# 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 which will hold two following conditions simultaneously:

- Sum of numbers between all N given ranges will be 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

01101First 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

1110000In 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 make the count of 1’s between all l and r to be max. We count the number of 1’s in the array and store 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 prefix sum of the array so that we get the positions where to fix the 1’s first. Then we run a loop in that prefix sum array from left. If we get any position with 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 left of the prefix sum array if we find an index having 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:

`# Python implementation of the approach ` `def` `arrange(s): ` ` ` `cc ` `=` `0` ` ` ` ` `# Storing the count of 1's in the string ` ` ` `for` `i ` `in` `range` `(` `len` `(s)): ` ` ` `if` `(s[i]` `=` `=` `"1"` `): ` ` ` `cc` `+` `=` `1` ` ` `a ` `=` `[` `0` `]` `*` `(` `len` `(s)` `+` `1` `) ` ` ` ` ` `# Query of l and r ` ` ` `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` ` ` ` ` `# Applying range update technique. ` ` ` `a[r ` `+` `1` `]` `-` `=` `1` ` ` ` ` `for` `i ` `in` `range` `(` `1` `, ` `len` `(a)): ` ` ` ` ` `# Taking prefix sum to get the range update values ` ` ` `a[i]` `=` `a[i]` `+` `a[i` `-` `1` `] ` ` ` ` ` `# Final array which will store the arranged string ` ` ` `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 after maximizing the ranges any 1 is left ` ` ` `# then we maximize the string lexicographically. ` ` ` `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` `) ` |

*chevron_right*

*filter_none*

**Output:**

01101

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