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# Solution class containing the logic
class Solution:
    # Function to print characters that occur more than once
    def print_duplicates(self, string):
        # List to store frequency of 26 lowercase letters
        freq = [0] * 26

        # Loop through the string and count character frequency
        for ch in string:
            freq[ord(ch) - ord('a')] += 1

        # Print characters that appear more than once
        for i in range(26):
            if freq[i] > 1:
                print(f"{chr(i + ord('a'))} - {freq[i]}")

# Driver code
if __name__ == "__main__":
    # Sample input string
    string = "sinstriiintng"

    # Create object of Solution class
    sol = Solution()

    # Call the function to print duplicates
    sol.print_duplicates(string)

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// Solution class containing the logic
class Solution {
    // Function to print characters that occur more than once
    printDuplicates(str) {
        // Array to store frequency of 26 lowercase letters
        const freq = new Array(26).fill(0);

        // Loop through the string and count character frequency
        for (let i = 0; i < str.length; i++) {
            freq[str.charCodeAt(i) - 'a'.charCodeAt(0)]++;
        }

        // Print characters that appear more than once
        for (let i = 0; i < 26; i++) {
            if (freq[i] > 1) {
                console.log(String.fromCharCode(i + 'a'.charCodeAt(0)) + " - " + freq[i]);
            }
        }
    }
}

// Driver code
const str = "sinstriiintng";

// Create object of Solution class
const sol = new Solution();

// Call the function to print duplicates
sol.printDuplicates(str);

Complexity Analysis

Time Complexity: O(N), we traverse each character in the string, and accessing the array takes only constant time.
Space Complexity: O(1), since the string has only characters from ‘a’ to ‘z’, the maximum size of the array is 26.