LeetCode #1864 — MEDIUM

Minimum Number of Swaps to Make the Binary String Alternating

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

The Problem

Problem Statement

Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible.

The string is called alternating if no two adjacent characters are equal. For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

Any two characters may be swapped, even if they are not adjacent.

Example 1:

Input: s = "111000"
Output: 1
Explanation: Swap positions 1 and 4: "111000" -> "101010"
The string is now alternating.

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating, no swaps are needed.

Example 3:

Input: s = "1110"
Output: -1

Constraints:

1 <= s.length <= 1000
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible. The string is called alternating if no two adjacent characters are equal. For example, the strings "010" and "1010" are alternating, while the string "0100" is not. Any two characters may be swapped, even if they are not adjacent.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"111000"

Example 2

"010"

Example 3

"1110"

Step 02

Core Insight

What unlocks the optimal approach

Think about all valid strings of length n.
Try to count the mismatched positions with each valid string of length n.

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 #1864: Minimum Number of Swaps to Make the Binary String Alternating
class Solution {
    private char[] s;

    public int minSwaps(String s) {
        this.s = s.toCharArray();
        int n1 = 0;
        for (char c : this.s) {
            n1 += (c - '0');
        }
        int n0 = this.s.length - n1;
        if (Math.abs(n0 - n1) > 1) {
            return -1;
        }
        if (n0 == n1) {
            return Math.min(calc(0), calc(1));
        }
        return calc(n0 > n1 ? 0 : 1);
    }

    private int calc(int c) {
        int cnt = 0;
        for (int i = 0; i < s.length; ++i) {
            int x = s[i] - '0';
            if ((i & 1 ^ c) != x) {
                ++cnt;
            }
        }
        return cnt / 2;
    }
}

// Accepted solution for LeetCode #1864: Minimum Number of Swaps to Make the Binary String Alternating
func minSwaps(s string) int {
	n0 := strings.Count(s, "0")
	n1 := len(s) - n0
	if abs(n0-n1) > 1 {
		return -1
	}
	calc := func(c int) int {
		cnt := 0
		for i, ch := range s {
			x := int(ch - '0')
			if i&1^c != x {
				cnt++
			}
		}
		return cnt / 2
	}
	if n0 == n1 {
		return min(calc(0), calc(1))
	}
	if n0 > n1 {
		return calc(0)
	}
	return calc(1)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1864: Minimum Number of Swaps to Make the Binary String Alternating
class Solution:
    def minSwaps(self, s: str) -> int:
        def calc(c: int) -> int:
            return sum((c ^ i & 1) != x for i, x in enumerate(map(int, s))) // 2

        n0 = s.count("0")
        n1 = len(s) - n0
        if abs(n0 - n1) > 1:
            return -1
        if n0 == n1:
            return min(calc(0), calc(1))
        return calc(0 if n0 > n1 else 1)

// Accepted solution for LeetCode #1864: Minimum Number of Swaps to Make the Binary String Alternating
/**
 * [1864] Minimum Number of Swaps to Make the Binary String Alternating
 *
 * Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible.
 * The string is called alternating if no two adjacent characters are equal. For example, the strings "010" and "1010" are alternating, while the string "0100" is not.
 * Any two characters may be swapped, even if they are not adjacent.
 *  
 * Example 1:
 *
 * Input: s = "111000"
 * Output: 1
 * Explanation: Swap positions 1 and 4: "1<u>1</u>10<u>0</u>0" -> "1<u>0</u>10<u>1</u>0"
 * The string is now alternating.
 *
 * Example 2:
 *
 * Input: s = "010"
 * Output: 0
 * Explanation: The string is already alternating, no swaps are needed.
 *
 * Example 3:
 *
 * Input: s = "1110"
 * Output: -1
 *
 *  
 * Constraints:
 *
 * 	1 <= s.length <= 1000
 * 	s[i] is either '0' or '1'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-binary-string-alternating/
// discuss: https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-binary-string-alternating/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_swaps(s: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1864_example_1() {
        let s = "111000".to_string();

        let result = 1;

        assert_eq!(Solution::min_swaps(s), result);
    }

    #[test]
    #[ignore]
    fn test_1864_example_2() {
        let s = "010".to_string();

        let result = 0;

        assert_eq!(Solution::min_swaps(s), result);
    }

    #[test]
    #[ignore]
    fn test_1864_example_3() {
        let s = "1110".to_string();

        let result = -1;

        assert_eq!(Solution::min_swaps(s), result);
    }
}

// Accepted solution for LeetCode #1864: Minimum Number of Swaps to Make the Binary String Alternating
function minSwaps(s: string): number {
    const n0 = (s.match(/0/g) || []).length;
    const n1 = s.length - n0;
    if (Math.abs(n0 - n1) > 1) {
        return -1;
    }
    const calc = (c: number): number => {
        let cnt = 0;
        for (let i = 0; i < s.length; i++) {
            const x = +s[i];
            if (((i & 1) ^ c) !== x) {
                cnt++;
            }
        }
        return Math.floor(cnt / 2);
    };
    if (n0 === n1) {
        return Math.min(calc(0), calc(1));
    }
    return calc(n0 > n1 ? 0 : 1);
}

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.