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+---
+id: sort-characters-by-frequency
+title: Sort Characters By Frequency (LeetCode)
+sidebar_label: 0451-Sort Characters By Frequency
+tags:
+  - Hash Table
+  - String
+  - Sorting
+  - Heap(Priority Queue)
+  - Bucket Sort
+  - Counting
+description: Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.
+sidebar_position: 0451
+---
+
+## Problem Description
+
+Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.
+
+Return the sorted string. If there are multiple answers, return any of them.
+
+Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
+
+### Example 1
+
+- **Input:** ` s = "tree" `
+- **Output:** `"eert"`
+- **Explanation:** 'e' appears twice while 'r' and 't' both appear once.So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+
+### Example 2
+
+- **Input:** ` s = "cccaaa" `
+- **Output:** `"aaaccc"`
+- **Explanation:** Both 'c' and 'a' appear three times, so both "cccaaa" and "aaaccc" are valid answers.Note that "cacaca" is incorrect, as the same characters must be together.
+
+### Example 3
+
+- **Input:** ` s = "Aabb" `
+- **Output:** `"bbAa"`
+- **Explanation:** "bbaA" is also a valid answer, but "Aabb" is incorrect.Note that 'A' and 'a' are treated as two different characters.
+
+
+
+### Constraints
+
+- `1 <= s.length <= 5 * 105`
+- `s consists of uppercase and lowercase English letters and digits.`
+
+## Approach
+
+  - 1. Count the occurrences: We need to go through the string and count the occurrences of each character. This can be done efficiently by using a hash table or a counter data structure where each character is a key, and its count is the value.
+      
+  - 2. Sort based on counts: Once we have the counts, the next step is to sort the characters by these counts in descending order. We want the characters with higher counts to come first.
+      
+  - 3. With these counts, we can construct a new string.
+
+
+
+### Solution Code
+
+#### C++
+
+```c++
+class Solution {
+public:
+    string frequencySort(string s) {
+        unordered_map<char, int> frequencyMap;
+        for (char ch : s) {
+            ++frequencyMap[ch];
+        }
+        vector<char> uniqueChars;
+        for (auto& keyValue : frequencyMap) {
+            uniqueChars.push_back(keyValue.first);
+        }
+        sort(uniqueChars.begin(), uniqueChars.end(), [&](char a, char b) {
+            return frequencyMap[a] > frequencyMap[b];
+        });
+        string result;
+        for (char ch : uniqueChars) {
+            result += string(frequencyMap[ch], ch);
+        }
+        return result;
+    }
+};
+
+```
+
+#### java
+```java
+class Solution {
+    public String frequencySort(String s) {
+        Map<Character, Integer> frequencyMap = new HashMap<>(52);
+        for (int i = 0; i < s.length(); ++i) {
+            frequencyMap.merge(s.charAt(i), 1, Integer::sum);
+        }
+        List<Character> characters = new ArrayList<>(frequencyMap.keySet());
+        characters.sort((a, b) -> frequencyMap.get(b) - frequencyMap.get(a));
+        StringBuilder sortedString = new StringBuilder();
+        for (char c : characters) {
+            for (int frequency = frequencyMap.get(c); frequency > 0; --frequency) {
+                sortedString.append(c);
+            }
+        }
+        return sortedString.toString();
+    }
+}
+
+```
+
+#### Python
+```python
+class Solution:
+    def frequencySort(self, s: str) -> str:
+        char_frequency = Counter(s)
+        sorted_characters = sorted(char_frequency.items(), key=lambda item: -item[1])
+        frequency_sorted_string = ''.join(character * frequency for character, frequency in sorted_characters)
+        return frequency_sorted_string
+```
+
+#### Conclusion
+The above solutions use two-pointers approach to reverse the string.
+  - 1. Time complexity: O(n logn)
+  - 2. Space complexity: O(n)