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++ 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)