1. Sort the string in ascending order and compare the character array i with i-1, if its same, store it in a map. Time O(nlgn) 2. Use a map and put the values to the map. If getting a value to map is not null, its a duplicate character and maintain count in map values. Time :O(n) 3. Without using collections.Create an int array of size 26, where 0 denotes a and 25 denotes z. Increment the value by 1 when we get a character. Again go for a for loop to find the count of each character, if its greater than 0, print values of character and count. Time :O(n+k) ,k=26