LeetCode #3750 — EASY

Minimum Number of Flips to Reverse Binary String

Build confidence with an intuition-first walkthrough focused on math fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a positive integer n.

Let s be the binary representation of n without leading zeros.

The reverse of a binary string s is obtained by writing the characters of s in the opposite order.

You may flip any bit in s (change 0 → 1 or 1 → 0). Each flip affects exactly one bit.

Return the minimum number of flips required to make s equal to the reverse of its original form.

Example 1:

Input: n = 7

Output: 0

Explanation:

The binary representation of 7 is "111". Its reverse is also "111", which is the same. Hence, no flips are needed.

Example 2:

Input: n = 10

Output: 4

Explanation:

The binary representation of 10 is "1010". Its reverse is "0101". All four bits must be flipped to make them equal. Thus, the minimum number of flips required is 4.

Constraints:

1 <= n <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer n. Let s be the binary representation of n without leading zeros. The reverse of a binary string s is obtained by writing the characters of s in the opposite order. You may flip any bit in s (change 0 → 1 or 1 → 0). Each flip affects exactly one bit. Return the minimum number of flips required to make s equal to the reverse of its original form.

Baseline thinking

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

Pattern signal: Math · Two Pointers · Bit Manipulation

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

Generate the reverse of the binary string and use two pointers from the ends to determine where flips are needed.

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 #3750: Minimum Number of Flips to Reverse Binary String
class Solution {
    public int minimumFlips(int n) {
        String s = Integer.toBinaryString(n);
        int m = s.length();
        int cnt = 0;
        for (int i = 0; i < m / 2; i++) {
            if (s.charAt(i) != s.charAt(m - i - 1)) {
                cnt++;
            }
        }
        return cnt * 2;
    }
}

// Accepted solution for LeetCode #3750: Minimum Number of Flips to Reverse Binary String
func minimumFlips(n int) int {
    s := strconv.FormatInt(int64(n), 2)
    m := len(s)
    cnt := 0
    for i := 0; i < m/2; i++ {
        if s[i] != s[m-i-1] {
            cnt++
        }
    }
    return cnt * 2
}

# Accepted solution for LeetCode #3750: Minimum Number of Flips to Reverse Binary String
class Solution:
    def minimumFlips(self, n: int) -> int:
        s = bin(n)[2:]
        m = len(s)
        return sum(s[i] != s[m - i - 1] for i in range(m // 2)) * 2

// Accepted solution for LeetCode #3750: Minimum Number of Flips to Reverse Binary String
fn minimum_flips(n: i32) -> i32 {
    let s = format!("{n:b}");
    let cs : Vec<_> = s.chars().collect();
    let reverse : Vec<_> = s.chars().rev().collect();

    cs.into_iter().zip(reverse.into_iter()).filter(|(a, b)| a != b).count() as i32
}

fn main() {
    let ret = minimum_flips(10);
    println!("ret={ret}");
}

#[test]
fn test() {
    assert_eq!(minimum_flips(7), 0);
    assert_eq!(minimum_flips(10), 4);
}

// Accepted solution for LeetCode #3750: Minimum Number of Flips to Reverse Binary String
function minimumFlips(n: number): number {
    const s = n.toString(2);
    const m = s.length;
    let cnt = 0;
    for (let i = 0; i < m / 2; i++) {
        if (s[i] !== s[m - i - 1]) {
            cnt++;
        }
    }
    return cnt * 2;
}

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(log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.