
        // If state if already calculated
        // then return its answer
        if (dp[start,bucket] != -1)
        {
            return dp[start, bucket];
        }
        int zeroes = 0;
        int ones = 0;
        int ans = Int32.MaxValue;
          
        // Start filling zeroes and ones which to be
        // accommodated in jth bucket then ans for
        // current arrangement will be
        // ones*zeroes + recur(i+1, buket+1)
        for(int i = start; i < N; ++i)
        {
            if (str[i] == '1')
            {
                ones++;
            }
            else
            {
                zeroes++;
            }
            if (ones * zeroes > ans)
            {
                break;
            }
            int temp = solveUtil(i + 1, bucket + 1, str, K);
              
            // If this arrangement is not possible then
            // don't calculate further
            if (temp != Int32.MaxValue)
            {
                ans = Math.Min(ans, temp + (ones * zeroes));
            }
        }
        return dp[start, bucket] = ans;
    }
      
    // Function to initialze the dp and call
    // solveUtil() method to get the answer
    static int solve(string str, int K)
    {
        int N = str.Length;
        dp = new int[N, K];       
        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < K; j++)
            {
                dp[i, j] = -1;
            }
        }
          
        // Start with 0-th index and 1 bucket
        int ans = solveUtil(0, 0, str, K);
        return ans == Int32.MaxValue ? -1 : ans;
    }
 
  // Driver code
  static void Main()
  {
    string S = "0101";
      
    // K buckets
    int K = 2;
      
    Console.WriteLine(solve(S, K));
  }
}
 
// This code is contributed by divyeshrabadiya07.

输出：

2

时间复杂度： O(N ³ )
空间复杂度： O(N ² )
该解决方案仍未优化，因为它多次调用同一状态。因此，现在来看优化的自下而上的DP方法。

自下而上的动态方法：让我们尝试首先考虑最终状态。这里的变量是存储桶的数量和字符串的索引。令dp [i] [j]是字符串元素0到j-1和i桶的乘积的最小和。现在定义过渡函数，我们必须从后面开始，并考虑在每个可能的位置k处进行分区。因此，我们的转换函数如下所示：

dp [i][j] = for all k = 0 to j min(dp[i][k-1] + numberOfZeroes * numberOfOnes)

for i = 0 (single partition) simple count number of 0's and 1's and do the multiplication. 
And if number of buckets is more than string length till now ans is -1 as we cant fill all 
the available buckets.

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the minimum required
// sum using dynamic programming
int solve(string str, int K)
{
    int n = str.length();
 
    // dp[i][j] = minimum val of accommodation
    // till j'th index of the string
    // using i+1 number of buckets.
    // Final ans will be in dp[n-1][K-1]
    long long dp[K][n];
 
    // Initialise dp with all states as 0
    memset(dp, 0, sizeof dp);
 
    // Corner cases
    if (n < K)
        return -1;
    else if (n == K)
        return 0;
 
    // Filling first row, if only 1 bucket then simple count
    // number of zeros and ones and do the multiplication
    long long zeroes = 0, ones = 0;
 
    for (int i = 0; i < n; i++) {
        if (str[i] == '0')
            zeroes++;
        else
            ones++;
 
        dp[0][i] = ones * zeroes;
    }
 
    for (int s = 1; s < K; s++) {
        for (int i = 0; i < n; i++) {
 
            dp[s][i] = INT_MAX;
 
            ones = 0;
            zeroes = 0;
            for (int k = i; k >= 0; k--) {
                if (str[k] == '0')
                    zeroes++;
                else
                    ones++;
 
                // If k = 0 then this arrangement is not possible
                dp[s][i] = min(dp[s][i],
                               +((k - 1 >= 0)
                                     ? ones * zeroes + dp[s - 1][k - 1]
                                     : INT_MAX));
            }
        }
    }
 
    // If no arrangement is possible then
    // our answer will remain INT_MAX so return -1
    return (dp[K - 1][n - 1] == INT_MAX) ? -1 : dp[K - 1][n - 1];
}
 
// Driver code
int main()
{
    string S = "0101";
 
    // K buckets
    int K = 2;
 
    cout << solve(S, K) << endl;
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
    // Function to find the minimum required
    // sum using dynamic programming
    static long solve(String str, int K)
    {
        int n = str.length();
     
        // dp[i][j] = minimum val of accommodation
        // till j'th index of the string
        // using i+1 number of buckets.
        // Final ans will be in dp[n-1][K-1]
        long dp[][] = new long[K][n];
     
     
        // Corner cases
        if (n < K)
            return -1;
        else if (n == K)
            return 0;
     
        // Filling first row, if only 1 bucket then simple count
        // number of zeros and ones and do the multiplication
        long zeroes = 0, ones = 0;
     
        for (int i = 0; i < n; i++)
        {
            if (str.charAt(i) == '0')
                zeroes++;
            else
                ones++;
     
            dp[0][i] = ones * zeroes;
        }
     
        for (int s = 1; s < K; s++)
        {
            for (int i = 0; i < n; i++)
            {
     
                dp[s][i] = Integer.MAX_VALUE;
     
                ones = 0;
                zeroes = 0;
                for (int k = i; k >= 0; k--)
                {
                    if (str.charAt(k) == '0')
                        zeroes++;
                    else
                        ones++;
     
                    // If k = 0 then this arrangement is not possible
                    dp[s][i] = Math.min(dp[s][i],
                                +((k - 1 >= 0)
                                        ? ones * zeroes + dp[s - 1][k - 1]
                                        : Integer.MAX_VALUE));
                }
            }
        }
     
        // If no arrangement is possible then
        // our answer will remain INT_MAX so return -1
        return (dp[K - 1][n - 1] == Integer.MAX_VALUE) ? -1 : dp[K - 1][n - 1];
    }
     
    // Driver code
    public static void main (String[] args)
    {
     
        String S = "0101";
     
        // K buckets
        int K = 2;
     
        System.out.println(solve(S, K));
    }
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of the approach
import sys
 
# Function to find the minimum required
# sum using dynamic programming
def solve(Str, K):
    n = len(Str)
     
    # dp[i][j] = minimum val of accommodation
    # till j'th index of the string
    # using i+1 number of buckets.
    # Final ans will be in dp[n-1][K-1]
    # Initialise dp with all states as 0
    dp = [[0 for i in range(n)] for j in range(K)]
 
    # Corner cases
    if(n < K):
        return -1
    elif(n == K):
        return 0
 
    # Filling first row, if only 1 bucket then simple count
    # number of zeros and ones and do the multiplication
    zeroes = 0
    ones = 0
 
    for i in range(n):
        if(Str[i] == '0'):
            zeroes += 1
        else:
            ones += 1
        dp[0][i] = ones * zeroes
 
    for s in range(1, K):
        for i in range(n):
            dp[s][i] = sys.maxsize
            ones = 0
            zeroes = 0
 
            for k in range(i, -1, -1):
                if(Str[k] == '0'):
                    zeroes += 1
                else:
                    ones += 1
                 
                # If k = 0 then this arrangement
                # is not possible
                temp = 0
                if(k - 1 >= 0):
                    temp = ones * zeroes + dp[s - 1][k - 1]
                else:
                    temp = sys.maxsize
                dp[s][i] = min(dp[s][i], temp)
     
    # If no arrangement is possible then
    # our answer will remain INT_MAX so return -1
    if(dp[K - 1][n - 1] == sys.maxsize):
        return -1
    else:
        return dp[K - 1][n - 1]
       
# Driver code
S = "0101"
 
# K buckets
K = 2
print(solve(S, K))
 
# This code is contributed by avanitrachhadiya2155

C＃

// C# implementation of the approach
using System;
 
class GFG
{
    // Function to find the minimum required
    // sum using dynamic programming
    static long solve(string str, int K)
    {
        int n = str.Length;
     
        // dp[i, j] = minimum val of accommodation
        // till j'th index of the string
        // using i+1 number of buckets.
        // Final ans will be in dp[n-1, K-1]
        long [, ] dp = new long[K, n];
     
     
        // Corner cases
        if (n < K)
            return -1;
        else if (n == K)
            return 0;
     
        // Filling first row, if only 1 bucket then simple count
        // number of zeros and ones and do the multiplication
        long zeroes = 0, ones = 0;
     
        for (int i = 0; i < n; i++)
        {
            if (str[i] == '0')
                zeroes++;
            else
                ones++;
     
            dp[0, i] = ones * zeroes;
        }
     
        for (int s = 1; s < K; s++)
        {
            for (int i = 0; i < n; i++)
            {
     
                dp[s, i] = Int32.MaxValue;
     
                ones = 0;
                zeroes = 0;
                for (int k = i; k >= 0; k--)
                {
                    if (str[k] == '0')
                        zeroes++;
                    else
                        ones++;
     
                    // If k = 0 then this arrangement is not possible
                    dp[s, i] = Math.Min(dp[s, i],
                                +((k - 1 >= 0)
                                        ? ones * zeroes + dp[s - 1, k - 1]
                                        : Int32.MaxValue));
                }
            }
        }
     
        // If no arrangement is possible then
        // our answer will remain INT_MAX so return -1
        return (dp[K - 1, n - 1] == Int32.MaxValue) ? -1 : dp[K - 1, n - 1];
    }
     
    // Driver code
    public static void Main (string[] args)
    {
     
        string S = "0101";
     
        // K buckets
        int K = 2;
     
        Console.WriteLine(solve(S, K));
     
    }
         
}
 
// This code is contributed by ihritik

输出：

2

时间复杂度： O(N ³ )
空间复杂度： O(N ² )

要从最佳影片策划和实践问题去学习，检查了C++基础课程为基础，以先进的C++和C++ STL课程基础加上STL。要完成从学习语言到DS Algo等的更多准备工作，请参阅“完整面试准备课程” 。