Minimizing Binary statements: Transforming a Binary String to all Zeroes
Last Updated :
07 Nov, 2023
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 xk/2+1, statements 2 and xk/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++
#include <bits/stdc++.h>
using namespace std;
int IsPossibleToMakeAllFalse(string s, int n)
{
int one = count(s.begin(), s.end(), '1');
if (one & 1)
return -1;
bool adj = 0;
for (int i = 0; i < n - 1; i++)
if (s[i] == '1' and s[i] == s[i + 1])
adj = 1;
if (one != 2 or !adj)
return one / 2;
int x = 0;
while (s[x] != '1')
x++;
if (n == 3)
return -1;
if (n == 4 and s == "0110")
return 3;
if (x >= 3)
return 2;
if (x + 1 <= n - 2)
return 2;
return -1;
}
int main()
{
int n = 6;
string s = "101101";
cout << IsPossibleToMakeAllFalse(s, n) << endl;
return 0;
}
|
Java
import java.util.Scanner;
public class Main {
public static int isPossibleToMakeAllFalse(String s, int n) {
int one = 0;
for (char c : s.toCharArray()) {
if (c == '1') {
one++;
}
}
if (one % 2 != 0) {
return -1;
}
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 (one != 2 || !adj) {
return one / 2;
}
int x = 0;
while (s.charAt(x) != '1') {
x++;
}
if (n == 3) {
return -1;
}
if (n == 4 && s.equals("0110")) {
return 3;
}
if (x >= 3) {
return 2;
}
if (x + 1 <= n - 2) {
return 2;
}
return -1;
}
public static void main(String[] args) {
int n = 6;
String s = "101101";
System.out.println(isPossibleToMakeAllFalse(s, n));
}
}
|
Python3
def IsPossibleToMakeAllFalse(s, n):
one = s.count('1')
if one & 1:
return -1
adj = False
for i in range(n - 1):
if s[i] == '1' and s[i] == s[i + 1]:
adj = True
if one != 2 or not adj:
return one // 2
x = 0
while s[x] != '1':
x += 1
if n == 3:
return -1
if n == 4 and s == "0110":
return 3
if x >= 3:
return 2
if x + 1 <= n - 2:
return 2
return -1
n = 6
s = "101101"
print(IsPossibleToMakeAllFalse(s, n))
|
C#
using System;
class Program
{
static int IsPossibleToMakeAllFalse(string s, int n)
{
int one = s.Split('1').Length - 1;
if (one % 2 != 0)
return -1;
bool adj = false;
for (int i = 0; i < n - 1; i++)
{
if (s[i] == '1' && s[i] == s[i + 1])
{
adj = true;
break;
}
}
if (one != 2 || !adj)
return one / 2;
int x = 0;
while (s[x] != '1')
{
x++;
}
if (n == 3)
return -1;
if (n == 4 && s == "0110")
return 3;
if (x >= 3)
return 2;
if (x + 1 <= n - 2)
return 2;
return -1;
}
static void Main()
{
int n = 6;
string s = "101101";
Console.WriteLine(IsPossibleToMakeAllFalse(s, n));
}
}
|
Javascript
function isPossibleToMakeAllFalse(s, n) {
const one = s.split('1').length - 1;
if (one % 2 !== 0)
return -1;
let adj = false;
for (let i = 0; i < n - 1; i++) {
if (s[i] === '1' && s[i] === s[i + 1]) {
adj = true;
break;
}
}
if (one !== 2 || !adj)
return Math.floor(one / 2);
let x = 0;
while (s[x] !== '1') {
x++;
}
if (n === 3)
return -1;
if (n === 4 && s === "0110")
return 3;
if (x >= 3)
return 2;
if (x + 1 <= n - 2)
return 2;
return -1;
}
const n = 6;
const s = "101101";
console.log(isPossibleToMakeAllFalse(s, n));
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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