CSES Solutions – Playlist

Last Updated : 26 Mar, 2024

You are given a playlist of a radio station since its establishment. The playlist has a total of N songs. What is the longest sequence of successive songs where each song is unique?

Examples:

Input: N = 8, songs[] = {1, 2, 1, 3, 2, 7, 4, 2}
Output: 5
Explanation: The longest sequence of successive unique songs is: {1, 3, 2, 7, 4} which has 5 elements.

Input: N = 6, songs[] = {1, 2, 3, 1, 2, 3}
Output: 3
Explanation: The longest sequence of successive unique songs is: {1, 2, 3} which has 3 elements.

Approach: To solve the problem, follow the below idea:

The problem can be solved using a map to store the index of each character and two pointers: start and end to mark the starting and ending of the range of unique characters. Initially, we start from the first character and end at the first character. Then, we keep on shifting the end pointer and storing the index of each character in the map. If we encounter a character whose index > start pointer, then it means that in our window we already have that character. Then we will shrink the window by moving the start pointer to the previous index of the character + 1. This will maintain that all the character from start to end have unique characters.

Step-by-step algorithm:

Maintain two pointers, start = 0 and end = 0 to mark the starting and ending point of window.
Also maintain a map, say mp to store the last index of each character.
Iterate end from 0 to N – 1,
- If the character at index = end is not present in the map, we insert the character along with the current index.
- Otherwise, if the character is present in the map and its last occurrence is greater than the start of the window, it means that the current character is already present in the window. So, we need to shrink the window by moving start to last occurrence of that character + 1.
- Also check for the maximum length of window (end – start + 1) in each iteration.
After all the iterations, print the maximum length of a window.

Below is the implementation of the algorithm:

C++

#include <bits/stdc++.h>
#define ll long long
using namespace std;

ll solve(ll N, ll* songs)
{
    ll start = 0, ans = 0;
    // Map to store the last occurrence of all the
    // characters
    map<ll, ll> mp;

    // Increment end pointer character by character
    for (ll end = 0; end < N; end++) {
        // If the current song is not present in the map
        if (mp.find(songs[end]) == mp.end())
            mp.insert({ songs[end], end });
        else {
            // If the current song is present in the map and
            // is inside the window
            if (mp[songs[end]] >= start)
                start = mp[songs[end]] + 1;
            // Update the last occurrence of current
            // character to the current index
            mp[songs[end]] = end;
        }
        // Update ans to store the maximum length of range
        // of unique values
        ans = max(ans, end - start + 1);
    }
    return ans;
}

int main()
{
    // Sample Input
    ll N = 8;
    ll songs[N] = { 1, 2, 1, 3, 2, 7, 4, 2 };

    cout << solve(N, songs) << "\n";
}

Java

import java.util.HashMap;
import java.util.Map;

public class Main {

    static long solve(long N, long[] songs)
    {
        long start = 0, ans = 0;
        // Map to store the last occurrence of all the
        // characters
        Map<Long, Long> mp = new HashMap<>();

        // Increment end pointer character by character
        for (long end = 0; end < N; end++) {
            // If the current song is not present in the map
            if (!mp.containsKey(songs[(int)end]))
                mp.put(songs[(int)end], end);
            else {
                // If the current song is present in the map
                // and is inside the window
                if (mp.get(songs[(int)end]) >= start)
                    start = mp.get(songs[(int)end]) + 1;
                // Update the last occurrence of current
                // character to the current index
                mp.put(songs[(int)end], end);
            }
            // Update ans to store the maximum length of
            // range of unique values
            ans = Math.max(ans, end - start + 1);
        }
        return ans;
    }

    public static void main(String[] args)
    {
        // Sample Input
        long N = 8;
        long[] songs = { 1, 2, 1, 3, 2, 7, 4, 2 };

        System.out.println(solve(N, songs));
    }
}

using System;
using System.Collections.Generic;

public class MainClass {
    public static long Solve(int N, int[] songs)
    {
        int start = 0;
        long ans = 0;
        Dictionary<int, int> mp
            = new Dictionary<int, int>();

        for (int end = 0; end < N; end++) {
            if (!mp.ContainsKey(songs[end])) {
                mp.Add(songs[end], end);
            }
            else {
                if (mp[songs[end]] >= start) {
                    start = mp[songs[end]] + 1;
                }
                mp[songs[end]] = end;
            }
            ans = Math.Max(ans, end - start + 1);
        }
        return ans;
    }

    public static void Main(string[] args)
    {
        // Sample Input
        int N = 8;
        int[] songs = { 1, 2, 1, 3, 2, 7, 4, 2 };

        Console.WriteLine(Solve(N, songs));
    }
}

Javascript

function solve(N, songs) {
    let start = 0;
    let ans = 0;
    // Map to store the last occurrence of all the characters
    let mp = new Map();

    // Increment end pointer character by character
    for (let end = 0; end < N; end++) {
        // If the current song is not present in the map
        if (!mp.has(songs[end])) {
            mp.set(songs[end], end);
        } else {
            // If the current song is present in the map and is inside the window
            if (mp.get(songs[end]) >= start) {
                start = mp.get(songs[end]) + 1;
            }
            // Update the last occurrence of current character to the current index
            mp.set(songs[end], end);
        }
        // Update ans to store the maximum length of range of unique values
        ans = Math.max(ans, end - start + 1);
    }
    return ans;
}

// Sample Input
let N = 8;
let songs = [1, 2, 1, 3, 2, 7, 4, 2];

console.log(solve(N, songs));

Python3

def solve(N, songs):
    start = 0
    ans = 0
    # Dictionary to store the last occurrence index of each song
    mp = {}

    # Iterate through the songs list
    for end in range(N):
        # If the current song is not present in the dictionary
        if songs[end] not in mp:
            mp[songs[end]] = end
        else:
            # If the current song is present in the dictionary and within the window
            if mp[songs[end]] >= start:
                start = mp[songs[end]] + 1
            # Update the last occurrence of the current song to the current index
            mp[songs[end]] = end
        # Update 'ans' to store the maximum length of the range of unique values
        ans = max(ans, end - start + 1)
    return ans


def main():
    # Sample Input
    N = 8
    songs = [1, 2, 1, 3, 2, 7, 4, 2]

    print(solve(N, songs))


if __name__ == "__main__":
    main()