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

Last Updated : 09 Feb, 2022

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

Input: S = “0001”, K = 2
Output:
Explanation
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 occurrences 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 occurrence of` `// characters in them is minimized`   `#include ` `using` `namespace` `std;`   `// Function to return the minimum` `// sum of products of occurrences` `// of 0 and 1 in each segments` `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;`   `    ``vector<``int``> 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 code` `int` `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 occurrence of` `// characters in them is minimized` `import` `java.util.*;`   `class` `GFG{`   `// Function to return the minimum` `// sum of products of occurrences` `// of 0 and 1 in each segments` `static` `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 code` `public` `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 occurrence of` `# characters in them is minimized` `import` `sys`   `# Function to return the minimum` `# sum of products of occurrences` `# of 0 and 1 in each segments` `def` `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 ``=` `[``0``] ``*` `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 code` `if` `__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 occurrence of` `// characters in them is minimized` `using` `System;`   `class` `GFG{`   `// Function to return the minimum` `// sum of products of occurrences` `// of 0 and 1 in each segments` `static` `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 code` `public` `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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