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Maximum Consecutive Zeroes in Concatenated Binary String in C++
Suppose we have a binary string of length n, another value say k is given. We have to concatenate the binary string k times. Then we have to find the maximum number of consecutive 0s in the concatenated string. Suppose binary string is “0010010”, and k = 2, then after concatenating the string k times, it will be “00100100010010”. So the maximum number of consecutive 0s is 3.
The approach is simple. If the number has all 0s, then the answer will be n * k. If the string contains ones, then the result will be either the max length of a substring of string, containing all 0s, or the sum between the length of the maximum prefix of the string containing only 0s, and the length of the maximal suffix of a string containing only 0s.
Algorithm
max_zero_count (str, n, k) −
Begin total := 0 len := 0 for i in range 0 to n, do if str[i] = 0, then increase len else len := 0 total := maximum of total and len done if total = n, then return n * k prefix := length of maximal prefix with only 0 suffix:= length of maximal suffix with only 0 if k > 1, then total := max of total, and (prefix + suffix) return total End
Example
#include <iostream> using namespace std; int max_length_substring(string str, int n, int k) { int total_len = 0; int len = 0; for (int i = 0; i < n; ++i) { if (str[i] == '0') //if the current character is 0, increase len len++; else len = 0; total_len = max(total_len, len); } if (total_len == n) //if the whole string has 0 only return n * k; int prefix = 0, suffix = 0; for (int i = 0; str[i] == '0'; ++i, ++prefix) //find length of maximal prefix with only 0; for (int i = n - 1; str[i] == '0'; --i, ++suffix) //find length of maximal suffix with only 0; if (k > 1) total_len = max(total_len, prefix + suffix); return total_len; } int main() { int k = 3; string str = "0010010"; int res = max_length_substring(str, str.length(), k); cout << "Maximum length of 0s: " << res; }
Output
Maximum length of 0s: 3