LeetCode #2423 — EASY

Remove Letter To Equalize Frequency

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.

Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise.

Note:

The frequency of a letter x is the number of times it occurs in the string.
You must remove exactly one letter and cannot choose to do nothing.

Example 1:

Input: word = "abcc"
Output: true
Explanation: Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.

Example 2:

Input: word = "aazz"
Output: false
Explanation: We must delete a character, so either the frequency of "a" is 1 and the frequency of "z" is 2, or vice versa. It is impossible to make all present letters have equal frequency.

Constraints:

2 <= word.length <= 100
word consists of lowercase English letters only.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal. Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise. Note: The frequency of a letter x is the number of times it occurs in the string. You must remove exactly one letter and cannot choose to do nothing.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"abcc"

Example 2

"aazz"

Core Insight

What unlocks the optimal approach

Brute force all letters that could be removed.
Use a frequency array of size 26.

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 #2423: Remove Letter To Equalize Frequency
class Solution {
    public boolean equalFrequency(String word) {
        int[] cnt = new int[26];
        for (int i = 0; i < word.length(); ++i) {
            ++cnt[word.charAt(i) - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if (cnt[i] > 0) {
                --cnt[i];
                int x = 0;
                boolean ok = true;
                for (int v : cnt) {
                    if (v == 0) {
                        continue;
                    }
                    if (x > 0 && v != x) {
                        ok = false;
                        break;
                    }
                    x = v;
                }
                if (ok) {
                    return true;
                }
                ++cnt[i];
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #2423: Remove Letter To Equalize Frequency
func equalFrequency(word string) bool {
	cnt := [26]int{}
	for _, c := range word {
		cnt[c-'a']++
	}
	for i := range cnt {
		if cnt[i] > 0 {
			cnt[i]--
			x := 0
			ok := true
			for _, v := range cnt {
				if v == 0 {
					continue
				}
				if x > 0 && v != x {
					ok = false
					break
				}
				x = v
			}
			if ok {
				return true
			}
			cnt[i]++
		}
	}
	return false
}

# Accepted solution for LeetCode #2423: Remove Letter To Equalize Frequency
class Solution:
    def equalFrequency(self, word: str) -> bool:
        cnt = Counter(word)
        for c in cnt.keys():
            cnt[c] -= 1
            if len(set(v for v in cnt.values() if v)) == 1:
                return True
            cnt[c] += 1
        return False

// Accepted solution for LeetCode #2423: Remove Letter To Equalize Frequency
// 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 #2423: Remove Letter To Equalize Frequency
// class Solution {
//     public boolean equalFrequency(String word) {
//         int[] cnt = new int[26];
//         for (int i = 0; i < word.length(); ++i) {
//             ++cnt[word.charAt(i) - 'a'];
//         }
//         for (int i = 0; i < 26; ++i) {
//             if (cnt[i] > 0) {
//                 --cnt[i];
//                 int x = 0;
//                 boolean ok = true;
//                 for (int v : cnt) {
//                     if (v == 0) {
//                         continue;
//                     }
//                     if (x > 0 && v != x) {
//                         ok = false;
//                         break;
//                     }
//                     x = v;
//                 }
//                 if (ok) {
//                     return true;
//                 }
//                 ++cnt[i];
//             }
//         }
//         return false;
//     }
// }

// Accepted solution for LeetCode #2423: Remove Letter To Equalize Frequency
function equalFrequency(word: string): boolean {
    const cnt: number[] = new Array(26).fill(0);
    for (const c of word) {
        cnt[c.charCodeAt(0) - 97]++;
    }
    for (let i = 0; i < 26; ++i) {
        if (cnt[i]) {
            cnt[i]--;
            let x = 0;
            let ok = true;
            for (const v of cnt) {
                if (v === 0) {
                    continue;
                }
                if (x && v !== x) {
                    ok = false;
                    break;
                }
                x = v;
            }
            if (ok) {
                return true;
            }
            cnt[i]++;
        }
    }
    return false;
}

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 + C^2)

Space

O(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

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.