Kth lexicographically smallest distinct String from given Array of Strings
Last Updated :
10 Dec, 2021
Given an array arr having N strings and an integer K, the task is to find the lexicographically smallest Kth distinct string. Print an empty string if no such string exists.
Example:
Input: arr[]={"aa", "aa", "bb", "cc", "dd", "cc"}, K = 2
Output: dd
Explanation: Distinct strings are: "bb", "dd". 2nd smallest string among these is "dd"
Input: arr[]={"aa", "aa", "bb", "cc", "dd", "cc"}, K = 1
Output: bb
Approach: The given problem can be solved by first sorting the given array of strings, and then printing the Kth string with frequency 1. Follow the below steps to solve this problem:
- Sort the given array of strings
- Create a map to store the frequency of each string.
- Now, traverse the map and reduce the value of K each time a string having a frequency of one is found.
- When K becomes zero, print the next string having a frequency of 1.
Below is the implementation of the above approach:
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print lexicographically
// smallest Kth string
string KthDistinctString(vector<string>& arr, int K)
{
// Sorting the array of strings
sort(arr.begin(), arr.end());
// Map to store the strings
map<string, int> mp;
for (auto x : arr) {
mp[x]++;
}
for (auto x : mp) {
// Reducing K
if (x.second == 1) {
K--;
}
if (K == 0 and x.second == 1) {
return x.first;
}
}
return "";
}
// Driver Code
int main()
{
vector<string> a
= { "aa", "aa", "bb", "cc", "dd", "cc" };
int K = 2;
cout << KthDistinctString(a, K);
}
// Java code for the above approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
class GFG {
// Function to print lexicographically
// smallest Kth string
static String KthDistinctString(ArrayList<String> arr, int K) {
// Sorting the array of strings
Collections.sort(arr);
// Map to store the strings
HashMap<String, Integer> mp = new HashMap<String, Integer>();
for (String x : arr) {
int count = 0;
if (mp.containsKey(x)) {
count = mp.get(x);
}
mp.put(x, count + 1);
}
for (String x : mp.keySet()) {
// Reducing K
if (mp.get(x) == 1) {
K--;
}
if (K == 0 && mp.get(x) == 1) {
return x;
}
}
return "";
}
// Driver Code
public static void main(String args[]) {
ArrayList<String> a = new ArrayList<String>();
a.add("aa");
a.add("aa");
a.add("bb");
a.add("cc");
a.add("dd");
a.add("cc");
int K = 2;
System.out.println(KthDistinctString(a, K));
}
}
// This code is contributed by gfgking
# Python3 code for the above approach
# Function to print lexicographically
# smallest Kth string
def KthDistinctString(arr, K):
# Sorting the array of strings
arr.sort()
# Map to store the strings
mp = {}
for x in arr:
if x in mp:
mp[x] += 1
else:
mp[x] = 1
for x in mp:
# Reducing K
if (mp[x] == 1):
K -= 1
if (K == 0 and mp[x] == 1):
return x
return ""
# Driver Code
if __name__ == "__main__":
a = [ "aa", "aa", "bb", "cc", "dd", "cc" ]
K = 2
print(KthDistinctString(a, K))
# This code is contributed by rakeshsahni
// C# code for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function to print lexicographically
// smallest Kth string
static string KthDistinctString(ArrayList arr, int K)
{
// Sorting the array of strings
arr.Sort();
// Map to store the strings
Dictionary<string, int> mp =
new Dictionary<string, int>();
foreach (string x in arr) {
int count = 0;
if (mp.ContainsKey(x)) {
count = mp[x];
}
mp[x] = count + 1;
}
foreach (KeyValuePair<string, int> x in mp) {
// Reducing K
if (x.Value == 1) {
K--;
}
if (K == 0 && x.Value == 1) {
return x.Key;
}
}
return "";
}
// Driver Code
public static void Main()
{
ArrayList a = new ArrayList();
a.Add("aa");
a.Add("aa");
a.Add("bb");
a.Add("cc");
a.Add("dd");
a.Add("cc");
int K = 2;
Console.Write(KthDistinctString(a, K));
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript Program to implement
// the above approach
// Function to print lexicographically
// smallest Kth string
function KthDistinctString(arr, K) {
// Sorting the array of strings
arr.sort();
// Map to store the strings
let mp = new Map();
for (let x of arr) {
if (mp.has(x))
mp.set(x, mp.get(x) + 1);
else
mp.set(x, 1);
}
for (let [key, value] of mp)
{
// Reducing K
if (value == 1) {
K--;
}
if (K == 0 && value == 1) {
return key;
}
}
return "";
}
// Driver Code
let a
= ["aa", "aa", "bb", "cc", "dd", "cc"];
let K = 2;
document.write(KthDistinctString(a, K));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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