Minimize cost to make all characters of a Binary String equal to ‘1’ by reversing or flipping characters of substrings

Minimize cost to make all characters of a Binary String equal to ‘1’ by reversing or flipping characters of substrings

Given a binary string S, and two integers A, which denotes the cost of reversing a substring, and B, which denotes the cost of flipping all characters of a substring, the task is to find the minimum cost to reduce the string S to 1s only.

Examples:

Input: S = “01100”, A = 1, B = 5
Output: 6
Explanation: 
One possible way to make all characters equal to ‘1’ is as follows:

1. Reverse the substring {S[0], S[2]}. Cost = A (= 1), The string modifies to “11000”.
2. Flip the characters of substring {S[2], S[4]}. Cost of B (= 5). The string modifies to “11111”.

Therefore, the total cost = 5+1 = 6 (which is the minimum cost possible)

Input: S = “1111111”, A = 2, B = 3
Output: 0

Approach: The given problem can be solved based on the following observations: 

• Assuming there are P disjoint groups of continuous ‘0’s,
• If A is less than B, then it is optimal to convert P groups into ‘1’ group of continuous ‘0’s by performing the first type of operation for a cost of (P – 1) * A and then converting all the ‘0’s to ‘1’s for a cost of B.
• Otherwise, it is optimal to perform the second operation on each group separately, for a cost of B * P.

Follow the steps below to solve the problem:

• Initialize a variable say P with 0 value to store the count of disjoint groups of continuous 0s.
• Also, initialize a variable say count as 0 to store the temporary count of the number of 0s in a group.
• Iterate over the character of the string S and do the following:
  • If the current character is ‘0‘ then increment the count by 1.
  • Otherwise, if the count is greater than 0 then increment P by 1 and then assign 0 to count.
• If the count is greater than 0 then increment P by 1.
• Print the minimum cost obtained as (P-1)*A+B.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h> using namespace std;    void MinimumCost(string S, int A, int B) {          int count = 0;                  int group = 0;             for (int i = 0; i < S.size(); i++) {                     if (S[i] == '0') {             count += 1;         }         else {                                       if (count > 0) {                 group += 1;             }                             count = 0;         }     }                  if (count > 0)         group += 1;                  if (group == 0) {         cout << 0 << endl;     }     else {                  cout << min(A, B) * (group - 1) + B;     } }    int main() {             int A = 1;     int B = 5;     string S = "01100";             MinimumCost(S, A, B);     return 0; }

Time Complexity: O(N)
Auxiliary Space: O(1)

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