`"mississippi"` - **Output**: `{'m': 1, 'i': 4, 's': 4, 'p': 2}` - **Justification**: The string `"mississippi"` contains 'm' once, 'i' four times, 's' four times, and 'p' twice. ### Solution To solve this problem, we will use a dictionary to keep track of the frequency of each character in the string. We will iterate through the string, and for each character, we will update its count in the dictionary. This approach is efficient because it only requires a single pass through the string, and dictionary operations like adding and updating elements are generally fast. ### Step-by-Step Algorithm 1. **Initialize** a dictionary `freqDict` to store character frequencies. 2. **Iterate** through each character `char` in the string `s`. 3. **Check** if `char` is already in `freqDict`: - If **yes**, increment the count by 1. - If **no**, add `char` to `freqDict` with a count of 1. 4. **Return** `freqDict` as the final frequency dictionary. ### Algorithm Walkthrough Using the input `"hello"`: 27:T436,#include #include #include using namespace std; class Solution { public: unordered_map countFrequency(string s) { unordered_map freqDict; // Initialize frequency dictionary for (char c : s) { freqDict[c]++; // Update the count for the character } return freqDict; } }; int main() { Solution solution; // Test the method with example inputs unordered_map result1 = solution.countFrequency("hello"); unordered_map result2 = solution.countFrequency("apple"); unordered_map result3 = solution.countFrequency("mississippi"); // Print the results for (const auto &pair : result1) cout << pair.first << ": " << pair.second << " "; cout << endl; for (const auto &pair : result2) cout << pair.first << ": " << pair.second << " "; cout << endl; for (const auto &pair : result3) cout << pair.first << ": " << pair.