Given a binary string S, and an integer K, the task is to find the minimum number of flips required such that every substring of length K contains at least one ‘1’.
Examples:
Input: S = “10000001” K = 2
Output: 3
Explanation:
We need only 3 changes in string S ( at position 2, 4 and 6 ) so that the string contain at least one ‘1’ in every sub-string of length 2.
Input: S = “000000” K = 3
Output: 2
Explanation:
We need only 3 changes in string S ( at position 2 and 5 ) so that the string contain at least one ‘1’ in every sub-string of length 3.
Input: S = “111010111” K = 2
Output: 0
Naive Approach:
To solve the problem, the simplest approach is to iterate for each substring of length K and find the minimum number of flips required to satisfy the given condition.
Time complexity: O(N * K)
Space Complexity: O(1)
Efficient Approach:
The idea is to use Sliding Window Approach.
- Set a window size of K.
- Store the index of previous appearance of 1.
- If the current bit is unset and the difference between the current ith bit and the previous set bit exceeds K, set the current bit and store the current index as that of the previous set bit and proceed further.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count min flips int CountMinFlips(string s, int n,
int k)
{ // To store the count of minimum
// flip required
int cnt = 0;
// To store the position of last '1'
int prev = -1;
for ( int i = 0; i < k; i++) {
// Track last position of '1'
// in current window of size k
if (s[i] == '1' ) {
prev = i;
}
}
// If no '1' is present in the current
// window of size K
if (prev == -1) {
cnt++;
// Set last index of window '1'
s[k - 1] = '1' ;
// Track previous '1'
prev = k - 1;
}
// Traverse the given string
for ( int i = k; i < n; i++) {
// If the current bit is not set,
// then the condition for flipping
// the current bit
if (s[i] != '1' ) {
if (prev <= (i - k)) {
// Set i'th index to '1'
s[i] = '1' ;
// Track previous one
prev = i;
// Increment count
cnt++;
}
}
// Else update the prev set bit
// position to current position
else {
prev = i;
}
}
// Return the final count
return (cnt);
} // Driver Code int main()
{ // Given binary string
string str = "10000001" ;
// Size of given string str
int n = str.size();
// Window size
int k = 2;
// Function Call
cout << CountMinFlips(str, n, k)
<< endl;
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to count min flips static int CountMinFlips( char []s, int n,
int k)
{ // To store the count of minimum
// flip required
int cnt = 0 ;
// To store the position of last '1'
int prev = - 1 ;
for ( int i = 0 ; i < k; i++)
{
// Track last position of '1'
// in current window of size k
if (s[i] == '1' )
{
prev = i;
}
}
// If no '1' is present in the current
// window of size K
if (prev == - 1 )
{
cnt++;
// Set last index of window '1'
s[k - 1 ] = '1' ;
// Track previous '1'
prev = k - 1 ;
}
// Traverse the given String
for ( int i = k; i < n; i++)
{
// If the current bit is not set,
// then the condition for flipping
// the current bit
if (s[i] != '1' )
{
if (prev <= (i - k))
{
// Set i'th index to '1'
s[i] = '1' ;
// Track previous one
prev = i;
// Increment count
cnt++;
}
}
// Else update the prev set bit
// position to current position
else
{
prev = i;
}
}
// Return the final count
return (cnt);
} // Driver Code public static void main(String[] args)
{ // Given binary String
String str = "10000001" ;
// Size of given String str
int n = str.length();
// Window size
int k = 2 ;
// Function Call
System.out.print(CountMinFlips(
str.toCharArray(), n, k) + "\n" );
} } // This code is contributed by Rohit_ranjan |
# Python3 code to count minimum no. # of flips required such that # every substring of length K # contain at least one '1'. # Function to count min flips def CountMinFlips(s, n, k):
cnt = 0
prev = - 1
for i in range ( 0 , k):
# Track last position of '1'
# in current window of size k
if (s[i] = = '1' ):
prev = i;
# means no '1' is present
if (prev = = - 1 ):
cnt + = 1
# track previous '1'
prev = k - 1 ;
for i in range (k, n):
if (s[i] ! = '1' ):
if ( prev < = (i - k) ):
# track previous one
prev = i;
# increment count
cnt + = 1
else :
prev = i
return (cnt);
# Driver code s = "10000001"
n = len (s)
k = 2
print (CountMinFlips(s, n, k))
|
// C# program for the above approach using System;
class GFG{
// Function to count min flips static int CountMinFlips( char []s, int n,
int k)
{ // To store the count of minimum
// flip required
int cnt = 0;
// To store the position of last '1'
int prev = -1;
for ( int i = 0; i < k; i++)
{
// Track last position of '1'
// in current window of size k
if (s[i] == '1' )
{
prev = i;
}
}
// If no '1' is present in the current
// window of size K
if (prev == -1)
{
cnt++;
// Set last index of window '1'
s[k - 1] = '1' ;
// Track previous '1'
prev = k - 1;
}
// Traverse the given String
for ( int i = k; i < n; i++)
{
// If the current bit is not set,
// then the condition for flipping
// the current bit
if (s[i] != '1' )
{
if (prev <= (i - k))
{
// Set i'th index to '1'
s[i] = '1' ;
// Track previous one
prev = i;
// Increment count
cnt++;
}
}
// Else update the prev set bit
// position to current position
else
{
prev = i;
}
}
// Return the readonly count
return (cnt);
} // Driver Code public static void Main(String[] args)
{ // Given binary String
String str = "10000001" ;
// Size of given String str
int n = str.Length;
// Window size
int k = 2;
// Function Call
Console.Write(CountMinFlips(
str.ToCharArray(), n, k) + "\n" );
} } // This code is contributed by sapnasingh4991 |
<script> // javascript program for the above approach // Function to count min flips function CountMinFlips(s , n,k)
{ // To store the count of minimum
// flip required
var cnt = 0;
// To store the position of last '1'
var prev = -1;
for (i = 0; i < k; i++)
{
// Track last position of '1'
// in current window of size k
if (s[i] == '1' )
{
prev = i;
}
}
// If no '1' is present in the current
// window of size K
if (prev == -1)
{
cnt++;
// Set last index of window '1'
s[k - 1] = '1' ;
// Track previous '1'
prev = k - 1;
}
// Traverse the given String
for (i = k; i < n; i++)
{
// If the current bit is not set,
// then the condition for flipping
// the current bit
if (s[i] != '1' )
{
if (prev <= (i - k))
{
// Set i'th index to '1'
s[i] = '1';
// Track previous one
prev = i;
// Increment count
cnt++;
}
}
// Else update the prev set bit
// position to current position
else
{
prev = i;
}
}
// Return the final count
return (cnt);
} // Driver Code // Given binary String
var str = "10000001" ;
// Size of given String str
var n = str.length;
// Window size
var k = 2;
// Function Call
document.write(CountMinFlips(
str, n, k) );
// This code contributed by shikhasingrajput </script> |
3
Time complexity: O(N + K)
Auxiliary Space: O(1)