Minimizing Binary statements: Transforming a Binary String to all Zeroes

Given a set of N binary statements represented by a string S, where '1' represents a true statement and '0' represents a false one. You can perform the following operation multiple times: choose two indices i and j (1 ≤ i < j ≤ N) such that j - i ≥ 2 and toggle the states of statements i and j, the task is to determine whether it is possible to make all N statements false using these operations. If possible, calculate the minimum number of operations needed to achieve this. If it's not possible to make all statements false, return -1.

Examples:

Input: N = 6, S = 101101
Output: 2
Explanation: Here, we can perform the operation with ((i, j) = (1, 3) and then with (i, j) = (4, 6) to make all the statements false.

Input: N = 4, S = 1111
Output: -1
Explanation: Here, we can't make all statements false with any operation.

Approach: To solve the problem follow the below observations:

• Case I: In each operation, the count of 1  increases by 0, +2, or -2, so the objective is unachievable if the count of 1 is odd. In all the below cases we consider the count of 1 is even.
• Case II: If there is no 1, then nothing to do.
• Case III: If a count of 1 is greater than or equal to 4, then Let x1​, x2,…,xk(k is the count of 1) be the true statements from left to right. Then, one can achieve the objective in k/2 ​ operations by changing the states of statements 1 and x_k/2+1​, statements 2 and x_k/2+2, and so on.
• Case IV: The count of 1 is 2, If the true statements are not adjacent, the answer is 1. Below, assume that they are adjacent, and let X be the statement to the left and Y be the statement to the right.
  • If  X >=3 , one can achieve the objective in two operations by changing the states of staements 1 and X and then by changing the states of staements 1 and Y.
  • If X+1 ≤ N-2, one can again achieve the objective in two operations by changing the states of staements 1 and N and then by changing the states of staements X and Y.
• Case V: Neither of the above is satisfied if X < 3 and X+1>N-2, which happens only if N=3 or 4.
  • If N=3, then S is 011 or 110, in which case one can only flip coins 1 and 3, so the objective is unachievable.
  • If N=4, X<3, and X+1>N-2, then S is 0110, in which case the objective is achievable in three operations (0110→1111→0101→0000).

Steps to implement the above approach:

• Read the integer 'n', the number of statements, and the string 's', representing the current state of the statements.
• Count the number of '1's in the string 's' using the count() function. Let's call this count 'one'.
• Check if the count 'one' is odd:
  • If odd, return -1, indicating that it's not possible to make all statements false.
• Initialize a boolean variable 'adj' as false. This will be used to check if adjacent pairs of '1's exist.
• Iterate through the characters of the string 's' from index 0 to 'n-2':
  • If the current character and the next character are both '1's, set 'adj' to true.
• If 'one' is not equal to 2 or 'adj' is false, calculate the minimum number of operations needed:
  • If 'one' is greater than or equal to 4, or 'one' is 0, or 'adj' is false, return 'one / 2' as the minimum number of operations.
• If 'one' is equal to 2 and 'adj' is true, find the position of the leftmost '1' in the string 's'. Let's call this position 'x'.
• Check for special cases:
  • If 'n' is 3, return -1 as it's not possible to achieve the objective.
  • If 'n' is 4 and 's' is "0110", return 3 as it can be achieved in 3 operations.
• If none of the special cases apply, check if the position 'x' is greater than or equal to 3:
  • If yes, return 2 as the objective can be achieved by flipping statements 1 and 'x', then statements 1 and the position of the second '1'.
• If 'x+1' is less than or equal to 'n-2':
  • If yes, return 2 as the objective can be achieved by flipping statements 1 and 'n', then statements 'x' and the position of the second '1'.
• If none of the above conditions are satisfied, return -1 as the objective cannot be achieved.

Below is the implementation of the above approach:

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

// Function to solve the
// problem for a single test case
int IsPossibleToMakeAllFalse(string s, int n)
{

    // Count the number of
    //'1's in the string S
    int one = count(s.begin(), s.end(), '1');

    // If the count 'one' is odd,
    // it is not possible to make
    // all statements are false
    if (one & 1)
        return -1;

    // Check if the '1's in the
    // string S are adjacent or not
    bool adj = 0;
    for (int i = 0; i < n - 1; i++)
        if (s[i] == '1' and s[i] == s[i + 1])
            adj = 1;

    // If the count 'one' is not 2
    // or the '1's are not
    // adjacent, the objective is
    // achievable in 'one/2'
    // operations
    if (one != 2 or !adj)
        return one / 2;

    // If the '1's are adjacent,
    // find the position of the
    // leftmost '1'
    int x = 0;
    while (s[x] != '1')
        x++;

    // Check for special cases
    if (n == 3)
        return -1;

    if (n == 4 and s == "0110")
        return 3;

    // If 'x' is greater than or
    // equal to 3, perform two operations:

    // 1. Flip statements 1 and 'x'
    // 2. Flip statements 1 and 'y', where
    // 'y' is the position of the second '1'
    if (x >= 3)
        return 2;

    // If 'x+1' is less than or equal
    // to 'N-2', perform two operations:
    // 1. Flip statements 1 and 'N'
    // 2. Flip statements 'x' and 'y',
    // where 'y' is the
    // position of the second '1'
    if (x + 1 <= n - 2)
        return 2;

    // If neither of the above
    // conditions is satisfied,
    // return -1 (unachievable)
    return -1;
}

// Driver Code
int main()
{
    int n = 6;
    string s = "101101";

    // Function Call
    cout << IsPossibleToMakeAllFalse(s, n) << endl;

    return 0;
}

import java.util.Scanner;

public class Main {
    // Function to solve the problem for a single test case
    public static int isPossibleToMakeAllFalse(String s, int n) {
        // Count the number of '1's in the string s
        int one = 0;
        for (char c : s.toCharArray()) {
            if (c == '1') {
                one++;
            }
        }

        // If the count 'one' is odd, it is not possible to make all statements false
        if (one % 2 != 0) {
            return -1;
        }

        // Check if the '1's in the string s are adjacent or not
        boolean adj = false;
        for (int i = 0; i < n - 1; i++) {
            if (s.charAt(i) == '1' && s.charAt(i) == s.charAt(i + 1)) {
                adj = true;
                break;
            }
        }

        // If the count 'one' is not 2 or the '1's are not adjacent, 
        // the objective is achievable in 'one/2' operations
        if (one != 2 || !adj) {
            return one / 2;
        }

        // If the '1's are adjacent, find the position of the leftmost '1'
        int x = 0;
        while (s.charAt(x) != '1') {
            x++;
        }

        // Check for special cases
        if (n == 3) {
            return -1;
        }

        if (n == 4 && s.equals("0110")) {
            return 3;
        }

        // If 'x' is greater than or equal to 3, perform two operations:
        // 1. Flip statements 1 and 'x'
        // 2. Flip statements 1 and 'y', where 'y' is the position of the second '1'
        if (x >= 3) {
            return 2;
        }

        // If 'x+1' is less than or equal to 'N-2', perform two operations:
        // 1. Flip statements 1 and 'N'
        // 2. Flip statements 'x' and 'y', where 'y' is the position of the second '1'
        if (x + 1 <= n - 2) {
            return 2;
        }

        // If neither of the above conditions is satisfied, return -1 (unachievable)
        return -1;
    }

    public static void main(String[] args) {
        int n = 6;
        String s = "101101";

        // Function Call
        System.out.println(isPossibleToMakeAllFalse(s, n));
    }
}

# Function to solve the
# problem for a single test case
def IsPossibleToMakeAllFalse(s, n):
    # Count the number of
    #'1's in the string S
    one = s.count('1')
    # If the count 'one' is odd,
    # it is not possible to make
    # all statements are false
    if one & 1:
        return -1
    # Check if the '1's in the
    # string S are adjacent or not
    adj = False
    for i in range(n - 1):
        if s[i] == '1' and s[i] == s[i + 1]:
            adj = True
    # If the count 'one' is not 2
    # or the '1's are not
    # adjacent, the objective is
    # achievable in 'one/2'
    # operations
    if one != 2 or not adj:
        return one // 2
    # If the '1's are adjacent,
    # find the position of the
    # leftmost '1'
    x = 0
    while s[x] != '1':
        x += 1
    # Check for special cases
    if n == 3:
        return -1
    if n == 4 and s == "0110":
        return 3
    # If 'x' is greater than or
    # equal to 3, perform two operations:
    # 1. Flip statements 1 and 'x'
    # 2. Flip statements 1 and 'y', where
    # 'y' is the position of the second '1'
    if x >= 3:
        return 2
    # If 'x+1' is less than or equal
    # to 'N-2', perform two operations:
    # 1. Flip statements 1 and 'N'
    # 2. Flip statements 'x' and 'y',
    # where 'y' is the
    # position of the second '1'
    if x + 1 <= n - 2:
        return 2
    # If neither of the above
    # conditions is satisfied,
    # return -1 (unachievable)
    return -1

# Driver Code
n = 6
s = "101101"
# Function Call
print(IsPossibleToMakeAllFalse(s, n))

using System;

class Program
{
    // Function to solve the problem for a vsingle test case
    static int IsPossibleToMakeAllFalse(string s, int n)
    {
        // Count the number of '1's in the string s
        int one = s.Split('1').Length - 1;

        // If the count 'one' is odd, it is not possible to make all statements false
        if (one % 2 != 0)
            return -1;

        // Check if the '1's in the string s are adjacent or not
        bool adj = false;
        for (int i = 0; i < n - 1; i++)
        {
            if (s[i] == '1' && s[i] == s[i + 1])
            {
                adj = true;
                break;
            }
        }

        // If the count 'one' is not 2 or the '1's are not adjacent, the objective is achievable in 'one/2' operations
        if (one != 2 || !adj)
            return one / 2;

        // If the '1's are adjacent, find the position of the leftmost '1'
        int x = 0;
        while (s[x] != '1')
        {
            x++;
        }

        // Check for special cases
        if (n == 3)
            return -1;

        if (n == 4 && s == "0110")
            return 3;

        // If 'x' is greater than or equal to 3, perform two operations:
        // 1. Flip statements 1 and 'x'
        // 2. Flip statements 1 and 'y', where 'y' is the position of the second '1'
        if (x >= 3)
            return 2;

        // If 'x+1' is less than or equal to 'n-2', perform two operations:
        // 1. Flip statements 1 and 'n'
        // 2. Flip statements 'x' and 'y', where 'y' is the position of the second '1'
        if (x + 1 <= n - 2)
            return 2;

        // If neither of the above conditions is satisfied, return -1 (unachievable)
        return -1;
    }

    // Driver Code
    static void Main()
    {
        int n = 6;
        string s = "101101";

        // Function Call
        Console.WriteLine(IsPossibleToMakeAllFalse(s, n));
    }
}
//This code is Contributed by chinmaya121221

function isPossibleToMakeAllFalse(s, n) {
    // Count the number of '1's in the string s
    const one = s.split('1').length - 1;

    // If the count 'one' is odd, it is not possible to make all statements false
    if (one % 2 !== 0)
        return -1;

    // Check if the '1's in the string s are adjacent or not
    let adj = false;
    for (let i = 0; i < n - 1; i++) {
        if (s[i] === '1' && s[i] === s[i + 1]) {
            adj = true;
            break;
        }
    }

    // If the count 'one' is not 2 or the '1's are not adjacent, the objective is achievable in 'one/2' operations
    if (one !== 2 || !adj)
        return Math.floor(one / 2);

    // If the '1's are adjacent, find the position of the leftmost '1'
    let x = 0;
    while (s[x] !== '1') {
        x++;
    }

    // Check for special cases
    if (n === 3)
        return -1;

    if (n === 4 && s === "0110")
        return 3;

    // If 'x' is greater than or equal to 3, perform two operations:
    // 1. Flip statements 1 and 'x'
    // 2. Flip statements 1 and 'y', where 'y' is the position of the second '1'
    if (x >= 3)
        return 2;

    // If 'x+1' is less than or equal to 'n-2', perform two operations:
    // 1. Flip statements 1 and 'n'
    // 2. Flip statements 'x' and 'y', where 'y' is the position of the second '1'
    if (x + 1 <= n - 2)
        return 2;

    // If neither of the above conditions is satisfied, return -1 (unachievable)
    return -1;
}

// Test case
const n = 6;
const s = "101101";

// Function call
console.log(isPossibleToMakeAllFalse(s, n));

2

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