Remove Letter To Equalize Frequency

Remove Letter To Equalize Frequency - Problem

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.

Input & Output

Example 1 — Two Close Frequencies

$ Input: word = "abcc"

› Output: true

💡 Note: Remove one 'c' to get "abc": a:1, b:1, c:1 - all frequencies equal

Example 2 — Cannot Equalize

$ Input: word = "aab"

› Output: true

💡 Note: Remove one 'a' to get "ab": a:1, b:1 - all frequencies equal

Example 3 — Single Character Type

$ Input: word = "aaa"

› Output: true

💡 Note: Remove one 'a' to get "aa": only character 'a' with frequency 2

Constraints

1 ≤ word.length ≤ 10⁵
word consists of lowercase English letters
You must remove exactly one character

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8

The key insight is recognizing that only specific frequency patterns allow successful equalization after removing one character. The optimal approach uses frequency analysis to identify these patterns in O(n) time rather than testing each removal. Valid patterns include: all equal frequencies, two frequencies differing by 1, or one character with frequency 1 that can be eliminated.

Common Approaches

✓ Brute Force - Try Every Removal

⏱️ Time: O(n²) Space: O(n)

For each position in the string, simulate removing that character and count frequencies of remaining characters. Check if all frequencies are equal.

Frequency Analysis - Pattern Recognition

⏱️ Time: O(n) Space: O(1)

Count character frequencies once, then check if removing one character can create equal frequencies by analyzing frequency distribution patterns.

Brute Force - Try Every Removal — Algorithm Steps

Step 1: For each index i in word
Step 2: Create new string without character at index i
Step 3: Count frequencies in new string
Step 4: Check if all frequencies are equal

Visualization

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Step-by-Step Walkthrough

Try Each Position

Remove character at each index

Count Frequencies

Count remaining character frequencies

Check Equality

Verify all frequencies are equal

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* word) {
    int n = strlen(word);
    
    for (int i = 0; i < n; i++) {
        // Create string without character at index i
        char newWord[1001];
        int k = 0;
        for (int j = 0; j < n; j++) {
            if (j != i) {
                newWord[k++] = word[j];
            }
        }
        newWord[k] = '\0';
        
        if (k == 0) continue;
        
        // Count frequencies
        int freq[26] = {0};
        for (int j = 0; j < k; j++) {
            freq[newWord[j] - 'a']++;
        }
        
        // Check if all frequencies are equal
        int targetFreq = -1;
        bool allEqual = true;
        
        for (int j = 0; j < 26; j++) {
            if (freq[j] > 0) {
                if (targetFreq == -1) {
                    targetFreq = freq[j];
                } else if (freq[j] != targetFreq) {
                    allEqual = false;
                    break;
                }
            }
        }
        
        if (allEqual && targetFreq > 0) {
            return true;
        }
    }
    
    return false;
}

int main() {
    char word[1001];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    bool result = solution(word);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n characters, we create new string and count frequencies (O(n) each)

⚠ Quadratic Growth

Space Complexity

O(n)

Store temporary strings and frequency maps

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

856 Likes

Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try Every Removal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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