# Minimize cost to convert all characters of a binary string to 0s

Given a binary string, str, two integer arrays R[], and C[] of size N. Flipping all the characters from index i to R[i] requires C[i] cost. The task is to minimize the cost required to convert the given binary string to only 0s.

Examples:

Input: str = “1010”, R[] = {1, 2, 2, 3}, C[] = {3, 1, 2, 3}
Output: 4
Explanation:
Flipping all the characters from index 1 to 2, modifies str to “1100”. Therefore, cost = 1
Flipping all the characters from index 1 to 2, modifies str to “0000”. Therefore, cost = cost + 3 = 4
Therefore, minimum cost required is 4.

Input: str = “01100”, R[] = {1, 2, 3, 4, 5}, C[] = {1, 5, 5, 2, 3}
Output: 10

Approach: The problem can be solved using Greedy technique. The idea is to traverse the given string from left to right and check if the current character is a non-zero character or not. If found to be true, then flip all the characters from the current index (= i) to R[i]th index. Follow the steps below to solve the problem:

• Initialize a variable, say flip, to store the number of times current character can be flipped.
• Create a priority queue, say pq to store the of range of indexes of characters on the right side of current character whose flip value is greater than 0.
• Initialize a variable, say cost to store the minimum cost to obtain the required string.
• Traverse the given string and check if the current character is ‘1‘ or not. If found to be true, then flip all the characters in the range i to R[i] and increment the value of cost by C[i].
• Finally, print the value of cost.

Below is the implementation of the above approach:

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to get the minimum` `// Cost to convert all characters` `// of given string to 0s` `int` `minCost(string str, ``int` `N, ``int` `R[], ``int` `C[])` `{` `    ``// Stores the range of indexes` `    ``// of characters that need` `    ``// to be flipped` `    ``priority_queue<``int``, vector<``int``>, greater<``int``> > pq;`   `    ``// Stores the number of times` `    ``// current character is flipped` `    ``int` `flip = 0;`   `    ``// Stores minimum cost to get` `    ``// the required string` `    ``int` `cost = 0;`   `    ``// Traverse the given string` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{` `        ``// Remove all value from pq` `        ``// whose value is less than i` `        ``while` `(pq.size() > 0 and pq.top() < i) ` `        ``{` `            ``pq.pop();`   `            ``// Update flip` `            ``flip--;` `        ``}`   `        ``// If current character` `        ``// is flipped odd times` `        ``if` `(flip % 2 == 1)` `        ``{` `            ``str[i] = ``'1'` `- str[i] + ``'0'``;` `        ``}`   `        ``// If current character contains` `        ``// non-zero value` `        ``if` `(str[i] == ``'1'``)` `        ``{` `            ``// Update flip` `            ``flip++;`   `            ``// Update cost` `            ``cost += C[i];`   `            ``// Append R[i] into pq` `            ``pq.push(R[i]);` `        ``}` `    ``}` `    ``return` `cost;` `}`   `// Driver Code` `int` `main()` `{` `    ``string str = ``"1010"``;` `    ``int` `R[] = { 1, 2, 2, 3 };` `    ``int` `C[] = { 3, 1, 2, 3 };` `    ``int` `N = str.length();` `  `  `    ``// Function call` `    ``cout << minCost(str, N, R, C);` `}`

 `// Java program to implement` `// the above approach`   `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`   `class` `GFG {`   `    ``// Function to get the minimum` `    ``// Cost to convert all characters` `    ``// of given string to 0s` `    ``public` `static` `int` `minCost(String s, ` `                              ``int` `R[], ` `                              ``int` `C[],` `                              ``int` `N)` `    ``{` `        ``char` `ch[] = s.toCharArray();`   `        ``// Stores the range of indexes` `        ``// of characters that need` `        ``// to be flipped` `        ``PriorityQueue pq = ``new` `PriorityQueue<>();`   `        ``// Stores the number of times` `        ``// current character is flipped` `        ``int` `flip = ``0``;`   `        ``// Stores minimum cost to get` `        ``// the required string` `        ``int` `cost = ``0``;`   `        ``// Traverse the given string` `        ``for` `(``int` `i = ``0``; i < N; i++) {`   `            ``// Remove all value from pq` `            ``// whose value is less than i`   `            ``while` `(pq.size() > ``0` `&& pq.peek() < i) ` `            ``{` `                ``pq.poll();`   `                ``// Update flip` `                ``flip--;` `            ``}`   `            ``// Get the current number` `            ``int` `cn = ch[i] - ``'0'``;`   `            ``// If current character` `            ``// is flipped odd times` `            ``if` `(flip % ``2` `== ``1``)` `                ``cn = ``1` `- cn;`   `            ``// If current character contains` `            ``// non-zero value` `            ``if` `(cn == ``1``) ` `            ``{` `                ``// Update flip` `                ``flip++;`   `                ``// Update cost` `                ``cost += C[i];`   `                ``// Append R[i] into pq` `                ``pq.add(R[i]);` `            ``}` `        ``}` `        ``return` `cost;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `N = ``4``;` `        ``String s = ``"1010"``;` `        ``int` `R[] = { ``1``, ``2``, ``2``, ``3` `};` `        ``int` `C[] = { ``3``, ``1``, ``2``, ``3` `};`   `        ``// Function call` `        ``System.out.println(minCost(s, R, C, N));` `    ``}` `}`

 `// C# program to implement` `// the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to get the minimum` `// Cost to convert all characters` `// of given string to 0s` `public` `static` `int` `minCost(String s, ``int` `[]R,` `                           ``int` `[]C, ``int` `N)` `{` `    ``char` `[]ch = s.ToCharArray();`   `    ``// Stores the range of indexes` `    ``// of characters that need` `    ``// to be flipped` `    ``Queue<``int``> pq = ``new` `Queue<``int``>();`   `    ``// Stores the number of times` `    ``// current character is flipped` `    ``int` `flip = 0;`   `    ``// Stores minimum cost to get` `    ``// the required string` `    ``int` `cost = 0;`   `    ``// Traverse the given string` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{` `        `  `        ``// Remove all value from pq` `        ``// whose value is less than i` `        ``while` `(pq.Count > 0 && pq.Peek() < i) ` `        ``{` `            ``pq.Dequeue();`   `            ``// Update flip` `            ``flip--;` `        ``}`   `        ``// Get the current number` `        ``int` `cn = ch[i] - ``'0'``;`   `        ``// If current character` `        ``// is flipped odd times` `        ``if` `(flip % 2 == 1)` `            ``cn = 1 - cn;`   `        ``// If current character contains` `        ``// non-zero value` `        ``if` `(cn == 1) ` `        ``{` `            `  `            ``// Update flip` `            ``flip++;`   `            ``// Update cost` `            ``cost += C[i];`   `            ``// Append R[i] into pq` `            ``pq.Enqueue(R[i]);` `            `  `        ``}` `    ``}` `    ``return` `cost;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `N = 4;` `    ``String s = ``"1010"``;` `    ``int` `[]R = { 1, 2, 2, 3 };` `    ``int` `[]C = { 3, 1, 2, 3 };`   `    ``// Function call` `    ``Console.WriteLine(minCost(s, R, C, N));` `}` `}`   `// This code is contributed by Amit Katiyar`

Output
```4

```

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

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