LeetCode #2914 — MEDIUM

Minimum Number of Changes to Make Binary String Beautiful

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed binary string s having an even length.

A string is beautiful if it's possible to partition it into one or more substrings such that:

Each substring has an even length.
Each substring contains only 1's or only 0's.

You can change any character in s to 0 or 1.

Return the minimum number of changes required to make the string s beautiful.

Example 1:

Input: s = "1001"
Output: 2
Explanation: We change s[1] to 1 and s[3] to 0 to get string "1100".
It can be seen that the string "1100" is beautiful because we can partition it into "11|00".
It can be proven that 2 is the minimum number of changes needed to make the string beautiful.

Example 2:

Input: s = "10"
Output: 1
Explanation: We change s[1] to 1 to get string "11".
It can be seen that the string "11" is beautiful because we can partition it into "11".
It can be proven that 1 is the minimum number of changes needed to make the string beautiful.

Example 3:

Input: s = "0000"
Output: 0
Explanation: We don't need to make any changes as the string "0000" is beautiful already.

Constraints:

2 <= s.length <= 10⁵
s has an even length.
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed binary string s having an even length. A string is beautiful if it's possible to partition it into one or more substrings such that: Each substring has an even length. Each substring contains only 1's or only 0's. You can change any character in s to 0 or 1. Return the minimum number of changes required to make the string s beautiful.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"1001"

Example 2

"10"

Example 3

"0000"

Step 02

Core Insight

What unlocks the optimal approach

For any valid partition, since each part consists of an even number of the same characters, we can further partition each part into lengths of exactly <code>2</code>.
After noticing the first hint, we can decompose the whole string into disjoint blocks of size <code>2</code> and find the minimum number of changes required to make those blocks beautiful.

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 #2914: Minimum Number of Changes to Make Binary String Beautiful
class Solution {
    public int minChanges(String s) {
        int ans = 0;
        for (int i = 1; i < s.length(); i += 2) {
            if (s.charAt(i) != s.charAt(i - 1)) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2914: Minimum Number of Changes to Make Binary String Beautiful
func minChanges(s string) (ans int) {
	for i := 1; i < len(s); i += 2 {
		if s[i] != s[i-1] {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2914: Minimum Number of Changes to Make Binary String Beautiful
class Solution:
    def minChanges(self, s: str) -> int:
        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))

// Accepted solution for LeetCode #2914: Minimum Number of Changes to Make Binary String Beautiful
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2914: Minimum Number of Changes to Make Binary String Beautiful
// class Solution {
//     public int minChanges(String s) {
//         int ans = 0;
//         for (int i = 1; i < s.length(); i += 2) {
//             if (s.charAt(i) != s.charAt(i - 1)) {
//                 ++ans;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2914: Minimum Number of Changes to Make Binary String Beautiful
function minChanges(s: string): number {
    let ans = 0;
    for (let i = 1; i < s.length; i += 2) {
        if (s[i] !== s[i - 1]) {
            ++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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.