LeetCode #1417 — EASY

Reformat The String

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

The Problem

Problem Statement

You are given an alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Constraints:

1 <= s.length <= 500
s consists of only lowercase English letters and/or digits.

Step 01

Brute Force Baseline

Problem summary: You are given an alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits). You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type. Return the reformatted string or return an empty string if it is impossible to reformat the string.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"a0b1c2"

Example 2

"leetcode"

Example 3

"1229857369"

Step 02

Core Insight

What unlocks the optimal approach

Count the number of letters and digits in the string. if cntLetters - cntDigits has any of the values [-1, 0, 1] we have an answer, otherwise we don't have any answer.
Build the string anyway as you wish. Keep in mind that you need to start with the type that has more characters if cntLetters ≠ cntDigits.

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 #1417: Reformat The String
class Solution {
    public String reformat(String s) {
        StringBuilder a = new StringBuilder();
        StringBuilder b = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (Character.isDigit(c)) {
                a.append(c);
            } else {
                b.append(c);
            }
        }
        int m = a.length(), n = b.length();
        if (Math.abs(m - n) > 1) {
            return "";
        }
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < Math.min(m, n); ++i) {
            if (m > n) {
                ans.append(a.charAt(i));
                ans.append(b.charAt(i));
            } else {
                ans.append(b.charAt(i));
                ans.append(a.charAt(i));
            }
        }
        if (m > n) {
            ans.append(a.charAt(m - 1));
        }
        if (m < n) {
            ans.append(b.charAt(n - 1));
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #1417: Reformat The String
func reformat(s string) string {
	a := []byte{}
	b := []byte{}
	for _, c := range s {
		if unicode.IsLetter(c) {
			a = append(a, byte(c))
		} else {
			b = append(b, byte(c))
		}
	}
	if len(a) < len(b) {
		a, b = b, a
	}
	if len(a)-len(b) > 1 {
		return ""
	}
	var ans strings.Builder
	for i := range b {
		ans.WriteByte(a[i])
		ans.WriteByte(b[i])
	}
	if len(a) > len(b) {
		ans.WriteByte(a[len(a)-1])
	}
	return ans.String()
}

# Accepted solution for LeetCode #1417: Reformat The String
class Solution:
    def reformat(self, s: str) -> str:
        a = [c for c in s if c.islower()]
        b = [c for c in s if c.isdigit()]
        if abs(len(a) - len(b)) > 1:
            return ''
        if len(a) < len(b):
            a, b = b, a
        ans = []
        for x, y in zip(a, b):
            ans.append(x + y)
        if len(a) > len(b):
            ans.append(a[-1])
        return ''.join(ans)

// Accepted solution for LeetCode #1417: Reformat The String
struct Solution;
use std::mem::swap;

impl Solution {
    fn reformat(s: String) -> String {
        let mut chars: Vec<char> = vec![];
        let mut digits: Vec<char> = vec![];
        let mut res: Vec<char> = vec![];
        for c in s.chars() {
            if c.is_digit(10) {
                digits.push(c);
            } else {
                chars.push(c);
            }
        }
        let mut iter;
        let mut next_iter;
        if digits.len() >= chars.len() {
            if digits.len() > chars.len() + 1 {
                return "".to_string();
            } else {
                iter = digits.iter();
                next_iter = chars.iter();
            }
        } else {
            if chars.len() > digits.len() + 1 {
                return "".to_string();
            } else {
                iter = chars.iter();
                next_iter = digits.iter();
            }
        }
        while let Some(c) = iter.next() {
            res.push(*c);
            swap(&mut iter, &mut next_iter);
        }
        res.into_iter().collect()
    }
}

#[test]
fn test() {
    let s = "a0b1c2".to_string();
    let res = "0a1b2c".to_string();
    assert_eq!(Solution::reformat(s), res);
    let s = "leetcode".to_string();
    let res = "".to_string();
    assert_eq!(Solution::reformat(s), res);
    let s = "1229857369".to_string();
    let res = "".to_string();
    assert_eq!(Solution::reformat(s), res);
    let s = "covid2019".to_string();
    let res = "c2o0v1i9d".to_string();
    assert_eq!(Solution::reformat(s), res);
    let s = "ab123".to_string();
    let res = "1a2b3".to_string();
    assert_eq!(Solution::reformat(s), res);
}

// Accepted solution for LeetCode #1417: Reformat The String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1417: Reformat The String
// class Solution {
//     public String reformat(String s) {
//         StringBuilder a = new StringBuilder();
//         StringBuilder b = new StringBuilder();
//         for (char c : s.toCharArray()) {
//             if (Character.isDigit(c)) {
//                 a.append(c);
//             } else {
//                 b.append(c);
//             }
//         }
//         int m = a.length(), n = b.length();
//         if (Math.abs(m - n) > 1) {
//             return "";
//         }
//         StringBuilder ans = new StringBuilder();
//         for (int i = 0; i < Math.min(m, n); ++i) {
//             if (m > n) {
//                 ans.append(a.charAt(i));
//                 ans.append(b.charAt(i));
//             } else {
//                 ans.append(b.charAt(i));
//                 ans.append(a.charAt(i));
//             }
//         }
//         if (m > n) {
//             ans.append(a.charAt(m - 1));
//         }
//         if (m < n) {
//             ans.append(b.charAt(n - 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)

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.