Flip String To Monotone Increasing - Leetcode Solution

Problem Statement: A string of '0's and '1's is given. We need to flip the minimum number of '0's to '1's such that the resulting string is monotone increasing.

Monotone increasing string means that every character to the right of the current character should be greater than or equal to the current character.

Input: "00110" Output: 1 Explanation: We need to flip the 1st zero to '1', resulting in "10110", which is monotone increasing.

Solution Approach: Let's start by understanding the problem statement. We need to convert a string to monotone increasing by flipping the minimum number of '0's to '1's. We can achieve this by iterating the string from left to right and keeping track of the number of '0's encountered so far. For each '1' encountered, we check how many '0's are encountered before it. If the number of '0's is less than the current minimum, we update the minimum. We also keep track of the number of '1's encountered so far. After iterating through the string, we get the total number of '0's and '1's in the string. There are two cases:

• If the number of '0's is 0, then the string is already monotone increasing, and we return 0.
• If the number of '0's is greater than 0, then we need to flip the minimum number of '0's to '1's such that the resulting string is monotone increasing. We can achieve this by flipping all the '0's to the left of the minimum '0' we encountered during iteration. Thus, the number of '0's flipped will be the minimum of the total number of '0's and the number of '1's encountered before the minimum '0' during iteration.

Let's implement the above approach in code:

public int minFlipsMonoIncr(String S) { int len = S.length(); int[] zeros = new int[len]; int[] ones = new int[len]; int countZeros = 0, countOnes = 0; for(int i=0; i<len; i++) { zeros[i] = countZeros; ones[i] = countOnes; if(S.charAt(i) == '0') { countZeros++; } else { countOnes++; } } int minFlips = Integer.MAX_VALUE; for(int i=0; i<len; i++) { int flips = zeros[i] + ones[len-1] - ones[i]; minFlips = Math.min(minFlips, flips); } return minFlips; }

First, we initialize two arrays 'zeros' and 'ones' of length 'len'. We also initialize two counters to keep track of the number of '0's and '1's encountered so far. We iterate through the string, updating the 'zeros' and 'ones' arrays with the count of '0's and '1's encountered so far. Next, we iterate through the 'zeros' array and for each element, we calculate the total number of '0's that need to be flipped to the right of the current element and the total number of '1's encountered up to the current element. We add these two values to get the total number of '0's that need to be flipped to make the string monotone increasing. Finally, we return the minimum number of flips required.

Time Complexity: O(N), where N is the length of the input string. Space Complexity: O(N), where N is the length of the input string.

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