Sponsored
Sponsored
This approach utilizes prefix sums to calculate the minimum number of flips required to make a binary string monotone increasing. The idea is to traverse the string while maintaining a prefix count of '1's and suffix count of '0's. For each position i, calculate how many flips would be necessary to make all characters before i '0' + make all characters from i '1'.
Time Complexity: O(n) where n is the length of the string.
Space Complexity: O(n) due to the prefix ones array.
1public class Solution {
2 public int minFlipsMonoIncr(String s) {
3 int n = s.length();
4 int[] prefixOnes = new int[n + 1];
5 for (int i = 0; i < n; ++i) {
6 prefixOnes[i + 1] = prefixOnes[i] + (s.charAt(i) == '1' ? 1 : 0);
7 }
8 int minFlips = Integer.MAX_VALUE;
9 for (int i = 0; i <= n; ++i) {
10 int flips = prefixOnes[i] + (n - i - (prefixOnes[n] - prefixOnes[i]));
11 minFlips = Math.min(minFlips, flips);
12 }
13 return minFlips;
14 }
15}The Java implementation uses arrays to keep track of the prefix sums of '1's. It then calculates the optimal place to flip based on these sums, returning the minimum number of flips required.
This approach leverages dynamic programming to find the minimum flips. Two states are maintained for any position: one representing the minimal flips to have a monotone ending with '0' and the other with '1'. Update these states as we traverse the string.
Time Complexity: O(n) as we traverse across the string once.
Space Complexity: O(1) since only fixed states are maintained.
1public classThe Java-based solution adopts a dynamic programming method to selectively compute necessary bit flips ensuring efficiency by utilizing two rolling state variables.