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‎FlipStringToMonotoneIncreasing/flip_string_to_monotone_increasing.dart

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/*
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-* 926. Flip String to Monotone Increasing *-
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A binary string is monotone increasing if it consists of some number of 0's (possibly none), followed by some number of 1's (also possibly none).
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You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0.
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Return the minimum number of flips to make s monotone increasing.
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Example 1:
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Input: s = "00110"
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Output: 1
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Explanation: We flip the last digit to get 00111.
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Example 2:
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Input: s = "010110"
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Output: 2
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Explanation: We flip to get 011111, or alternatively 000111.
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Example 3:
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Input: s = "00011000"
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Output: 2
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Explanation: We flip to get 00000000.
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Constraints:
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1 <= s.length <= 105
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s[i] is either '0' or '1'.
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*/
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import 'dart:math';
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class A {
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int minFlipsMonoIncr(String s) {
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// Initialize variables to store the minimum number of flips and the number of flips currently needed
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int ans = 0;
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int noOfFlip = 0;
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// Iterate through each character in the input string
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for (int i = 0; i < s.length; i++) {
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// If the current character is a '0', update the minimum number of flips and add 1 to the current number of flips
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if (s.codeUnitAt(i) == '0')
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ans = min(noOfFlip, ans + 1);
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// If the current character is a '1', increment the number of flips
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else
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noOfFlip++;
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}
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// Return the minimum number of flips
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return ans;
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}
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}
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class B {
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int minFlipsMonoIncr(String s) {
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int n = s.length;
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int ans = 100000;
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List<int> oneCount = List.filled(n, 0);
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List<int> zeroCount = List.filled(n, 0);
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oneCount[0] = 0;
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zeroCount[n - 1] = 0;
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for (int i = 1; i < n; i++) {
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if (s[i - 1] == '1')
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oneCount[i] = oneCount[i - 1] + 1;
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else
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oneCount[i] = oneCount[i - 1];
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if (s[n - i] == '0')
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zeroCount[n - i - 1] = zeroCount[n - i] + 1;
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else
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zeroCount[n - i - 1] = zeroCount[n - i];
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}
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for (int i = 0; i < n; i++) {
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ans = min(ans, zeroCount[i] + oneCount[i]);
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}
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return ans;
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}
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}
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class C {
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//TLE
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int minFlipsMonoIncr(String s) {
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int n = s.length;
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int mini = 10000;
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List<int> ans = List.filled(n, 0);
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for (int i = 0; i < n; i++) {
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ans[i] += oneCount(s.substring(0, i));
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ans[i] += zeroCount(s.substring(i + 1));
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mini = min(mini, ans[i]);
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}
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return mini;
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}
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int oneCount(String s) {
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int n = s.length;
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int cnt = 0;
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for (int i = 0; i < n; i++) {
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if (s[i] == '1') cnt++;
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}
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return cnt;
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}
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int zeroCount(String s) {
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int n = s.length;
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int cnt = 0;
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for (int i = 0; i < n; i++) {
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if (s[i] == '0') cnt++;
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}
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return cnt;
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}
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}

‎FlipStringToMonotoneIncreasing/flip_string_to_monotone_increasing.go

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package main
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// func minFlipsMonoIncr(s string) int {
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// var answer int = 0
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// var numberOfFlips = 0
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// for i := 0; i < len(s); i++ {
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// if s[i] == '0' {
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// answer = min(numberOfFlips, answer+1)
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// } else {
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// numberOfFlips++
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// }
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// }
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// return answer
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// }
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func min(a int, b int) int {
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if a < b {
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return a
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} else {
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return b
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}
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}
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func minFlipsMonoIncr(s string) int {
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var n int = len(s)
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var ans int = 100000
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var oneCount []int = make([]int, n)
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var zeroCount []int = make([]int, n)
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oneCount[0] = 0
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zeroCount[n-1] = 0
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for i := 1; i < n; i++ {
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if s[i-1] == '1' {
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oneCount[i] = oneCount[i-1] + 1
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} else {
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oneCount[i] = oneCount[i-1]
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}
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if s[n-i] == '0' {
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zeroCount[n-i-1] = zeroCount[n-i] + 1
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} else {
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zeroCount[n-i-1] = zeroCount[n-i]
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}
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for i := 0; i < n; i++ {
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ans = min(ans, zeroCount[i]+oneCount[i])
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}
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}
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return ans
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}
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/*
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int minFlipsMonoIncr(String s) {
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int n = s.length;
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int ans = 100000;
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List<int> oneCount = List.filled(n, 0);
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List<int> zeroCount = List.filled(n, 0);
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oneCount[0] = 0;
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zeroCount[n - 1] = 0;
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for (int i = 1; i < n; i++) {
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if (s[i - 1] == '1')
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oneCount[i] = oneCount[i - 1] + 1;
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else
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oneCount[i] = oneCount[i - 1];
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if (s[n - i] == '0')
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zeroCount[n - i - 1] = zeroCount[n - i] + 1;
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else
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zeroCount[n - i - 1] = zeroCount[n - i];
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}
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for (int i = 0; i < n; i++) {
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ans = min(ans, zeroCount[i] + oneCount[i]);
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}
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return ans;
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}
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*/

‎FlipStringToMonotoneIncreasing/flip_string_to_monotone_increasing.md

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# 🔥 3 Approaches 🔥 || Simple Fast and Easy || with Explanation
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## Solution - 1
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### Approach
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Skip the loop until we find the first '1'
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after finding first occurrence of '1' we have to keep a count of number of '1's in countOnes
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if we find any '0' after occurrence of '1' it means turning that '0' to one may be one possible solution so we have to keep number of flip needed in countFlips variable
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if value of countFlips is greater than countOnes(number of ones) then converting all '1' to '0' would be better option so assign value of countOnes to the countFlips
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in the end we will return countFlips
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### Complexity
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#### Time complexity
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O(n)
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#### Space complexity
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O(1)
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```dart
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class Solution {
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int minFlipsMonoIncr(String s) {
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// Initialize variables to store the minimum number of flips and the number of flips currently needed
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int ans = 0;
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int noOfFlip = 0;
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// Iterate through each character in the input string
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for (int i = 0; i < s.length; i++) {
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// If the current character is a '0', update the minimum number of flips and add 1 to the current number of flips
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if (s.codeUnitAt(i) == '0')
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ans = min(noOfFlip, ans + 1);
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// If the current character is a '1', increment the number of flips
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else
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noOfFlip++;
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}
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// Return the minimum number of flips
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return ans;
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}
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}
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```
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## Solution - 2
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### Intuition
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Counting the no. of 1's to flip on left side + no. of 0's to flip on right side of each index i and then select minimum flips from it.
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### Approach
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Maintain a prefix and suffix array to count 1 on left side and 0 on right side.
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#### Complexity
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Time complexity:
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O(n)
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#### Space complexity
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O(n)
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```dart
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class Solution {
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int minFlipsMonoIncr(String s) {
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int n = s.length;
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int ans = 100000;
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List<int> oneCount = List.filled(n, 0);
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List<int> zeroCount = List.filled(n, 0);
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oneCount[0] = 0;
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zeroCount[n - 1] = 0;
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for (int i = 1; i < n; i++) {
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if (s[i - 1] == '1')
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oneCount[i] = oneCount[i - 1] + 1;
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else
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oneCount[i] = oneCount[i - 1];
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if (s[n - i] == '0')
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zeroCount[n - i - 1] = zeroCount[n - i] + 1;
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else
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zeroCount[n - i - 1] = zeroCount[n - i];
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}
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for (int i = 0; i < n; i++) {
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ans = min(ans, zeroCount[i] + oneCount[i]);
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}
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return ans;
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}
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}
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```
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### Solution - 3
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```dart
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class Solution {
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//TLE
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int minFlipsMonoIncr(String s) {
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int n = s.length;
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int mini = 10000;
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List<int> ans = List.filled(n, 0);
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for (int i = 0; i < n; i++) {
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ans[i] += oneCount(s.substring(0, i));
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ans[i] += zeroCount(s.substring(i + 1));
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mini = min(mini, ans[i]);
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}
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return mini;
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}
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int oneCount(String s) {
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int n = s.length;
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int cnt = 0;
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for (int i = 0; i < n; i++) {
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if (s[i] == '1') cnt++;
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}
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return cnt;
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}
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int zeroCount(String s) {
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int n = s.length;
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int cnt = 0;
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for (int i = 0; i < n; i++) {
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if (s[i] == '0') cnt++;
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}
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return cnt;
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}
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}
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```

‎README.md

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- [**1061.** Lexicographically Smallest Equivalent String](LexicographicallySmallestEquivalentString/lexicographically_smallest_equivalent_string.dart)
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- [**2421.** Number of Good Paths](NumberOfGoodPaths/number_of_good_paths.dart)
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- [**57.** Insert Interval](InsertInterval/insert_interval.dart)
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- [**926.** Flip String to Monotone Increasing](FlipStringToMonotoneIncreasing/flip_string_to_monotone_increasing.dart)
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## Reach me via
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