LeetCode #1758 — EASY

Minimum Changes To Make Alternating Binary String

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa.

The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not.

Return the minimum number of operations needed to make s alternating.

Example 1:

Input: s = "0100"
Output: 1
Explanation: If you change the last character to '1', s will be "0101", which is alternating.

Example 2:

Input: s = "10"
Output: 0
Explanation: s is already alternating.

Example 3:

Input: s = "1111"
Output: 2
Explanation: You need two operations to reach "0101" or "1010".

Constraints:

1 <= s.length <= 10⁴
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa. The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not. Return the minimum number of operations needed to make s alternating.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"0100"

Example 2

"10"

Example 3

"1111"

Core Insight

What unlocks the optimal approach

Think about how the final string will look like.
It will either start with a '0' and be like '010101010..' or with a '1' and be like '10101010..'
Try both ways, and check for each way, the number of changes needed to reach it from the given string. The answer is the minimum of both ways.

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 #1758: Minimum Changes To Make Alternating Binary String
class Solution {
    public int minOperations(String s) {
        int cnt = 0, n = s.length();
        for (int i = 0; i < n; ++i) {
            cnt += (s.charAt(i) != "01".charAt(i & 1) ? 1 : 0);
        }
        return Math.min(cnt, n - cnt);
    }
}

// Accepted solution for LeetCode #1758: Minimum Changes To Make Alternating Binary String
func minOperations(s string) int {
	cnt := 0
	for i, c := range s {
		if c != []rune("01")[i&1] {
			cnt++
		}
	}
	return min(cnt, len(s)-cnt)
}

# Accepted solution for LeetCode #1758: Minimum Changes To Make Alternating Binary String
class Solution:
    def minOperations(self, s: str) -> int:
        cnt = sum(c != '01'[i & 1] for i, c in enumerate(s))
        return min(cnt, len(s) - cnt)

// Accepted solution for LeetCode #1758: Minimum Changes To Make Alternating Binary String
impl Solution {
    pub fn min_operations(s: String) -> i32 {
        let n = s.len();
        let s = s.as_bytes();
        let cs = [b'0', b'1'];
        let mut count = 0;
        for i in 0..n {
            count += if s[i] != cs[i & 1] { 1 } else { 0 };
        }
        count.min(n - count) as i32
    }
}

// Accepted solution for LeetCode #1758: Minimum Changes To Make Alternating Binary String
function minOperations(s: string): number {
    const n = s.length;
    let count = 0;
    for (let i = 0; i < n; i++) {
        count += s[i] !== '01'[i & 1] ? 1 : 0;
    }
    return Math.min(count, n - count);
}

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.