Split a binary string into K subsets minimizing sum of products of occurrences of 0 and 1

• Difficulty Level : Hard
• Last Updated : 11 Jun, 2021

Given a binary string S, the task is to partition the sequence into K non-empty subsets such that the sum of products of occurences of 0 and 1 for all subsets is minimum. If impossible print -1.
Examples:

Input: S = “0001”, K = 2
Output:
Explaination
We have 3 choices {0, 001}, {00, 01}, {000, 1}
The respective sum of products are {1*0 + 2*1 = 2}, {2*0 + 1*1 = 1}, {3*0 + 0*1 = 0}.
Input: S = “1011000110110100”, K = 5
Output:
Explanation: The subsets {10, 11, 000, 11011, 0100} minimizes the sum of product { 1*1 + 0*2 + 3*0 + 1*4 + 3*1 = 8 }.

Approach: In order to solve this problem we are using bottom-up dynamic programming

• We calculate the minimum sum of products for all subsets and then, for every subset we use this value to compute the minimum sum of products for all sizes of that subset.
• For a subset starting at index i and ending at index j the value will be minimum of dp[i-1] + (count_zero * count_one) where count_zero and count_one are the occurences of 0 and 1 between i and j respectively.
• For each index j, we find the minimum value amongst all possible values of i.
• Return dp[N – 1] for the final answer.

Below code is the implementation of the above approach:

C++

 // C++ Program to split a given string// into K segments such that the sum// of product of occurence of// characters in them is minimized #include using namespace std; // Function to return the minimum// sum of products of occurences// of 0 and 1 in each segmentsint minSumProd(string S, int K){    // Store the length of    // the string    int len = S.length();     // Not possible to    // generate subsets    // greater than the    // length of string    if (K > len)        return -1;     // If the number of subsets    // equals the length    if (K == len)        return 0;     vector dp(len);    int count_zero = 0, count_one = 0;     // Precompute the sum of    // products for all index    for (int j = 0; j < len; j++) {         (S[j] == '0')            ? count_zero++            : count_one++;        dp[j] = count_zero * count_one;    }     // Calculate the minimum sum of    // products for K subsets    for (int i = 1; i < K; i++) {         for (int j = len; j >= i; j--) {             count_zero = 0, count_one = 0;            dp[j] = INT_MAX;             for (int k = j; k >= i; k--) {                 (S[k] == '0') ? count_zero++                              : count_one++;                dp[j]                    = min(                        dp[j],                        count_zero * count_one                            + dp[k - 1]);            }        }    }     return dp[len - 1];} // Driver codeint main(){    string S = "1011000110110100";    int K = 5;    cout << minSumProd(S, K) << '\n';    return 0;}

Java

 // Java Program to split a given String// into K segments such that the sum// of product of occurence of// characters in them is minimizedimport java.util.*; class GFG{ // Function to return the minimum// sum of products of occurences// of 0 and 1 in each segmentsstatic int minSumProd(String S, int K){    // Store the length of    // the String    int len = S.length();     // Not possible to    // generate subsets    // greater than the    // length of String    if (K > len)        return -1;     // If the number of subsets    // equals the length    if (K == len)        return 0;     int []dp = new int[len];    int count_zero = 0, count_one = 0;     // Precompute the sum of    // products for all index    for (int j = 0; j < len; j++)    {        if(S.charAt(j) == '0')            count_zero++;        else            count_one++;        dp[j] = count_zero * count_one;    }     // Calculate the minimum sum of    // products for K subsets    for (int i = 1; i < K; i++)    {        for (int j = len-1; j >= i; j--)        {            count_zero = 0;            count_one = 0;            dp[j] = Integer.MAX_VALUE;             for (int k = j; k >= i; k--)            {                if(S.charAt(k) == '0')                    count_zero++;                else                    count_one++;                dp[j] = Math.min(dp[j], count_zero *                                        count_one +                                        dp[k - 1]);            }        }    }         return dp[len - 1];} // Driver codepublic static void main(String[] args){    String S = "1011000110110100";    int K = 5;    System.out.print(minSumProd(S, K));}} // This code is contributed by 29AjayKumar

Python3

 # Python3 program to split a given String# into K segments such that the sum# of product of occurence of# characters in them is minimizedimport sys # Function to return the minimum# sum of products of occurences# of 0 and 1 in each segmentsdef minSumProd(S, K):         # Store the length of    # the String    Len = len(S);     # Not possible to    # generate subsets    # greater than the    # length of String    if (K > Len):        return -1;     # If the number of subsets    # equals the length    if (K == Len):        return 0;     dp =  * Len;    count_zero = 0;    count_one = 0;     # Precompute the sum of    # products for all index    for j in range(0, Len, 1):        if (S[j] == '0'):            count_zero += 1;        else:            count_one += 1;        dp[j] = count_zero * count_one;     # Calculate the minimum sum of    # products for K subsets    for i in range(1, K):        for j in range(Len - 1, i - 1, -1):            count_zero = 0;            count_one = 0;            dp[j] = sys.maxsize;             for k in range(j, i - 1, -1):                if (S[k] == '0'):                    count_zero += 1;                else:                    count_one += 1;                 dp[j] = min(dp[j], count_zero *                                   count_one +                                   dp[k - 1]);    return dp[Len - 1]; # Driver codeif __name__ == '__main__':     S = "1011000110110100";    K = 5;         print(minSumProd(S, K)); # This code is contributed by 29AjayKumar

C#

 // C# program to split a given String// into K segments such that the sum// of product of occurence of// characters in them is minimizedusing System; class GFG{ // Function to return the minimum// sum of products of occurences// of 0 and 1 in each segmentsstatic int minSumProd(string S, int K){         // Store the length of    // the String    int len = S.Length;     // Not possible to    // generate subsets    // greater than the    // length of String    if (K > len)        return -1;     // If the number of subsets    // equals the length    if (K == len)        return 0;     int []dp = new int[len];    int count_zero = 0, count_one = 0;     // Precompute the sum of    // products for all index    for(int j = 0; j < len; j++)    {       if(S[j] == '0')       {           count_zero++;       }       else       {           count_one++;       }       dp[j] = count_zero * count_one;    }     // Calculate the minimum sum    // of products for K subsets    for(int i = 1; i < K; i++)    {       for(int j = len - 1; j >= i; j--)       {          count_zero = 0;          count_one = 0;          dp[j] = Int32.MaxValue;                     for(int k = j; k >= i; k--)          {             if(S[k] == '0')             {                 count_zero++;             }             else             {                 count_one++;             }             dp[j] = Math.Min(dp[j], count_zero *                                      count_one +                                      dp[k - 1]);          }       }    }    return dp[len - 1];} // Driver codepublic static void Main(string[] args){    string S = "1011000110110100";    int K = 5;         Console.Write(minSumProd(S, K));}} // This code is contributed by rutvik_56

Javascript


Output:
8

Time Complexity: O(K*N*N)
Auxiliary Space Complexity: O(N)

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