Minimum Suffix Flips solution and explanation
6 more videos on this problem
1529. Bulb Switcher IV | Leetcode Medium | GreedyChhavi Bansal · 8:23 · 2K views
LeetCode 1529. Minimum Suffix FlipsNitro Nick · 8:23 · 1K views
JPMorgan Interview Question - Minimum Suffix Flips- Leetcode 1529- PythonSephorus · 4:56 · 883 views
1529. Bulb Switcher IV (Leetcode Medium)Programming Live with Larry · 3:33 · 492 views
Bulb Switcher IV | Leetcode 1529Competitive Coding · 7:45 · 307 views
LeetCode 1529 | Minimum Suffix Flips | JavaSleepy Cracker · 3:23 · 216 views
ข้อ 1529. Minimum Suffix FlipsThakorn Jareonrat · 6:40 · 7 views
The problem
You are given a 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target.
In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'.
Return the minimum number of operations needed to make s equal to target.
Example 1:
Input: target = "10111" Output: 3 Explanation: Initially, s = "00000". Choose index i = 2: "00000" -> "00111" Choose index i = 0: "00111" -> "11000" Choose index i = 1: "11000" -> "10111" We need at least 3 flip operations to form target.
Example 2:
Input: target = "101" Output: 3 Explanation: Initially, s = "000". Choose index i = 0: "000" -> "111" Choose index i = 1: "111" -> "100" Choose index i = 2: "100" -> "101" We need at least 3 flip operations to form target.
Example 3:
Input: target = "00000" Output: 0 Explanation: We do not need any operations since the initial s already equals target.
Constraints:
n == target.length1 <= n <= 105target[i]is either'0'or'1'.
Approach overview
Problem Overview: You start with a binary string of all 0s and want to transform it into a target binary string. In one operation, you can flip every bit from an index i to the end of the string (a suffix flip). The task is to compute the minimum number of such flips needed to match the target.
Approach 1: Track Transitions in Target String (Greedy) (Time: O(n), Space: O(1))
The key observation is that a flip only matters when the desired bit changes relative to the current effective state. Start with the assumption that the string is all 0s. As you iterate through the target string, track whether the current state matches the desired character. Every time the desired bit differs from the current effective value, you must perform a suffix flip starting at that index. After a flip, the effective value toggles for the rest of the scan. This reduces the problem to counting transitions where the expected bit differs from the actual target bit. The algorithm performs a single pass through the string while applying a greedy decision at each position.
Approach 2: Pattern Matching from Index 0 (Simulation) (Time: O(n), Space: O(1))
Another way to think about the process is by simulating the flips logically without modifying the string. Maintain a variable representing the current flipped state (0 or 1). Iterate from index 0 to n-1. If the effective bit after considering previous flips differs from the target bit, perform a new flip and toggle the state variable. This effectively simulates applying the suffix flip while keeping the algorithm linear. Instead of physically flipping characters, you just track whether the sequence has been flipped an odd or even number of times so far.
Both approaches rely on the same core insight: the only moments that matter are when the required bit differs from the current effective value. Once a flip happens, the state of all remaining characters is inverted, so the algorithm simply keeps track of that inversion logically.
Recommended for interviews: The transition-count greedy approach is what most interviewers expect. It demonstrates that you recognized the flipping pattern and reduced the problem to counting state changes instead of simulating expensive operations. Explaining the reasoning behind why transitions correspond to flips shows strong understanding of greedy decision making on a string traversal.
Approach 1: Track Transitions in Target String
Whenever the current bit in `target` is different from the last established state of a similar position of the string `s`, a move (flip) must be made. Start with `s` as all zeros, which initially is '000...0'. As you iterate through `target`, every time a current bit is different from the last flip to match target, a new operation is needed.
def min_flips(target):
current = '0'
flips = 0
for char in target:
if char != current:
flips += 1
current = char
return flips
# Example Usage
print(min_flips("10111")) # Output: 3public class MinimumFlips {
public static int minFlips(String target) {
char current = '0';
int flips = 0;
for (char c : target.toCharArray()) {
if (c != current) {
flips++;
current = c;
}
}
return flips;
}
public static void main(String[] args) {
String target = "10111";
System.out.println(minFlips(target)); // Output: 3
}
}#include <iostream>
#include <string>
int minFlips(const std::string& target) {
char current = '0';
int flips = 0;
for (auto bit : target) {
if (bit != current) {
flips++;
current = bit;
}
}
return flips;
}
int main() {
std::string target = "10111";
std::cout << minFlips(target) << std::endl; // Output: 3
return 0;
}function minFlips(target) {
let current = '0';
let flips = 0;
for (let char of target) {
if (char !== current) {
flips++;
current = char;
}
}
return flips;
}
// Example usage
console.log(minFlips("10111")); // Output: 3#include <stdio.h>
#include <string.h>
int minFlips(char* target) {
char current = '0';
int flips = 0;
for (int i = 0; target[i] != '\0'; i++) {
if (target[i] != current) {
flips++;
current = target[i];
}
}
return flips;
}
int main() {
char target[] = "10111";
printf("%d\n", minFlips(target)); // Output: 3
return 0;
}using System;
public class MinimumFlips {
public static int MinFlips(string target) {
char current = '0';
int flips = 0;
foreach (char c in target) {
if (c != current) {
flips++;
current = c;
}
}
return flips;
}
public static void Main() {
string target = "10111";
Console.WriteLine(MinFlips(target)); // Output: 3
}
}The Python solution directly evaluates and changes a control variable, `current`, when a transition occurs in the target string. It counts the changes and returns this count.
Time Complexity: O(n), where n is the length of the target string.
Space Complexity: O(1) as only fixed-sized data is used.
Approach 2: Pattern Matching from Index 0
Directly compare the initial `s` state to `target`, beginning with the simplest match from index 0 and capturing changes between segments entirely when flipping results in mismatch. The initial states switch whenever they are inconsistent with the target.
def min_flips_pattern(target):
s = '0'
flips = 0
for char in target:
if char != s:
flips += 1
s = char
return flips
# Example Usage
print(min_flips_pattern("101")) # Output: 3public class MinimumFlipsPattern {
public static int minFlipsPattern(String target) {
char s = '0';
int flips = 0;
for (char c : target.toCharArray()) {
if (c != s) {
flips++;
s = c;
}
}
return flips;
}
public static void main(String[] args) {
String target = "101";
System.out.println(minFlipsPattern(target)); // Output: 3
}
}#include <iostream>
#include <string>
int minFlipsPattern(const std::string& target) {
char s = '0';
int flips = 0;
for (auto bit : target) {
if (bit != s) {
flips++;
s = bit;
}
}
return flips;
}
int main() {
std::string target = "101";
std::cout << minFlipsPattern(target) << std::endl; // Output: 3
return 0;
}function minFlipsPattern(target) {
let s = '0';
let flips = 0;
for (let char of target) {
if (char !== s) {
flips++;
s = char;
}
}
return flips;
}
// Example usage
console.log(minFlipsPattern("101")); // Output: 3#include <stdio.h>
#include <string.h>
int minFlipsPattern(char* target) {
char s = '0';
int flips = 0;
for (int i = 0; target[i] != '\0'; i++) {
if (target[i] != s) {
flips++;
s = target[i];
}
}
return flips;
}
int main() {
char target[] = "101";
printf("%d\n", minFlipsPattern(target)); // Output: 3
return 0;
}using System;
public class MinimumFlipsPattern {
public static int MinFlipsPattern(string target) {
char s = '0';
int flips = 0;
foreach (char c in target) {
if (c != s) {
flips++;
s = c;
}
}
return flips;
}
public static void Main() {
string target = "101";
Console.WriteLine(MinFlipsPattern(target)); // Output: 3
}
}Python's implementation straightforwardly contrasts between a preset character compared to `target`, iteratively counting and flipping when facing distinctions.
Time Complexity: O(n), covering every character in `target`.
Space Complexity: O(1), constrained to constant auxiliary storage.
Approach 3: Greedy
We traverse the string target from left to right, using a variable ans to record the number of flips. When we reach index i, if the parity of the current flip count ans is different from target[i], we need to perform a flip operation at index i and increment ans by 1.
The time complexity is O(n), where n is the length of the string. The space complexity is O(1).
function minFlips(target: string): number {
let ans = 0;
for (const c of target) {
if (ans % 2 !== +c) {
++ans;
}
}
return ans;
}func minFlips(target string) int {
ans := 0
for _, c := range target {
v := int(c - '0')
if ((ans & 1) ^ v) != 0 {
ans++
}
}
return ans
}impl Solution {
pub fn min_flips(target: String) -> i32 {
let mut ans = 0;
for c in target.chars() {
let bit = (c as u8 - b'0') as i32;
if ans % 2 != bit {
ans += 1;
}
}
ans
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Track Transitions in Target String | Time Complexity: O(n), where n is the length of the target string. |
| Pattern Matching from Index 0 | Time Complexity: O(n), covering every character in `target`. |
| Greedy | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Track Transitions in Target String (Greedy) | O(n) | O(1) | Best general solution. Counts transitions to determine flips without simulation. |
| Pattern Matching from Index 0 (Flip Simulation) | O(n) | O(1) | Useful when explaining the logic step‑by‑step by tracking the current flipped state. |
Frequently asked questions
Is Minimum Suffix Flips easy or hard?
Minimum Suffix Flips is classified as a Medium problem on LeetCode. The implementation is short, but the challenge lies in recognizing the greedy insight that flips correspond to transitions in the target string.
How to solve Minimum Suffix Flips in O(n)?
Iterate through the target string while maintaining the current effective bit after previous flips. If the effective value differs from the target character, increment the flip counter and toggle the state. This greedy strategy ensures each index is processed once, giving O(n) time complexity.
What is the best approach for Minimum Suffix Flips?
The optimal approach is a greedy transition-count method. Traverse the target string and track the current effective bit after previous flips. Whenever the desired bit differs from the current state, perform a flip and toggle the state. This solves the problem in O(n) time and O(1) space.
What data structure is used in Minimum Suffix Flips?
The problem mainly uses string traversal with a greedy strategy. No complex data structures are required. A simple counter and a variable tracking the current flip state are enough to implement the optimal solution.
What is the time complexity of Minimum Suffix Flips?
The optimal solution runs in O(n) time because it scans the target string once. Each character is processed with constant-time checks. The space complexity is O(1) since only a few variables are used to track the current flip state.
Minimum Suffix Flips Python or Java solution approach?
Both Python and Java implementations follow the same logic: iterate through the string, compare the effective bit with the target bit, and increment the flip counter when they differ. The algorithm runs in O(n) time and uses constant extra space.
Is Minimum Suffix Flips asked at Google, Amazon, or Meta interviews?
Greedy string problems like Minimum Suffix Flips appear in interviews at large tech companies including Amazon and Google. The problem tests your ability to recognize patterns in state transitions and avoid unnecessary simulation of operations.