LeetCode #767 — MEDIUM

Reorganize String

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

The Problem

Problem Statement

Given a string s, rearrange the characters of s so that any two adjacent characters are not the same.

Return any possible rearrangement of s or return "" if not possible.

Example 1:

Input: s = "aab"
Output: "aba"

Example 2:

Input: s = "aaab"
Output: ""

Constraints:

1 <= s.length <= 500
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, rearrange the characters of s so that any two adjacent characters are not the same. Return any possible rearrangement of s or return "" if not possible.

Baseline thinking

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

Pattern signal: Hash Map · Greedy

Example 1

"aab"

Example 2

"aaab"

Core Insight

What unlocks the optimal approach

Alternate placing the most common letters.

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 #767: Reorganize String
class Solution {
    public String reorganizeString(String s) {
        int[] cnt = new int[26];
        int mx = 0;
        for (char c : s.toCharArray()) {
            int t = c - 'a';
            ++cnt[t];
            mx = Math.max(mx, cnt[t]);
        }
        int n = s.length();
        if (mx > (n + 1) / 2) {
            return "";
        }
        int k = 0;
        for (int v : cnt) {
            if (v > 0) {
                ++k;
            }
        }
        int[][] m = new int[k][2];
        k = 0;
        for (int i = 0; i < 26; ++i) {
            if (cnt[i] > 0) {
                m[k++] = new int[] {cnt[i], i};
            }
        }
        Arrays.sort(m, (a, b) -> b[0] - a[0]);
        k = 0;
        StringBuilder ans = new StringBuilder(s);
        for (int[] e : m) {
            int v = e[0], i = e[1];
            while (v-- > 0) {
                ans.setCharAt(k, (char) ('a' + i));
                k += 2;
                if (k >= n) {
                    k = 1;
                }
            }
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #767: Reorganize String
func reorganizeString(s string) string {
	cnt := make([]int, 26)
	for _, c := range s {
		t := c - 'a'
		cnt[t]++
	}
	mx := slices.Max(cnt)
	n := len(s)
	if mx > (n+1)/2 {
		return ""
	}
	m := [][]int{}
	for i, v := range cnt {
		if v > 0 {
			m = append(m, []int{v, i})
		}
	}
	sort.Slice(m, func(i, j int) bool {
		return m[i][0] > m[j][0]
	})
	ans := make([]byte, n)
	k := 0
	for _, e := range m {
		v, i := e[0], e[1]
		for v > 0 {
			ans[k] = byte('a' + i)
			k += 2
			if k >= n {
				k = 1
			}
			v--
		}
	}
	return string(ans)
}

# Accepted solution for LeetCode #767: Reorganize String
class Solution:
    def reorganizeString(self, s: str) -> str:
        n = len(s)
        cnt = Counter(s)
        mx = max(cnt.values())
        if mx > (n + 1) // 2:
            return ''
        i = 0
        ans = [None] * n
        for k, v in cnt.most_common():
            while v:
                ans[i] = k
                v -= 1
                i += 2
                if i >= n:
                    i = 1
        return ''.join(ans)

// Accepted solution for LeetCode #767: Reorganize String
use std::collections::{BinaryHeap, HashMap, VecDeque};

impl Solution {
    #[allow(dead_code)]
    pub fn reorganize_string(s: String) -> String {
        let mut map = HashMap::new();
        let mut pq = BinaryHeap::new();
        let mut ret = String::new();
        let mut queue = VecDeque::new();
        let n = s.len();

        // Initialize the HashMap
        for c in s.chars() {
            map.entry(c)
                .and_modify(|e| {
                    *e += 1;
                })
                .or_insert(1);
        }

        // Initialize the binary heap
        for (k, v) in map.iter() {
            if 2 * *v - 1 > n {
                return "".to_string();
            } else {
                pq.push((*v, *k));
            }
        }

        while !pq.is_empty() {
            let (v, k) = pq.pop().unwrap();
            ret.push(k);
            queue.push_back((v - 1, k));
            if queue.len() == 2 {
                let (v, k) = queue.pop_front().unwrap();
                if v != 0 {
                    pq.push((v, k));
                }
            }
        }

        if ret.len() == n {
            ret
        } else {
            "".to_string()
        }
    }
}

// Accepted solution for LeetCode #767: Reorganize String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #767: Reorganize String
// class Solution {
//     public String reorganizeString(String s) {
//         int[] cnt = new int[26];
//         int mx = 0;
//         for (char c : s.toCharArray()) {
//             int t = c - 'a';
//             ++cnt[t];
//             mx = Math.max(mx, cnt[t]);
//         }
//         int n = s.length();
//         if (mx > (n + 1) / 2) {
//             return "";
//         }
//         int k = 0;
//         for (int v : cnt) {
//             if (v > 0) {
//                 ++k;
//             }
//         }
//         int[][] m = new int[k][2];
//         k = 0;
//         for (int i = 0; i < 26; ++i) {
//             if (cnt[i] > 0) {
//                 m[k++] = new int[] {cnt[i], i};
//             }
//         }
//         Arrays.sort(m, (a, b) -> b[0] - a[0]);
//         k = 0;
//         StringBuilder ans = new StringBuilder(s);
//         for (int[] e : m) {
//             int v = e[0], i = e[1];
//             while (v-- > 0) {
//                 ans.setCharAt(k, (char) ('a' + i));
//                 k += 2;
//                 if (k >= n) {
//                     k = 1;
//                 }
//             }
//         }
//         return ans.toString();
//     }
// }

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.