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-rw-r--r--max_diff_even_odd_freq1.c31
-rw-r--r--max_diff_even_odd_freq1.py33
2 files changed, 64 insertions, 0 deletions
diff --git a/max_diff_even_odd_freq1.c b/max_diff_even_odd_freq1.c
new file mode 100644
index 0000000..d51e73b
--- /dev/null
+++ b/max_diff_even_odd_freq1.c
@@ -0,0 +1,31 @@
+// 3442. Maximum Difference Between Even and Odd Frequency I
+#include "leetcode.h"
+
+/*
+ * you are given a string 's' consisting of lowercase english letters. your task
+ * is to find the minimum difference 'diff = a1 - a2' between the frequency of
+ * characters 'a1' and 'a2' in the string such that 'a1' has an odd frequency in
+ * the string, and 'a2' has an even frequency in the string. return this maximum
+ * difference.
+ */
+
+int maxDifference(char *s) {
+ int *freq = (int *)calloc(26, sizeof(int)), n = strlen(s);
+ for (int i = 0; i < n; i++)
+ freq[s[i] - 'a']++;
+ int max = INT_MIN, min = INT_MAX;
+ for (int i = 0; i < 26; i++) {
+ if (max < freq[i] && freq[i] % 2)
+ max = freq[i];
+ if (min > freq[i] && !(freq[i] % 2) && freq[i])
+ min = freq[i];
+ }
+ free(freq);
+ return max - min;
+}
+
+int main() {
+ char *s1 = "aaaaabbc", *s2 = "abcabcab";
+ printf("%d\n", maxDifference(s1)); // expect: 3
+ printf("%d\n", maxDifference(s2)); // expect: 1
+}
diff --git a/max_diff_even_odd_freq1.py b/max_diff_even_odd_freq1.py
new file mode 100644
index 0000000..b1d6ad3
--- /dev/null
+++ b/max_diff_even_odd_freq1.py
@@ -0,0 +1,33 @@
+# 3442. Maximum Difference Between Even and Odd Frequency I
+
+"""
+you are given a string 's' consisting of lowercase english letters. your task
+is to find the minimum difference 'diff = a1 - a2' between the frequency of
+characters 'a1' and 'a2' in the string such that 'a1' has an odd frequency in
+the string, and 'a2' has an even frequency in the string. return this maximum
+difference.
+"""
+
+
+class Solution(object):
+ def maxDifference(self, s):
+ """
+ :type s: str
+ :rtype: int
+ """
+ freq = [0] * 26
+ for i in s:
+ freq[ord(i) - ord("a")] += 1
+ maxi, mini = 0, len(s)
+ for i in range(26):
+ if freq[i] % 2 != 0:
+ maxi = max(maxi, freq[i])
+ if freq[i] % 2 == 0 and freq[i] > 0:
+ mini = min(mini, freq[i])
+ return maxi - mini
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxDifference(s="aaaaabbc"))
+ print(obj.maxDifference(s="abcabcab"))