Move from brute-force thinking to an efficient approach using dynamic programming strategy.
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).
You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0.
Return the minimum number of flips to make s monotone increasing.
Example 1:
Input: s = "00110" Output: 1 Explanation: We flip the last digit to get 00111.
Example 2:
Input: s = "010110" Output: 2 Explanation: We flip to get 011111, or alternatively 000111.
Example 3:
Input: s = "00011000" Output: 2 Explanation: We flip to get 00000000.
Constraints:
1 <= s.length <= 105s[i] is either '0' or '1'.Problem summary: 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). You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0. Return the minimum number of flips to make s monotone increasing.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"00110"
"010110"
"00011000"
minimum-cost-to-make-all-characters-equal)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #926: Flip String to Monotone Increasing
class Solution {
public int minFlipsMonoIncr(String s) {
int n = s.length();
int tot = 0;
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == '0') {
++tot;
}
}
int ans = tot, cur = 0;
for (int i = 1; i <= n; ++i) {
if (s.charAt(i - 1) == '0') {
++cur;
}
ans = Math.min(ans, i - cur + tot - cur);
}
return ans;
}
}
// Accepted solution for LeetCode #926: Flip String to Monotone Increasing
func minFlipsMonoIncr(s string) int {
tot := strings.Count(s, "0")
ans, cur := tot, 0
for i, c := range s {
if c == '0' {
cur++
}
ans = min(ans, i+1-cur+tot-cur)
}
return ans
}
# Accepted solution for LeetCode #926: Flip String to Monotone Increasing
class Solution:
def minFlipsMonoIncr(self, s: str) -> int:
tot = s.count("0")
ans, cur = tot, 0
for i, c in enumerate(s, 1):
cur += int(c == "0")
ans = min(ans, i - cur + tot - cur)
return ans
// Accepted solution for LeetCode #926: Flip String to Monotone Increasing
impl Solution {
pub fn min_flips_mono_incr(s: String) -> i32 {
let (mut res, mut count_one) = (0, 0);
for ch in s.chars() {
if ch == '1' {
count_one += 1;
} else {
res = i32::min(res + 1, count_one);
}
}
res
}
}
// Accepted solution for LeetCode #926: Flip String to Monotone Increasing
function minFlipsMonoIncr(s: string): number {
let tot = 0;
for (const c of s) {
tot += c === '0' ? 1 : 0;
}
let [ans, cur] = [tot, 0];
for (let i = 1; i <= s.length; ++i) {
cur += s[i - 1] === '0' ? 1 : 0;
ans = Math.min(ans, i - cur + tot - cur);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.