Minimum cost to partition the given binary string

Last Updated : 01 Mar, 2022

Given a binary string str and an integer K, the task is to find the minimum cost required to partition the string into exactly K segments when the cost of each segment is the product of the number of set bits with the number of unset bits and total cost is sum of cost of all the individual segments.
Examples:

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

Approach: Write a function minCost(s, k, cost, i, n) where cost is the minimum cost so far, i is the starting index for the partition and k is the remaining segments to be partitioned. Now, starting from the i^th index calculate the cost of the current partition and call the same function recursively for the remaining substring. To memoize the result, a dp[][] array will be used where dp[i][j] will store the minimum cost of partitioning the string into j parts starting at the i^th index.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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# 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

<script>
    // 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.
</script>

Output:

Time Complexity: O((n - k) + (MAX * MAX))

Auxiliary Space: O(MAX * MAX)

Minimum cost to partition the given binary string

mayank77dh9

Improve

Article Tags :

Practice Tags :

Minimum cost to partition the given binary string

Similar Reads

Thank You!

What kind of Experience do you want to share?