分割给定二进制字符串的最小成本

给定一个二进制字符串str和一个整数K ，当每个段的成本是设置位数与未设置位数和总数的乘积时，任务是找到将字符串划分为恰好K 个段所需的最小成本成本是所有单个段的成本总和。
例子：

Input: str = “110101”, K = 3 
Output: 2 
11|0|101 is one of the possible partitions 
where the cost is 0 + 0 + 2 = 2
Input: str = “1000000”, K = 5 
Output: 0 
 

方法：编写一个函数minCost(s, k, cost, i, n)其中cost是到目前为止的最小成本， i是分区的起始索引， k是要分区的剩余段。现在，从第i^个索引开始计算当前分区的成本，并对剩余的子串递归调用相同的函数。为了记忆结果，将使用dp[][]数组，其中dp[i][j]将存储从第i^个索引开始将字符串划分为j 个部分的最小成本。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
const int MAX = 1001;
 
// dp[i][j] will store the minimum cost
// of partitioning the string into j
// parts starting at the ith index
int dp[MAX][MAX];
 
// Recursive function to find
// the minimum cost required
int minCost(string& s, int k, int cost, int i, int& n)
{
 
    // If the state has been solved before
    if (dp[i][k] != -1)
        return dp[i][k];
 
    // If only 1 part is left then the
    // remaining part of the string will
    // be considered as that part
    if (k == 1) {
 
        // To store the count of 0s and the
        // total characters of the string
        int count_0 = 0, total = n - i;
 
        // Count the 0s
        while (i < n)
            if (s[i++] == '0')
                count_0++;
 
        // Memoize and return the updated cost
        dp[i][k] = cost + (count_0
                           * (total - count_0));
        return dp[i][k];
    }
 
    int curr_cost = INT_MAX;
    int count_0 = 0;
 
    // Check all the positions to
    // make the current partition
    for (int j = i; j < n - k + 1; j++) {
 
        // Count the numbers of 0s
        if (s[j] == '0')
            count_0++;
        int curr_part_length = j - i + 1;
 
        // Cost of partition is equal to
        // (no. of 0s) * (no. of 1s)
        int part_cost = (count_0
                         * (curr_part_length - count_0));
 
        // string, partitions, curr cost,
        // start index, length
        part_cost += minCost(s, k - 1, 0, j + 1, n);
 
        // Update the current cost
        curr_cost = min(curr_cost, part_cost);
    }
 
    // Memoize and return the updated cost
    dp[i][k] = (cost + curr_cost);
    return (cost + curr_cost);
}
 
// Driver code
int main()
{
    string s = "110101";
    int n = s.length();
    int k = 3;
 
    // Initialise the dp array
    for (int i = 0; i < MAX; i++) {
        for (int j = 0; j < MAX; j++)
            dp[i][j] = -1;
    }
 
    // string, partitions, curr cost,
    // start index, length
    cout << minCost(s, k, 0, 0, n);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG {
     
    final static int MAX = 1001;
     
    // dp[i][j] will store the minimum cost
    // of partitioning the string into j
    // parts starting at the ith index
    static int dp[][] = new int [MAX][MAX];
     
    // Recursive function to find
    // the minimum cost required
    static int minCost(String s, int k, int cost, int i, int n)
    {
     
        // If the state has been solved before
        if (dp[i][k] != -1)
            return dp[i][k];
     
        // If only 1 part is left then the
        // remaining part of the string will
        // be considered as that part
        if (k == 1) {
     
            // To store the count of 0s and the
            // total characters of the string
            int count_0 = 0, total = n - i;
     
            // Count the 0s
            while (i < n)
                if (s.charAt(i++) == '0')
                    count_0++;
     
            // Memoize and return the updated cost
            dp[i][k] = cost + (count_0
                            * (total - count_0));
            return dp[i][k];
        }
     
        int curr_cost = Integer.MAX_VALUE;
        int count_0 = 0;
     
        // Check all the positions to
        // make the current partition
        for (int j = i; j < n - k + 1; j++) {
     
            // Count the numbers of 0s
            if (s.charAt(j) == '0')
                count_0++;
            int curr_part_length = j - i + 1;
     
            // Cost of partition is equal to
            // (no. of 0s) * (no. of 1s)
            int part_cost = (count_0
                            * (curr_part_length - count_0));
     
            // string, partitions, curr cost,
            // start index, length
            part_cost += minCost(s, k - 1, 0, j + 1, n);
     
            // Update the current cost
            curr_cost = Math.min(curr_cost, part_cost);
        }
     
        // Memoize and return the updated cost
        dp[i][k] = (cost + curr_cost);
        return (cost + curr_cost);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String s = "110101";
        int n = s.length();
        int k = 3;
     
        // Initialise the dp array
        for (int i = 0; i < MAX; i++) {
            for (int j = 0; j < MAX; j++)
                dp[i][j] = -1;
        }
     
        // string, partitions, curr cost,
        // start index, length
        System.out.println(minCost(s, k, 0, 0, n));
     
     
    }
}
 
// This code is contributed by AnkitRai01

Python 3

# Python 3 implementation of the approach
import sys
MAX = 1001
 
# dp[i][j] will store the minimum cost
# of partitioning the string into j
# parts starting at the ith index
dp = [[0 for i in range(MAX)]
         for j in range(MAX)]
 
# Recursive function to find
# the minimum cost required
def minCost(s, k, cost, i, n):
     
    # If the state has been solved before
    if (dp[i][k] != -1):
        return dp[i][k]
 
    # If only 1 part is left then the
    # remaining part of the string will
    # be considered as that part
    if (k == 1):
         
        # To store the count of 0s and the
        # total characters of the string
        count_0 = 0
        total = n - i
 
        # Count the 0s
        while (i < n):
            if (s[i] == '0'):
                count_0 += 1
            i += 1
 
        # Memoize and return the updated cost
        dp[i][k] = cost + (count_0 *
                          (total - count_0))
        return dp[i][k]
 
    curr_cost = sys.maxsize
    count_0 = 0
 
    # Check all the positions to
    # make the current partition
    for j in range(i, n - k + 1, 1):
         
        # Count the numbers of 0s
        if (s[j] == '0'):
            count_0 += 1
        curr_part_length = j - i + 1
 
        # Cost of partition is equal to
        # (no. of 0s) * (no. of 1s)
        part_cost = (count_0 *
                    (curr_part_length - count_0))
 
        # string, partitions, curr cost,
        # start index, length
        part_cost += minCost(s, k - 1, 0, j + 1, n)
 
        # Update the current cost
        curr_cost = min(curr_cost, part_cost)
 
    # Memoize and return the updated cost
    dp[i][k] = (cost + curr_cost)
    return (cost + curr_cost)
 
# Driver code
if __name__ == '__main__':
    s = "110101"
    n = len(s)
    k = 3
 
    # Initialise the dp array
    for i in range(MAX):
        for j in range(MAX):
            dp[i][j] = -1
 
    # string, partitions, curr cost,
    # start index, length
    print(minCost(s, k, 0, 0, n))
     
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
 
using System;
using System.Collections.Generic;
 
class GFG
{
     
    readonly static int MAX = 1001;
     
    // dp[i,j] will store the minimum cost
    // of partitioning the string into j
    // parts starting at the ith index
    static int [,]dp = new int [MAX, MAX];
     
    // Recursive function to find
    // the minimum cost required
    static int minCost(String s, int k, int cost, int i, int n)
    {
     
        int count_0 = 0;
        // If the state has been solved before
        if (dp[i, k] != -1)
            return dp[i, k];
     
        // If only 1 part is left then the
        // remaining part of the string will
        // be considered as that part
        if (k == 1)
        {
     
            // To store the count of 0s and the
            // total characters of the string
             
            int total = n - i;
     
            // Count the 0s
            while (i < n)
                if (s[i++] == '0')
                    count_0++;
     
            // Memoize and return the updated cost
            dp[i, k] = cost + (count_0
                            * (total - count_0));
            return dp[i, k];
        }
     
        int curr_cost = int.MaxValue;
        count_0 = 0;
     
        // Check all the positions to
        // make the current partition
        for (int j = i; j < n - k + 1; j++)
        {
     
            // Count the numbers of 0s
            if (s[j] == '0')
                count_0++;
            int curr_partlength = j - i + 1;
     
            // Cost of partition is equal to
            // (no. of 0s) * (no. of 1s)
            int part_cost = (count_0
                            * (curr_partlength - count_0));
     
            // string, partitions, curr cost,
            // start index,.Length
            part_cost += minCost(s, k - 1, 0, j + 1, n);
     
            // Update the current cost
            curr_cost = Math.Min(curr_cost, part_cost);
        }
     
        // Memoize and return the updated cost
        dp[i, k] = (cost + curr_cost);
        return (cost + curr_cost);
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        String s = "110101";
        int n = s.Length;
        int k = 3;
     
        // Initialise the dp array
        for (int i = 0; i < MAX; i++)
        {
            for (int j = 0; j < MAX; j++)
                dp[i, j] = -1;
        }
     
        // string, partitions, curr cost,
        // start index,.Length
        Console.WriteLine(minCost(s, k, 0, 0, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

    // Javascript implementation of the approach
     
    let MAX = 1001;
       
    // dp[i][j] will store the minimum cost
    // of partitioning the string into j
    // parts starting at the ith index
    let dp = new Array(MAX);
       
    // Recursive function to find
    // the minimum cost required
    function minCost(s, k, cost, i, n)
    {
       
        // If the state has been solved before
        if (dp[i][k] != -1)
            return dp[i][k];
       
        // If only 1 part is left then the
        // remaining part of the string will
        // be considered as that part
        if (k == 1) {
       
            // To store the count of 0s and the
            // total characters of the string
            let count_0 = 0, total = n - i;
       
            // Count the 0s
            while (i < n)
                if (s[i++] == '0')
                    count_0++;
       
            // Memoize and return the updated cost
            dp[i][k] = cost + (count_0
                            * (total - count_0));
            return dp[i][k];
        }
       
        let curr_cost = Number.MAX_VALUE;
        let count_0 = 0;
       
        // Check all the positions to
        // make the current partition
        for (let j = i; j < n - k + 1; j++) {
       
            // Count the numbers of 0s
            if (s[j] == '0')
                count_0++;
            let curr_part_length = j - i + 1;
       
            // Cost of partition is equal to
            // (no. of 0s) * (no. of 1s)
            let part_cost = (count_0 * (curr_part_length - count_0));
       
            // string, partitions, curr cost,
            // start index, length
            part_cost += minCost(s, k - 1, 0, j + 1, n);
       
            // Update the current cost
            curr_cost = Math.min(curr_cost, part_cost);
        }
       
        // Memoize and return the updated cost
        dp[i][k] = (cost + curr_cost);
        return (cost + curr_cost);
    }
     
    let s = "110101";
    let n = s.length;
    let k = 3;
 
    // Initialise the dp array
    for (let i = 0; i < MAX; i++) {
      dp[i] = new Array(MAX);
      for (let j = 0; j < MAX; j++)
        dp[i][j] = -1;
    }
 
    // string, partitions, curr cost,
    // start index, length
    document.write(minCost(s, k, 0, 0, n));
 
// This code is contributed by divyeshrabadiya07.

输出：

2

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