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Maximum Difference Between Even and Odd Frequency I

· Easy

LeetCode problem 3442

This question is like quite straightforward I reckon.

Just iterate through the string s, keeping track of each letter's frequency as you iterate through it using a hashmap.

Afterwards, just iterate through all the letters and find the:

  • maxOdd the maximum odd count of a letter
  • minEven the minimum even count of a letter

Then, simply return maxOdd - minEven as the solution.

code

class Solution {
public:
    int maxDifference(string s) {
        int alphabetFrequency[26] = {0};

        for (char c : s) {
            ++alphabetFrequency[c-'a'];
        }

        int minEven = INT32_MAX;
        int maxOdd = 0;
        for (int i = 0; i < 26; ++i) {
            if (alphabetFrequency[i] != 0) {
                if (alphabetFrequency[i] % 2 == 0) {
                    minEven = min(minEven, alphabetFrequency[i]);
                } else {
                    maxOdd = max(maxOdd, alphabetFrequency[i]);
                }
            }
        }
        return maxOdd - minEven;
    }
};

complexity

  • Time: O(n) where n is the size of the string s
  • Space: O(1) since there are always a maximum of 26 letters we have to store the count for