Minimize the cost to convert given string to either type XYXY... or XXYY...

Last Updated : 09 Feb, 2022

Given a binary string s and an array ct[] of even size. The task is to minimize the cost of operations to:

Convert string s to a string of either type XYXYXYXYXY... or XXYYXXYYXXYY...
In one operation it is allowed to flip i^th character with cost ct[i].

Examples

Input: s = "1011" ct = {1, 2, 1, 3}
Output: 1
Explanation: Follow operations are performed to convert s to given format:
Flip the 0th bit from 1 to 0, the updated s = "0011" which is in the form XXYY.
Hence, 1 is the answer which is minimum possible.

Input: "1010"
Output: 0
Explanation: The string is already in a given format.

Approach: This problem is observation-based. Follow the steps below to solve the given problem.

There are only four types of binary strings, according to the problem.
- 010101010101.......
- 101010101010.......
- 001100110011........
- 110011001100........
We have to check for these four different strings only.
Pass the string along with cost and the first four expected characters.
If the expected characters do not match with the string's actual characters then add the cost corresponding to the ith value.
Repeat the procedure for all four expected sequences and choose the minimum cost out of it.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

int solve_util(string s, int c[],
               char x, char y,
               char z, char w)
{
    int ans = 0;
    for (int i = 0; i < s.length(); i += 4) {
        if (s[i] != x)
            ans += c[i];
        if (i + 1 < s.length()
            && s[i + 1] != y)
            ans += c[i + 1];
        if (i + 2 < s.length()
            && s[i + 2] != z)
            ans += c[i + 2];
        if (i + 3 < s.length()
            && s[i + 3] != w)
            ans += c[i + 3];
    }
    return ans;
}

int solve_util2(string s, int c[],
                char x, char y)
{
    int ans = 0;
    if (s[0] != x)
        ans += c[0];
    if (s[1] != y)
        ans += c[1];
    return ans;
}

// Function to convert given
// string to required form
int minOperations(int N, string S, int C[])
{
    // code here
    if (S.length() == 2) {
        int x = solve_util2(
            S, C, '0', '1');
        int y = solve_util2(
            S, C, '1', '0');
        int z = solve_util2(
            S, C, '1', '1');
        int w = solve_util2(
            S, C, '0', '0');
        return min({ x, y, z, w });
    }
    int x = solve_util(
        S, C, '0', '1', '0', '1');
    int y = solve_util(
        S, C, '1', '0', '1', '0');
    int z = solve_util(
        S, C, '1', '1', '0', '0');
    int w = solve_util(
        S, C, '0', '0', '1', '1');
    return min({ x, y, z, w });
}

// Driver Code
int main()
{
    int N = 4;
    string s = "1011";
    int ct[] = { 1, 2, 1, 3 };

    cout << minOperations(N, s, ct) << "\n";
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {


  static int solve_util(char s[], int c[],
                        char x, char y,
                        char z, char w)
  {
    int ans = 0;
    for (int i = 0; i < s.length; i += 4) {
      if (s[i] != x)
        ans += c[i];
      if (i + 1 < s.length
          && s[i + 1] != y)
        ans += c[i + 1];
      if (i + 2 < s.length
          && s[i + 2] != z)
        ans += c[i + 2];
      if (i + 3 < s.length
          && s[i + 3] != w)
        ans += c[i + 3];
    }
    return ans;
  }

  static int solve_util2(char s[], int c[],
                         char x, char y)
  {
    int ans = 0;
    if (s[0] != x)
      ans += c[0];
    if (s[1] != y)
      ans += c[1];
    return ans;
  }

  // Function to convert given
  // string to required form
  static int minOperations(int N, char S[], int C[])
  {
    // code here
    if (S.length == 2) {
      int x = solve_util2(
        S, C, '0', '1');
      int y = solve_util2(
        S, C, '1', '0');
      int z = solve_util2(
        S, C, '1', '1');
      int w = solve_util2(
        S, C, '0', '0');
      return Math.min(x, Math.min( y, Math.min(z, w )));
    }
    int x = solve_util(
      S, C, '0', '1', '0', '1');
    int y = solve_util(
      S, C, '1', '0', '1', '0');
    int z = solve_util(
      S, C, '1', '1', '0', '0');
    int w = solve_util(
      S, C, '0', '0', '1', '1');
    return Math.min(x, Math.min( y, Math.min(z, w )));
  }

  // Driver Code
  public static void main (String[] args) {
    int N = 4;
    char s[] = {'1', '0', '1', '1'};
    int ct[] = { 1, 2, 1, 3 };

    System.out.print(minOperations(N, s, ct));
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python program for above approach
def solve_util(s, c, x, y, z, w):
    ans = 0
    for i in range(0, len(s), 4):
        if (s[i] != x):
            ans += c[i]
        if (i + 1 < len(s)
                and s[i + 1] != y):
            ans += c[i + 1]
        if (i + 2 < len(s)
                and s[i + 2] != z):
            ans += c[i + 2]
        if (i + 3 < len(s)
                and s[i + 3] != w):
            ans += c[i + 3]

    return ans

def solve_util2(s, c, x, y):
    ans = 0
    if (s[0] != x):
        ans += c[0]
    if (s[1] != y):
        ans += c[1]
    return ans

# Function to convert given
# string to required form
def minOperations(N, S, C):

    # code here
    if (len(S) == 2):
        x = solve_util2(S, C, '0', '1')
        y = solve_util2(S, C, '1', '0')
        z = solve_util2(S, C, '1', '1')
        w = solve_util2(S, C, '0', '0')
        print(f"{x},{y},{z},{w}")
        return min([x, y, z, w])

    x = solve_util(S, C, '0', '1', '0', '1')
    y = solve_util(S, C, '1', '0', '1', '0')
    z = solve_util(S, C, '1', '1', '0', '0')
    w = solve_util(S, C, '0', '0', '1', '1')

    return min([x, y, z, w])

# Driver Code
N = 4
s = "1011"
ct = [1, 2, 1, 3]

print(minOperations(N, s, ct))

# This code is contributed by gfgking

// C# program for the above approach
using System;
class GFG
{

  static int solve_util(char []s, int []c,
                        char x, char y,
                        char z, char w)
  {
    int ans = 0;
    for (int i = 0; i < s.Length; i += 4) {
      if (s[i] != x)
        ans += c[i];
      if (i + 1 < s.Length
          && s[i + 1] != y)
        ans += c[i + 1];
      if (i + 2 < s.Length
          && s[i + 2] != z)
        ans += c[i + 2];
      if (i + 3 < s.Length
          && s[i + 3] != w)
        ans += c[i + 3];
    }
    return ans;
  }

  static int solve_util2(char []s, int []c,
                         char x, char y)
  {
    int ans = 0;
    if (s[0] != x)
      ans += c[0];
    if (s[1] != y)
      ans += c[1];
    return ans;
  }

  // Function to convert given
  // string to required form
  static int minOperations(int N, char []S, int []C)
  {
    // code here
    if (S.Length == 2) {
      int x = solve_util2(
        S, C, '0', '1');
      int y = solve_util2(
        S, C, '1', '0');
      int z = solve_util2(
        S, C, '1', '1');
      int w = solve_util2(
        S, C, '0', '0');
      return Math.Min(x, Math.Min( y, Math.Min(z, w )));
    }
    
    else {
      int x = solve_util(
        S, C, '0', '1', '0', '1');
      int y = solve_util(
        S, C, '1', '0', '1', '0');
      int z = solve_util(
        S, C, '1', '1', '0', '0');
      int w = solve_util(
        S, C, '0', '0', '1', '1');
      return Math.Min(x, Math.Min( y, Math.Min(z, w )));
    }
  }

  // Driver Code
  public static void Main () {
    int N = 4;
    char []s = {'1', '0', '1', '1'};
    int []ct = { 1, 2, 1, 3 };

    Console.Write(minOperations(N, s, ct));
  }
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

    <script>
        // JavaScript program for above approach

        const solve_util = (s, c, x, y, z, w) => {
            let ans = 0;
            for (let i = 0; i < s.length; i += 4) {
                if (s[i] != x)
                    ans += c[i];
                if (i + 1 < s.length
                    && s[i + 1] != y)
                    ans += c[i + 1];
                if (i + 2 < s.length
                    && s[i + 2] != z)
                    ans += c[i + 2];
                if (i + 3 < s.length
                    && s[i + 3] != w)
                    ans += c[i + 3];
            }
            return ans;
        }

        const solve_util2 = (s, c, x, y) => {
            let ans = 0;
            if (s[0] != x)
                ans += c[0];
            if (s[1] != y)
                ans += c[1];
            return ans;
        }

        // Function to convert given
        // string to required form
        const minOperations = (N, S, C) => {
        
            // code here
            if (S.length == 2) {
                let x = solve_util2(S, C, '0', '1');
                let y = solve_util2(S, C, '1', '0');
                let z = solve_util2(S, C, '1', '1');
                let w = solve_util2(S, C, '0', '0');
                document.write(`${x},${y},${z},${w}`);
                return Math.min(...[x, y, z, w]);
            }
            let x = solve_util(S, C, '0', '1', '0', '1');
            let y = solve_util(S, C, '1', '0', '1', '0');
            let z = solve_util(S, C, '1', '1', '0', '0');
            let w = solve_util(S, C, '0', '0', '1', '1');

            return Math.min(...[x, y, z, w]);
        }

        // Driver Code
        let N = 4;
        let s = "1011";
        let ct = [1, 2, 1, 3];

        document.write(minOperations(N, s, ct));

        // This code is contributed by rakeshsahni

    </script>

Output

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

Minimize cost to convert given two integers to zero using given operations

harshalkhond

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Minimize the cost to convert given string to either type XYXY... or XXYY...

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