LeetCode #1529 — MEDIUM

Minimum Suffix Flips

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

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.length
1 <= n <= 10⁵
target[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"10111"

Example 2

"101"

Example 3

"00000"

Core Insight

What unlocks the optimal approach

Consider a strategy where the choice of bulb with number i is increasing. In such a strategy, you no longer need to worry about bulbs that have been set to the left.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1529: Minimum Suffix Flips
class Solution {
    public int minFlips(String target) {
        int ans = 0;
        for (int i = 0; i < target.length(); ++i) {
            int v = target.charAt(i) - '0';
            if (((ans & 1) ^ v) != 0) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1529: Minimum Suffix Flips
func minFlips(target string) int {
	ans := 0
	for _, c := range target {
		v := int(c - '0')
		if ((ans & 1) ^ v) != 0 {
			ans++
		}
	}
	return ans
}

# Accepted solution for LeetCode #1529: Minimum Suffix Flips
class Solution:
    def minFlips(self, target: str) -> int:
        ans = 0
        for v in target:
            if (ans & 1) ^ int(v):
                ans += 1
        return ans

// Accepted solution for LeetCode #1529: Minimum Suffix Flips
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
    }
}

// Accepted solution for LeetCode #1529: Minimum Suffix Flips
function minFlips(target: string): number {
    let ans = 0;
    for (const c of target) {
        if (ans % 2 !== +c) {
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.