Given an integer N, the task is to find the length of the longest substring containing only 4’s from the first N characters of the infinite string str.
The string str is generated by concatenating the numbers formed by only 4’s and 5’s in increasing order. For example 4, 5, 44, 45, 54, 55 and so on. Therefore the string str looks like “4544455455444445454455…”.
Examples:
Input : N = 4 Output : 2 First 4 characters of str are "4544". Therefore the required length is 2. Input : N = 10 Output : 3 First 10 characters of str are "4544455455". Therefore the required length is 3.
Approach: The problem can be solved easily by observing the pattern. The task is to count the maximum consecutive 4’s appearing in the string. So, there is no need to generate the whole string.
We can observe a pattern if we divide the string into different groups as the first group will have 2 characters, the second group will have 4 characters, the third group will have 8 characters, and so on…
For Example:
Group 1 -> 45
Group 2 -> 44455455
Group 3 -> 444445454455544545554555
.
.
.
and, so on…
Now, the task reduces to finding the group in which N lies, and how many characters it covers in that group from start.
Here,
- If N falls in group 2, the answer will be at least 3. That is if, length = 4 then the answer will be 2 as with length 4, the string will cover only up to the 2nd 4 in the group, and if length = 5 answer will be 3.
- Similarly, if length covers at least the first 5 “4’s” from group 3, the answer is 5.
Now,
Group 1 has 1 * 2^1 characters
Group 2 has 2 * 2^2 characters
Generally, group K has K * 2^K characters. So the problem reduces to finding to which group, the given N belongs to. This can be easily found by using the prefix sum array pre[] where the ith element contains the sum of a number of characters up to the ith group.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define MAXN 30 // Function to return the length of longest // contiguous string containing only 4’s from // the first N characters of the string int countMaxLength( int N)
{ // Initialize result
int res;
// Initialize prefix sum array of
// characters and product variable
int pre[MAXN], p = 1;
// Preprocessing of prefix sum array
pre[0] = 0;
for ( int i = 1; i < MAXN; i++) {
p *= 2;
pre[i] = pre[i - 1] + i * p;
}
// Initialize variable to store the
// string length where N belongs to
int ind;
// Finding the string length where
// N belongs to
for ( int i = 1; i < MAXN; i++) {
if (pre[i] >= N) {
ind = i;
break ;
}
}
int x = N - pre[ind - 1];
int y = 2 * ind - 1;
if (x >= y)
res = min(x, y);
else
res = max(x, 2 * (ind - 2) + 1);
return res;
} // Driver Code int main()
{ int N = 25;
cout << countMaxLength(N);
return 0;
} |
// Java implementation of the approach class GFG
{ static int MAXN = 30 ;
// Function to return the length of longest // contiguous string containing only 4's from // the first N characters of the string static int countMaxLength( int N)
{ // Initialize result
int res;
// Initialize prefix sum array of
// characters and product variable
int pre[] = new int [MAXN];
int p = 1 ;
// Preprocessing of prefix sum array
pre[ 0 ] = 0 ;
for ( int i = 1 ; i < MAXN; i++)
{
p *= 2 ;
pre[i] = pre[i - 1 ] + i * p;
}
// Initialize variable to store the
// string length where N belongs to
int ind = 0 ;
// Finding the string length where
// N belongs to
for ( int i = 1 ; i < MAXN; i++)
{
if (pre[i] >= N)
{
ind = i;
break ;
}
}
int x = N - pre[ind - 1 ];
int y = 2 * ind - 1 ;
if (x >= y)
res = Math.min(x, y);
else
res = Math.max(x, 2 * (ind - 2 ) + 1 );
return res;
} // Driver Code public static void main(String[] args)
{ int N = 25 ;
System.out.println(countMaxLength(N));
} } // This code is contributed by Code_Mech. |
# Python 3 implementation of the approach MAXN = 30
# Function to return the length of longest # contiguous string containing only 4’s from # the first N characters of the string def countMaxLength(N):
# Initialize result
# Initialize prefix sum array of
# characters and product variable
pre = [ 0 for i in range (MAXN)]
p = 1
# Preprocessing of prefix sum array
pre[ 0 ] = 0
for i in range ( 1 , MAXN, 1 ):
p * = 2
pre[i] = pre[i - 1 ] + i * p
# Initialize variable to store the
# string length where N belongs to
# Finding the string length where
# N belongs to
for i in range ( 1 , MAXN, 1 ):
if (pre[i] > = N):
ind = i
break
x = N - pre[ind - 1 ]
y = 2 * ind - 1
if (x > = y):
res = min (x, y)
else :
res = max (x, 2 * (ind - 2 ) + 1 )
return res
# Driver Code if __name__ = = '__main__' :
N = 25
print (countMaxLength(N))
# This code is contributed by # Surendra_Gangwar |
// C# implementation of the approach using System;
class GFG
{ static int MAXN = 30;
// Function to return the length of longest // contiguous string containing only 4's from // the first N characters of the string static int countMaxLength( int N)
{ // Initialize result
int res;
// Initialize prefix sum array of
// characters and product variable
int [] pre = new int [MAXN];
int p = 1;
// Preprocessing of prefix sum array
pre[0] = 0;
for ( int i = 1; i < MAXN; i++)
{
p *= 2;
pre[i] = pre[i - 1] + i * p;
}
// Initialize variable to store the
// string length where N belongs to
int ind = 0;
// Finding the string length where
// N belongs to
for ( int i = 1; i < MAXN; i++)
{
if (pre[i] >= N)
{
ind = i;
break ;
}
}
int x = N - pre[ind - 1];
int y = 2 * ind - 1;
if (x >= y)
res = Math.Min(x, y);
else
res = Math.Max(x, 2 * (ind - 2) + 1);
return res;
} // Driver Code public static void Main()
{ int N = 25;
Console.WriteLine(countMaxLength(N));
} } // This code is contributed by Code_Mech. |
<?php // PHP implementation of the approach $MAXN = 30;
// Function to return the length of longest // contiguous string containing only 4’s from // the first N characters of the string function countMaxLength( $N )
{ // Initialize result
$res = 0;
// Initialize prefix sum array of
// characters and product variable
$pre = array ();
$p = 1;
// Preprocessing of prefix sum array
$pre [0] = 0;
for ( $i = 1; $i < $GLOBALS [ 'MAXN' ]; $i ++)
{
$p *= 2;
$pre [ $i ] = $pre [ $i - 1] + $i * $p ;
}
// Initialize variable to store the
// string length where N belongs to
$ind = 0;
// Finding the string length where
// N belongs to
for ( $i = 1; $i < $GLOBALS [ 'MAXN' ]; $i ++)
{
if ( $pre [ $i ] >= $N )
{
$ind = $i ;
break ;
}
}
$x = $N - $pre [ $ind - 1];
$y = 2 * $ind - 1;
if ( $x >= $y )
$res = min( $x , $y );
else
$res = max( $x , 2 * ( $ind - 2) + 1);
return $res ;
} // Driver Code $N = 25;
echo countMaxLength( $N );
// This code is contributed by Ryuga ?> |
<script> // Javascript implementation of the approach var MAXN = 30;
// Function to return the length of longest // contiguous string containing only 4’s from // the first N characters of the string function countMaxLength(N)
{ // Initialize result
var res;
// Initialize prefix sum array of
// characters and product variable
var pre = Array(MAXN), p = 1;
// Preprocessing of prefix sum array
pre[0] = 0;
for ( var i = 1; i < MAXN; i++)
{
p *= 2;
pre[i] = pre[i - 1] + i * p;
}
// Initialize variable to store the
// string length where N belongs to
var ind;
// Finding the string length where
// N belongs to
for ( var i = 1; i < MAXN; i++)
{
if (pre[i] >= N)
{
ind = i;
break ;
}
}
var x = N - pre[ind - 1];
var y = 2 * ind - 1;
if (x >= y)
res = Math.min(x, y);
else
res = Math.max(x, 2 * (ind - 2) + 1);
return res;
} // Driver Code var N = 25;
document.write(countMaxLength(N)); // This code is contributed by noob2000 </script> |
5
Time Complexity: O(MAXN)
Auxiliary Space: O(MAXN), where MAXN is a predefined value here. MAXN = 30