Maximum contiguous 1 possible in a binary string after k rotations
Given a binary string, you can rotate any substring of this string. For Example, let string be denoted by s. Let the first element of string be represented by s[0], second element be represented by s[1] and so on.
s = “100110111”
Suppose, we rotate the substring starting from s[2] and ending at s[4]. Then the string after this operation will be:
Resultant String = “101100111”
Now, you are allowed to do at most k operations to rotate any substring. You have to tell the maximum number of contiguous 1 you can make in this string in k or less than k rotations of substring.
Examples:
Input : 100011001
k = 1
Output : 3
Explanation:
k is 1, hence you can rotate only once. Rotate the substring starting from s[1] and ending at s[5]. The resultant string will be : 111000001. Hence, maximum contiguous 1 are 3.Input : 001100111000110011100
k = 2
Output : 8
Explanation:
k is 2, hence you can rotate twice. Rotate the substring starting at s[6] and ending at s[15]. Resultant string after first rotation : 001100001100011111100. Then, rotate the substring starting at s[8] and ending at s[12]. Resultant string after second rotation : 001100000001111111100. Hence, maximum number of contiguous 1 are 8.
Concept For Solving:
In order to solve this problem, we will maintain the frequency of 1’s in portion of contiguous 1’s in the original string in a multiset. Then on each rotation, we will rotate that substring such that, 2 portions of contiguous 1(with maximum frequency) in the string come together. We will do this, by sorting the multiset from greatest to smallest element. We will take out top 2 elements of multiset and insert their sum back into the multiset. We will continue to do this until k rotations are completed or number of elements in multiset is reduced to 1.
// C++ program to calculate maximum contiguous// ones in string#include <bits/stdc++.h>using namespace std; // function to calculate maximum contiguous onesint maxContiguousOnes(string s, int k){ int i, j, a, b, count; // multiset is used to store frequency of // 1's of each portion of contiguous 1 in // string in decreasing order multiset<int, greater<int> > m; // this loop calculate all the frequency // and stores them in multiset for (i = 0; i < s.length(); i++) { if (s[i] == '1') { count = 0; j = i; while (s[j] == '1' && j < s.length()) { count++; j++; } m.insert(count); i = j - 1; } } // if their is no 1 in string, then return 0 if (m.size() == 0) return 0; // calculates maximum contiguous 1's on // doing rotations while (k > 0 && m.size() != 1) { // Delete largest two elements a = *(m.begin()); m.erase(m.begin()); b = *(m.begin()); m.erase(m.begin()); // insert their sum back into the multiset m.insert(a + b); k--; } // return maximum contiguous ones // possible after k rotations return *(m.begin());} // Driver codeint main(){ string s = "10011110011"; int k = 1; cout << maxContiguousOnes(s, k); return 0;} |
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