# Minimum cost to partition the given binary string

• Difficulty Level : Medium
• Last Updated : 10 Jun, 2021

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

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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 ith 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 ith index.
Below is the implementation of the above approach:

## 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

 ``
Output:
`2`

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