Find total possible ways to generate from given string

Question

I have a binary string of 0's and 1's. Now I want to select a string of length 3 such that no consecutive characters are same.

Example:

Input : 01011010

Output: 20

Explanation: All possible combinations are:

[0, 1, 0]:: indices : 0,1,2 :: valid
[0, 1, 1]:: indices : 0,1,3
[0, 1, 1]:: indices : 0,1,4
[0, 1, 0]:: indices : 0,1,5 :: valid
[0, 1, 1]:: indices : 0,1,6
[0, 1, 0]:: indices : 0,1,7 :: valid
[0, 0, 1]:: indices : 0,2,3
[0, 0, 1]:: indices : 0,2,4
[0, 0, 0]:: indices : 0,2,5
[0, 0, 1]:: indices : 0,2,6
[0, 0, 0]:: indices : 0,2,7
[0, 1, 1]:: indices : 0,3,4
[0, 1, 0]:: indices : 0,3,5 :: valid
[0, 1, 1]:: indices : 0,3,6
[0, 1, 0]:: indices : 0,3,7 :: valid
[0, 1, 0]:: indices : 0,4,5 :: valid
[0, 1, 1]:: indices : 0,4,6
[0, 1, 0]:: indices : 0,4,7 :: valid
[0, 0, 1]:: indices : 0,5,6
[0, 0, 0]:: indices : 0,5,7
[0, 1, 0]:: indices : 0,6,7 :: valid

[1, 0, 1]:: indices : 1,2,3 :: valid
[1, 0, 1]:: indices : 1,2,4 :: valid
[1, 0, 0]:: indices : 1,2,5
[1, 0, 1]:: indices : 1,2,6 :: valid
[1, 0, 0]:: indices : 1,2,7
[1, 1, 1]:: indices : 1,3,4
[1, 1, 0]:: indices : 1,3,5
[1, 1, 1]:: indices : 1,3,6
[1, 1, 0]:: indices : 1,3,7
[1, 1, 0]:: indices : 1,4,5
[1, 1, 1]:: indices : 1,4,6
[1, 1, 0]:: indices : 1,4,7
[1, 0, 1]:: indices : 1,5,6 :: valid
[1, 0, 0]:: indices : 1,5,7
[1, 1, 0]:: indices : 1,6,7

[0, 1, 1]:: indices : 2,3,4
[0, 1, 0]:: indices : 2,3,5 :: valid
[0, 1, 1]:: indices : 2,3,6
[0, 1, 0]:: indices : 2,3,7 :: valid
[0, 1, 0]:: indices : 2,4,5 :: valid
[0, 1, 1]:: indices : 2,4,6
[0, 1, 0]:: indices : 2,4,7 :: valid
[0, 0, 1]:: indices : 2,5,6
[0, 0, 0]:: indices : 2,5,7
[0, 1, 0]:: indices : 2,6,7 :: valid

[1, 1, 0]:: indices : 3,4,5
[1, 1, 1]:: indices : 3,4,6
[1, 1, 0]:: indices : 3,4,7
[1, 0, 1]:: indices : 3,5,6 :: valid
[1, 0, 0]:: indices : 3,5,7
[1, 1, 0]:: indices : 3,6,7

[1, 0, 1]:: indices : 4,5,6 :: valid
[1, 0, 0]:: indices : 4,5,7
[1, 1, 0]:: indices : 4,6,7

[0, 1, 0]:: indices : 5,6,7 :: valid

Of these only 20 combinations are valid so output is 20.

Here is my program.

public static long process(String s) {
    long result = 0;
    int n = s.length();
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (s.charAt(i) == s.charAt(j)) {
                continue;
            }
            for (int k = j + 1; k < n; k++) {
                if (s.charAt(j) != s.charAt(k)) {
                    result++;
                }
            }
        }
    }
    return result;
}

Constaints:

length of input is 1 to 2*10^5.

How to reduce time complexity of this program.

Answer 1

Let's first look at all possible binary strings of length 3 such that no consecutive characters are same:

• 101
• 010

With that in mind the problem can be reformulated to:

for each 0 in input, find all combinations of 1 on either side, plus
for each 1 in input, find all combinations of 0 on either side

For any given 0 , the number of combinations of 1 on either side is equal to the number of 1 to the left times the number of 1 to the right (and vice-versa).

Finally, we can combine these two for loops into one by switching between the two cases based on the current character. We should also keep track of the 0 and 1 to the left and right, as we don't want to recalculate them all the time. This leaves me with the following algorithm:

var leftZeroes = 0
var leftOnes = 0
var rightZeroes = input.count { it == '0' }
var rightOnes = input.count { it == '1' }
var output = 0
for (char in input) {
    if (char == '0') {
        output += leftOnes * rightOnes
        leftZeroes++
        rightZeroes--
    } else {
        output += leftZeroes * rightZeroes
        leftOnes++
        rightOnes--
    }
}
return output

Time complexity: O(n) ( count and for are linear in input size, everything else is constant)

Additional space complexity would be O(1) , as only those five integers are stored at any time