Maximum Difference Between Even and Odd Frequency I

Maximum Difference Between Even and Odd Frequency I - Problem

You are given a string s consisting of lowercase English letters. Your task is to find the maximum difference diff = freq(a₁) - freq(a₂) between the frequency of characters a₁ and a₂ in the string such that:

a₁ has an odd frequency in the string.
a₂ has an even frequency in the string.

Return this maximum difference.

Input & Output

Example 1 — Mixed Frequencies

$ Input: s = "aabbbcc"

› Output: 1

💡 Note: Frequencies: a=2 (even), b=3 (odd), c=2 (even). Max odd frequency is 3, min even frequency is 2. Difference = 3 - 2 = 1.

Example 2 — All Even Frequencies

$ Input: s = "abab"

› Output: 0

💡 Note: Frequencies: a=2 (even), b=2 (even). No odd frequencies exist, so we cannot form the required difference. Return 0.

Example 3 — Single Character

$ Input: s = "a"

› Output: 0

💡 Note: Frequency: a=1 (odd). Only odd frequency exists, no even frequencies to subtract from. Return 0.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to find the maximum difference between any odd frequency character and any even frequency character. The optimal approach counts character frequencies in O(n) time and tracks the maximum odd frequency and minimum even frequency. Best approach: Single pass frequency counting with simultaneous max/min tracking. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force Frequency Comparison

⏱️ Time: O(n + k²) Space: O(k)

First count the frequency of each character. Then iterate through all characters with odd frequencies and compare them against all characters with even frequencies to find the maximum difference.

One-Pass Optimal Solution

⏱️ Time: O(n) Space: O(k)

Count frequencies while simultaneously tracking the maximum odd frequency and minimum even frequency. This eliminates the need for a second pass and gives us the optimal solution in one iteration through the string.

Two-Pass Frequency Count

⏱️ Time: O(n + k) Space: O(k)

Count character frequencies in first pass. In second pass, track the maximum odd frequency and minimum even frequency. The difference gives us the answer directly without comparing all pairs.

Brute Force Frequency Comparison — Algorithm Steps

Count frequency of each character in the string
Find all characters with odd frequencies
Find all characters with even frequencies
Compare each odd frequency against each even frequency
Return the maximum difference found

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Count each character's frequency in the string

Separate Odd/Even

Group frequencies by odd and even

Compare All Pairs

Find maximum difference between any odd and even frequency

Code -

solution.c — C

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

int solution(char* s) {
    // Count frequencies of all characters
    int freq[26] = {0};
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        freq[s[i] - 'a']++;
    }
    
    // Separate odd and even frequencies
    int oddFreqs[26], evenFreqs[26];
    int oddCount = 0, evenCount = 0;
    
    for (int i = 0; i < 26; i++) {
        if (freq[i] > 0) {
            if (freq[i] % 2 == 1) {
                oddFreqs[oddCount++] = freq[i];
            } else {
                evenFreqs[evenCount++] = freq[i];
            }
        }
    }
    
    // If no odd or no even frequencies, return 0
    if (oddCount == 0 || evenCount == 0) {
        return 0;
    }
    
    // Find maximum difference by comparing all pairs
    int maxDiff = 0;
    for (int i = 0; i < oddCount; i++) {
        for (int j = 0; j < evenCount; j++) {
            int diff = oddFreqs[i] - evenFreqs[j];
            if (diff > maxDiff) {
                maxDiff = diff;
            }
        }
    }
    
    return maxDiff;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + k²)

O(n) to count frequencies, then O(k²) to compare all odd vs even frequencies where k ≤ 26

✓ Linear Growth

Space Complexity

O(k)

Hash map stores at most 26 character frequencies

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

450 Likes

Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Frequency Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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