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Problem Brief

Get Minimum Errors

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Amazon's database doesn’t support very large numbers, so numbers are stored as a string of binary characters, '0' and '1'. Accidentally, a '!' was entered at some positions and it is unknown whether they should be '0' or '1'.

The string of incorrect data is made up of the characters '0', '1' and '!' where '!' is the character that got entered incorrectly. '!' can be replaced with either '0' or '1'. Due to some internal faults, some errors are generated every time '0' and '1' occur together as '01' or '10' in any subsequence of the string. It is observed that the number of errors a subsequence '01' generates is x, while a subsequence '10' generates y errors.

Determine the minimum total errors generated. Since the answer can be very large, return it modulo 109+7.

Note: A subsequence of a string is obtained by omitting zero or more characters from the original string without changing their order.

Hint: It can be proved that (a + b) % c = ((a% c) + (b % c)) % c where a, b, and c are integers and % represents the modulo operation.

Function Description

Complete the function getMinErrors in the editor.

getMinErrors has the following parameter(s):

  1. String errorString: a string of characters '0', '1', and '!'
  2. int x: the number of errors generated for every occurrence of subsequence 01
  3. int y: the number of errors generated for every occurrence of subsequence 10

Returns

int: the minimum number of errors possible, modulo 109+7

♡ spike ໒꒱˚ deserves all the applause! 👏

1Example 1

Input
errorString = "101!1", x = 2, y = 3
Output
9
Explanation
If the '!' at index 3 is replaced with '0', the string is "10101". The number of times the subsequence 01 occurs is 3 at indices (1, 2), (1, 4), and (3, 4). The number of times the subsequence 10 occurs is also 3, indices (0, 1), (0, 3) and (2, 3). The number of errors is 3 * x + 3 * y = 6 + 9 = 15. If the '!' is replaced with '1', the string is "10111". The subsequence 01 occurs 3 times and 10 occurs 1 time. The number of errors is 3 * x + y = 9. The minimum number of errors is min(9, 15) modulo (109 + 7) = 9.

2Example 2

Input
errorString = "01!0", x = 2, y = 2
Output
6
Explanation
The better string is 0100 with one substring 01 at index (0,1) and two subsequence of 10 at indices (1,2) and (2,3) making total errors generate = 2 * 1 + 2 * 2 = 6. ࣪𓏲ּ ᥫ᭡ ₊ ⊹ Credit to 𓇼 Suat 𓇼 ˑ🌷 ִֶ 𓂃

3Example 3

Input
errorString = "!!!!!!!", x = 23, y = 27
Output
0
Explanation
There is a tie for the best string generated, 00000 or 11111, with zero substrings 01 or 10.

Constraints

Limits and guarantees your solution can rely on.

  • 1<= len (errorString)<=105
  • 0 <= x, y <= 105
  • s consists only of characters '0', '1', and 'l'
    • public int getMinErrors(String errorString, int x, int y) {
  // write your code here
}
    Input

    errorString

    "101!1"

    x

    2

    y

    3

    Output

    9

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