LeetCode 1529 - Minimum Suffix Flips

This problem is asking us to transform an initial binary string s (all zeros) into a target binary string target using a

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LeetCode Problem 1529

Difficulty: 🟡 Medium
Topics: String, Greedy

Solution

Problem Understanding

This problem is asking us to transform an initial binary string s (all zeros) into a target binary string target using a minimum number of suffix flip operations. A suffix flip operation involves picking an index i and flipping every bit from index i to the end of the string. Flipping a bit means turning '0' to '1' and '1' to '0'.

The input target is a string of length n containing only '0' and '1'. The output is the minimum number of flip operations needed to make s equal to target.

Constraints tell us 1 <= n <= 10^5, which means any solution with more than O(n) complexity is likely too slow. The input is guaranteed to be valid, containing only '0' and '1'.

Important edge cases include:

  • target contains all zeros: no operations needed.
  • target contains all ones: may require exactly one flip.
  • target alternates between zeros and ones: will require flips at every change in value.

Understanding these cases upfront ensures we do not overcomplicate or miss simple optimizations.

Approaches

Brute Force

A brute-force approach would simulate the string s and apply the flip operations for every possible index to see if it leads to target. This requires iterating over s and flipping suffixes repeatedly, potentially O(n^2) in time, because each flip operation may modify up to n characters. While correct, this approach is far too slow for the input constraints.

Optimal Approach

The key observation is that we only need to flip when the current state of s does not match the target at the current index. Since s starts with all zeros, we can iterate through target from left to right. Each time the current bit differs from the last "effective" state of s, we perform a flip.

We track the "flip parity" - the number of flips performed so far modulo 2 - to know whether the current bit in s has been flipped an even or odd number of times. This allows us to determine whether it matches target without explicitly modifying the string. Each mismatch triggers a flip, incrementing our operation count. This is O(n) time and O(1) space.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^2) O(n) Simulate all flips, too slow for n up to 10^5
Optimal O(n) O(1) Track parity of flips and count mismatches

Algorithm Walkthrough

  1. Initialize a variable flips to 0, which counts the number of flip operations performed.
  2. Initialize current_state to 0, representing the effective value of s[i] considering all flips performed so far (0 means no flip effect, 1 means flipped).
  3. Iterate over target from index 0 to n-1.
  4. At each index i, convert target[i] to integer bit (0 or 1).
  5. Compare bit with current_state. If they differ, a flip is required at this position.
  6. Increment flips and update current_state to 1 - current_state to reflect the effect of the flip for subsequent positions.
  7. Continue until the end of target. Return flips as the minimum number of operations.

Why it works:

Each flip changes all subsequent bits, so flipping exactly at the positions where the current effective state differs from the target ensures minimal operations. The parity tracking guarantees that we know the effective bit without simulating the entire string.

Python Solution

class Solution:
    def minFlips(self, target: str) -> int:
        flips = 0
        current_state = 0  # 0 means no flip, 1 means flipped

        for char in target:
            bit = int(char)
            if bit != current_state:
                flips += 1
                current_state = 1 - current_state

        return flips

The code initializes flips and current_state. Iterating through each character in target, it checks whether the effective bit matches. When it doesn't, we flip the suffix, increment the count, and update the state. This aligns perfectly with the algorithm above.

Go Solution

func minFlips(target string) int {
    flips := 0
    currentState := 0

    for _, ch := range target {
        bit := int(ch - '0')
        if bit != currentState {
            flips++
            currentState = 1 - currentState
        }
    }

    return flips
}

In Go, we iterate over the string using a for loop with range, converting each rune to an integer bit. The logic is identical to the Python version. Go handles strings and integer conversion carefully, subtracting '0' to get 0 or 1.

Worked Examples

Example 1: "10111"

Index Target[i] Current State Flip? Flips Count
0 1 0 Yes 1
1 0 1 Yes 2
2 1 0 Yes 3
3 1 1 No 3
4 1 1 No 3

Output: 3

Example 2: "101"

Index Target[i] Current State Flip? Flips Count
0 1 0 Yes 1
1 0 1 Yes 2
2 1 0 Yes 3

Output: 3

Example 3: "00000"

Index Target[i] Current State Flip? Flips Count
0 0 0 No 0
1 0 0 No 0
2 0 0 No 0
3 0 0 No 0
4 0 0 No 0

Output: 0

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through the string, comparing bits and updating counters
Space O(1) Only two integer variables used, independent of input size

The algorithm is efficient even for the maximum n = 10^5 and avoids unnecessary string operations.

Test Cases

# Provided examples
assert Solution().minFlips("10111") == 3  # Example 1
assert Solution().minFlips("101") == 3    # Example 2
assert Solution().minFlips("00000") == 0  # Example 3

# Edge cases
assert Solution().minFlips("1") == 1      # Single bit flip
assert Solution().minFlips("0") == 0      # Single bit, already 0
assert Solution().minFlips("11111") == 1  # All ones, single flip
assert Solution().minFlips("010101") == 6 # Alternating bits
assert Solution().minFlips("001100") == 3 # Multiple consecutive flips
Test Why
"10111" Checks typical mixed bits
"101" Short string, minimal flips
"00000" Already matching, no flips needed
"1" Single bit needing a flip
"0" Single bit, no flip
"11111" All ones, single flip from start
"010101" Alternating bits, flip at every change
"001100" Multiple consecutive bits requiring flips

Edge Cases

Single Character: When target has length 1, the answer is either 0 or 1. This tests the algorithm handles small input without errors.

All Zeros or All Ones: These cases test that unnecessary flips are not performed. For all zeros, flips remain 0. For all ones, a single flip at index 0 suffices.

Alternating Bits: Strings like "010101..." require flipping at every change. This tests the algorithm correctly tracks the flip parity and increments the counter at each mismatch without missing flips or double-counting.

Consecutive Blocks: Strings like "001100" verify that multiple consecutive changes are handled correctly, not just alternating bits, ensuring the logic works for sequences of zeros and ones.