Arrange a binary string to get maximum value within a range of indices
Given a string consisting of only 0’s and 1’s. Now you are given N non-intersecting ranges L, R ( L <= R), more specifically [L1, R1], [L2, R2], …, [LN, RN], No two of these intervals overlap — formally, for each valid i, j such that i!=j, either Ri<Lj or Rj<Li.
The task is to find a valid permutation that will hold two following conditions simultaneously:
- The sum of numbers between all N-given ranges will be the maximum.
- The string will be lexicographically largest. A string 1100 is lexicographically larger than string 1001.
Examples:
Input
11100
2
2 3
5 5
Output
01101
First we put 1’s in position 2 and 3 then in 5 as
there are no 1’s left, the string formed is 01101.
Input
0000111
2
1 1
1 2
Output
1110000
In the above example, we, 1st put 1 in 1st and 2nd position, then we have another ‘1’ left,
So, we use it to maximize the string lexicographically and we put it in the 3rd position and thus the rearrangement is complete.
Approach
- First priority is given to making the count of 1’s between all l and r be max. We count the number of 1’s in the array and store them in a variable.
- After taking input, we update the range of each l and r by 1 to just mark the position to be filled with 1 first.
- Then, we take the prefix sum of the array so that we get the position where to fix the 1’s first. Then we run a loop in that prefix sum array from the left. If we get any position with a value greater than 1, that means we have a l-r in that index. We continue to put 1’s in those indices until the count of 1 becomes zero.
- Now, after the maximization operation is finished and if there are some 1’s left, then we start the lexicographic maximization. We again start a loop from the left of the prefix sum array. If we find an index having the value 0 which indicates that there is no l-r having that index, then we put a 1 in that index and thus continue until all remaining 1’s are filled.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void arrange(string s)
{
int cc = 0;
for ( int i = 0; i < s.length(); i++)
{
if (s[i] == '1' ) cc++;
}
int a[s.length() + 1] = {0};
int qq[][2] = {{2, 3}, {5, 5}};
int n = sizeof (qq) / sizeof (qq[0]);
for ( int i = 0; i < n; i++)
{
int l = qq[i][0], r = qq[i][1];
l--, r--;
a[l]++;
a[r + 1]--;
}
int len_a = sizeof (a) / sizeof (a[0]);
for ( int i = 1; i < len_a; i++)
{
a[i] += a[i - 1];
}
int zz[s.length()] = {0};
for ( int i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break ;
}
if (cc == 0) break ;
}
if (cc > 0)
{
for ( int i = 0; i < s.length(); i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break ;
}
}
for ( int i = 0; i < s.length(); i++)
cout << zz[i];
cout << endl;
}
int main()
{
string str = "11100" ;
arrange(str);
return 0;
}
|
Java
class GFG
{
static void arrange(String s)
{
int cc = 0 ;
for ( int i = 0 ; i < s.length(); i++)
{
if (s.charAt(i) == '1' ) cc++;
}
int []a = new int [s.length() + 1 ];
int qq[][] = {{ 2 , 3 }, { 5 , 5 }};
int n = qq.length;
for ( int i = 0 ; i < n; i++)
{
int l = qq[i][ 0 ], r = qq[i][ 1 ];
l--; r--;
a[l]++;
a[r + 1 ]--;
}
int len_a = a.length;
for ( int i = 1 ; i < len_a; i++)
{
a[i] += a[i - 1 ];
}
int []zz = new int [s.length()];
for ( int i = 0 ; i < len_a - 1 ; i++)
{
if (a[i] > 0 )
{
if (cc > 0 )
{
zz[i] = 1 ;
cc--;
}
else
break ;
}
if (cc == 0 ) break ;
}
if (cc > 0 )
{
for ( int i = 0 ; i < s.length(); i++)
{
if (zz[i] == 0 )
{
zz[i] = 1 ;
cc--;
}
if (cc == 0 ) break ;
}
}
for ( int i = 0 ; i < s.length(); i++)
System.out.print(zz[i]);
System.out.println();
}
public static void main(String[] args)
{
String str = "11100" ;
arrange(str);
}
}
|
Python3
def arrange(s):
cc = 0
for i in range ( len (s)):
if (s[i] = = "1" ):
cc + = 1
a = [ 0 ] * ( len (s) + 1 )
qq = [( 2 , 3 ), ( 5 , 5 )]
n = len (qq)
for i in range (n):
l, r = qq[i][ 0 ], qq[i][ 1 ]
l - = 1
r - = 1
a[l] + = 1
a[r + 1 ] - = 1
for i in range ( 1 , len (a)):
a[i] = a[i] + a[i - 1 ]
zz = [ 0 ] * len (s)
for i in range ( len (a) - 1 ):
if (a[i]> 0 ):
if (cc> 0 ):
zz[i] = 1
cc - = 1
else :
break
if (cc = = 0 ):
break
if (cc> 0 ):
for i in range ( len (s)):
if (zz[i] = = 0 ):
zz[i] = 1
cc - = 1
if (cc = = 0 ):
break
print ( * zz, sep = "")
str = "11100"
arrange( str )
|
C#
using System;
class GFG
{
static void arrange(String s)
{
int cc = 0;
for ( int i = 0; i < s.Length; i++)
{
if (s[i] == '1' ) cc++;
}
int []a = new int [s.Length + 1];
int [,]qq = {{2, 3}, {5, 5}};
int n = qq.GetLength(0);
for ( int i = 0; i < n; i++)
{
int l = qq[i, 0], r = qq[i, 1];
l--; r--;
a[l]++;
a[r + 1]--;
}
int len_a = a.Length;
for ( int i = 1; i < len_a; i++)
{
a[i] += a[i - 1];
}
int []zz = new int [s.Length];
for ( int i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break ;
}
if (cc == 0) break ;
}
if (cc > 0)
{
for ( int i = 0; i < s.Length; i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break ;
}
}
for ( int i = 0; i < s.Length; i++)
Console.Write(zz[i]);
Console.WriteLine();
}
public static void Main(String[] args)
{
String str = "11100" ;
arrange(str);
}
}
|
Javascript
<script>
function arrange(s)
{
var cc = 0;
for ( var i = 0; i < s.length; i++)
{
if (s[i] == '1') cc++;
}
var a = Array(s.length+1).fill(0);
var qq = [[2, 3], [5, 5]];
var n = qq.length;
for ( var i = 0; i < n; i++)
{
var l = qq[i][0], r = qq[i][1];
l--, r--;
a[l]++;
a[r + 1]--;
}
var len_a = a.length;
for ( var i = 1; i < len_a; i++)
{
a[i] += a[i - 1];
}
var zz = Array(s.length).fill(0);
for ( var i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break ;
}
if (cc == 0) break ;
}
if (cc > 0)
{
for ( var i = 0; i < s.length; i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break ;
}
}
for ( var i = 0; i < s.length; i++)
document.write( zz[i]);
document.write( "<br>" );
}
var str = "11100" ;
arrange(str);
</script>
|
Time complexity: O(n)
Auxiliary space: O(n)
Last Updated :
16 Nov, 2022
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