Not All Flavours CodeChef Solution

Problem -Not All Flavours CodeChef Solution

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Not All Flavours CodeChef Solution in C++17

/*ॐ Arpit Blagan ॐ*/
#include<bits/stdc++.h>
using namespace std;
#define all(arr) arr.begin(),arr.end()
#define rep(i,s,e) for(int i=s;i<e;i++)
#define lli long long int
#define int long long
const long long INF=1e18;
const int mod=1e9+7,N=1e6+10;

int32_t main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t;cin>>t;
	while(t--){
		int n,k;cin>>n>>k;
		vector<int>arr(n);for(int i=0;i<n;i++){cin>>arr[i];}
		unordered_map<int,int>mp;
		int j=0;int ans=0;
		for(int i=0;i<n;i++){
			mp[arr[i]]++;
			if(mp.size()<k){
				ans=max(ans,i-j+1);
			}
			else{
				while(j<=i&&mp.size()==k){
					mp[arr[j]]--;
					if(mp[arr[j]]==0){mp.erase(arr[j]);}j++;
				}
				if(mp.size()<k){ans=max(ans,i-j+1);}
			}
		}cout<<ans<<"\n";		
	}
	return 0;
}

Not All Flavours CodeChef Solution in C++14

#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = 3.14159265358979323846;  /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 1010101;
const ll mod = 1e9 + 7;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef deque<ll> dqll;
typedef deque<pll> dqpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\n"
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n){cout<<v[i];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
void readf() {freopen("", "rt", stdin);}
template<typename T>
void readv(vector<T>& v){rep(i,0,sz(v)) cin>>v[i];}
template<typename T, typename U>
void readp(pair<T,U>& A) {cin>>A.first>>A.second;}
template<typename T, typename U>
void readvp(vector<pair<T,U>>& A) {rep(i,0,sz(A)) readp(A[i]); }
void readvall3(vall3& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2];}
void readvall5(vall5& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2]>>A[i][3]>>A[i][4];}
void readvvll(vvll& A) {rep(i,0,sz(A)) readv(A[i]);}

struct Combination {
    vll fac, inv;
    ll n, MOD;

    ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

    Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
        inv = fac = vll(n,1);
        rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
        inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
        rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
    }

    ll fact(ll n) {return fac[n];}
    ll nCr(ll n, ll r) {
        if(n < r or n < 0 or r < 0) return 0;
        return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
    }
};

struct Matrix {
    ll r,c;
    vvll matrix;
    Matrix(ll r, ll c, ll v = 0): r(r), c(c), matrix(vvll(r,vll(c,v))) {}

    Matrix operator*(const Matrix& B) const {
        Matrix res(r, B.c);
        rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
                    res.matrix[i][j] = (res.matrix[i][j] + matrix[i][k] * B.matrix[k][j] % mod) % mod;
                }
        return res;
    }

    Matrix copy() {
        Matrix copy(r,c);
        copy.matrix = matrix;
        return copy;
    }

    Matrix pow(ll n) {
        assert(r == c);
        Matrix res(r,r);
        Matrix now = copy();
        rep(i,0,r) res.matrix[i][i] = 1;
        while(n) {
            if(n & 1) res = res * now;
            now = now * now;
            n /= 2;
        }
        return res;
    }
};

// geometry data structures
template <typename T>
struct Point {
    T y,x;
    Point(T y, T x) : y(y), x(x) {}
    Point(pair<T,T> p) : y(p.first), x(p.second) {}
    Point() {}
    void input() {cin>>y>>x;}
    friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
    Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
    Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
    Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
    Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }
    bool operator<(const Point &other) const {if (x == other.x) return y < other.y;return x < other.x;}
    Point<T> rotate(Point<T> center, ld angle) {
        ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
        ld y = this->y - center.y;
        ld x = this->x - center.x;

        return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
    }
    ld distance(Point<T> other) {
        T dy = abs(this->y - other.y);
        T dx = abs(this->x - other.x);
        return sqrt(dy * dy + dx * dx);
    }

    T norm() { return x * x + y * y; }
};

template<typename T>
struct Line {
    Point<T> A, B;
    Line(Point<T> A, Point<T> B) : A(A), B(B) {}
    Line() {}

    void input() {
        A = Point<T>();
        B = Point<T>();
        A.input();
        B.input();
    }

    T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
        T res = a.x * b.y + b.x * c.y + c.x * a.y;
        res -= (a.x * c.y + b.x * a.y + c.x * b.y);
        return res;
    }

    bool isIntersect(Line<T> o) {
        T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
        T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
        if (p1p2 == 0 && p3p4 == 0) {
            pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
            if (p1 > p2) swap(p2, p1);
            if (p3 > p4) swap(p3, p4);
            return p3 <= p2 && p1 <= p4;
        }
        return p1p2 <= 0 && p3p4 <= 0;
    }

    pair<bool,Point<ld>> intersection(Line<T> o) {
        if(!this->intersection(o)) return {false, {}};
        ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
        ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
        return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
    }

    //@formula for : y = ax + b
    //@return {a,b};
    pair<ld, ld> formula() {
        T y1 = A.y, y2 = B.y;
        T x1 = A.x, x2 = B.x;
        if(y1 == y2) return {1e9, 0};
        if(x1 == x2) return {0, 1e9};
        ld a = 1. * (y2 - y1) / (x2 - x1);
        ld b = -x1 * a + y1;
        return {a, b};
    }
};

template<typename T>
struct Circle {
    Point<T> center;
    T radius;
    Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
    Circle(Point<T> center, T radius) : center(center), radius(radius) {}
    Circle() {}

    void input() {
        center = Point<T>();
        center.input();
        cin>>radius;
    }

    bool circumference(Point<T> p) {
        return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
    }

    bool intersect(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
    }

    bool include(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return d <= radius * radius;
    }
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll solve(vll A, ll k) {
    ll res = 0;
    umll freq;
    ll l = 0, r = 0, n = sz(A);
    while(r < n) {
        while(r < n and sz(freq) < k) {
            res = max(res, r - l);
            freq[A[r++]]++;
        }
        if(sz(freq) < k) res = max(res, r - l);
        while(r < n and sz(freq) >= k) {
            if(--freq[A[l]] == 0) freq.erase(A[l]);
            l += 1;
        }
    }
    return res;
}
int main() {
    Code By Sumfi
    cout.precision(12);
    ll tc = 1;
    cin>>tc;
    rep(i,1,tc+1) {
        ll n,k;
        cin>>n>>k;
        vll A(n);
        readv(A);
        print(solve(A,k));
    }
    return 0;
}

Not All Flavours CodeChef Solution in PYTH 3

for i in range(int(input())):
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    j,i,mx=0,0,0
    d=dict()
    while(j<n):
        if l[j] in d:
            d[l[j]]+=1 
        else:
            d[l[j]]=1 
        if(len(d)<=k-1):
            mx=max(mx,j-i+1)
            j+=1 
        else:
            while(len(d)>k-1):
                d[l[i]]-=1 
                if(d[l[i]]==0):
                    d.pop(l[i])
                i+=1 
            j+=1 
    print(mx)
            

Not All Flavours CodeChef Solution in C

#include <stdio.h>

int main(void) {
	int t;
	scanf("%d",&t);
	while(t--){
		int n,k;
		scanf("%d %d",&n,&k);
		int i,a[n];
		int b[k];
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		
		for(i=0;i<=k;i++)
		b[i]=0;
		int l=0,max=0,z=0,m=0;
		for(i=0;i<n;i++){
			if(b[a[i]] == 0)
			m++;
			l++;
			b[a[i]]++;
			if(m < k){
				if(l > max)
				max = l;
			}else{
				l--;
				if(b[a[z]] == 1)
				m--;
				b[a[z++]]--;
			}
		}
		
		printf("%d\n",max);
	}
	return 0;
}

Not All Flavours CodeChef Solution in JAVA

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		 Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){

            int n=sc.nextInt();
            int k=sc.nextInt();
            int a[]=new int[n];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
            }
            k=k-1;
            int diff=0;
            int start=0;
            int end=0;
            int curr_count=0;
            int prev_ele=0;
            int freq[]=new int[100001];
            for(int i=0;i<n;i++){
                freq[a[i]]+=1;
                if(freq[a[i]]==1){
                    curr_count=curr_count+1;
                }
                while(curr_count>k){
                    freq[a[prev_ele]]-=1;
                    if(freq[a[prev_ele]]==0){
                        curr_count=curr_count-1;
                    }
                    prev_ele = prev_ele + 1;
                }
                if(i-prev_ele+1>=end-start+1){
                    end=i;
                    start=prev_ele;
                }
            }
            System.out.println(end+1-start);
            /*
            int n=sc.nextInt();
            int a[]=new int[n];
            int freq[]=new int[100001];
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
            }
            int count=0;
            for(int i=0;i<n;i++){
                freq[a[i]]+=1;
                System.out.println(freq[a[i]]);
                if(freq[a[i]]==1){
                    System.out.println("hh");
                }
            }

            for(int i=0;i<n;i++) {
                System.out.println(freq[a[i]]);
            }
            */
        }
	}
}

Not All Flavours CodeChef Solution in PYPY 3

def main():
    t=int(input())
    while(t>0):
        n,k=map(int,input().split())
        arr=list(map(int,input().split()))
        freq=[0]*100001
        k=k-1
        st=0
        end=0
        currentcount=0
        prevelement=0
        for i in range(n):
            freq[arr[i]]+=1
            if(freq[arr[i]])==1:
                currentcount+=1
            while(currentcount>k):
                freq[arr[prevelement]]-=1
                if(freq[arr[prevelement]]==0):
                    currentcount-=1
                prevelement+=1
            if(i-prevelement+1>=end-st+1):
                end=i
                st=prevelement
        print(end-st+1)
    
        t=t-1
if __name__=='__main__':
    main()

Not All Flavours CodeChef Solution in PYTH


sinput = raw_input

def intput(s =''):
	''' To get data in the form of an integer'''
	return int(sinput(s) )

def lintput(s=''):
	''' To get data in the form of space separated integers and return a tuple containing them'''
	return map(int,sinput(s).split() )

def intputs(s=''):
	'''To get data in the form of space separated integers and return a tuple containing them'''
	return tuple(lintput(s) )

def lprint(l):
	for i in l:
		print i

'''Main structer for testcased problems'''
T = intput()
R = [0]*T
for Q in range(T):
	N,K = intputs()
	s = lintput()
	pre = [-1,]*(K+1) # Previous
	pre[0] = None
	mg = [0,]*(K+1) # Max gap
	mg[0] = None
	for i in range(N):
		#print mg[s[i]], i-pre[s[i]]
		mg[s[i]] = max(mg[s[i]], (i - pre[s[i]]))
		pre[s[i]] = i
	for i in range(1,K+1):
		mg[i] = max(mg[i], N-pre[i])
		
	#print mg
	if -1 in pre: # A flavour has not been used
		R[Q] = N
	else:
		R[Q] = max(mg)-1

lprint(R)

Not All Flavours CodeChef Solution in C#

using System;
using System.Collections.Generic;
using System.Linq;

public class Program
{
	public static void Main(string[] args)
	{
		NotAllFravours();
	}

	public static void NotAllFravours()
	{
		int n = int.Parse(Console.ReadLine());
		string[] NK;
		int nn;
		int k;

		for (int testCase = 0; testCase < n; testCase++)
		{
			NK = Console.ReadLine().Split(' ');
			nn = int.Parse(NK[0]);
			k = int.Parse(NK[1]);

			var integers = Console.ReadLine().Split(' ').ToList().Select(integ => int.Parse(integ)).ToList();

			var appearence = new List<int>();
			var set = new HashSet<int>();
			var order = new List<int>() { integers[0] };

			var lastChar = integers[0];
			set.Add(integers[0]);
			appearence.Add(1);

			var maxResult = 0;
			var currentResult = 1;

			for (int i = 1; i < integers.Count; i++)
			{
				if (integers[i] == lastChar)
				{
					appearence[appearence.Count - 1]++;
					currentResult++;
				}
				else
				{
					if (set.Count >= k - 1 && !set.Contains(integers[i]))
					{
						if (currentResult > maxResult)
						{
							maxResult = currentResult;
						}

						while (set.Count >= k - 1)
						{
							currentResult -= appearence[0];
							appearence.RemoveAt(0);
							var removedChar = order[0];
							order.RemoveAt(0);

							if (!order.Contains(removedChar))
							{
								set.Remove(removedChar);
							}
						}

						appearence.Add(1);
						set.Add(integers[i]);
						order.Add(integers[i]);
						currentResult++;
					}
					else
					{
						appearence.Add(1);
						set.Add(integers[i]);
						order.Add(integers[i]);
						currentResult++;
					}
					lastChar = integers[i];
				}
			}

			Console.WriteLine(Math.Max(currentResult, maxResult));
		}
	}

	public static void AtTheGate()
	{
		int numberOfTestCases = int.Parse(Console.ReadLine());

		for (int testCase = 0; testCase < numberOfTestCases; testCase++)
		{
			string[] NK = Console.ReadLine().Split(' ');
			int n = int.Parse(NK[0]);
			int k = int.Parse(NK[1]);

			string[] coinsLine = Console.ReadLine().Split(' ');

			int nrOfRolls = 0;
			bool currentHead = true;
			for (int j = n - 1; j >= n - k; j--)
			{
				if (coinsLine[j] == (currentHead ? "H" : "T"))
				{
					nrOfRolls++;
					currentHead = !currentHead;
				}
			}

			bool showHead = nrOfRolls % 2 == 0;
			int ans = 0;

			for (int l = 0; l < n - k; l++)
			{
				if (coinsLine[l] == (showHead ? "H" : "T"))
					ans++;
			}

			Console.WriteLine(ans);
		}
	}

	public static void SwappingToPlaindrome()
	{
		int numberOfTestCases = int.Parse(Console.ReadLine());

		for (int testCase = 0; testCase < numberOfTestCases; testCase++)
		{

		}
	}

	public static bool CheckIfCanBePalindrom(string s)
	{
		var characterCount = new Dictionary<char, int>();
		foreach (var c in s)
		{
			if (characterCount.ContainsKey(c))
			{
				characterCount[c]++;
			}
			else
			{
				characterCount.Add(c, 1);
			}
		}

		var numberOfOdds = 0;
		foreach (var value in characterCount.Values)
		{
			if (value % 2 != 0)
				numberOfOdds++;
			if (numberOfOdds > 1)
				return false;
		}

		return true;
	}
}

Not All Flavours CodeChef Solution in GO

package main

import (
	"bufio"
	"fmt"
	"os"
)

var reader = bufio.NewReader(os.Stdin)
var writer = bufio.NewWriter(os.Stdout)

func scanf(f string, a ...interface{})  { fmt.Fscanf(reader, f, a...) }
func printf(f string, a ...interface{}) { fmt.Fprintf(writer, f, a...) }

func main() {
	// STDOUT MUST BE FLUSHED MANUALLY!!!
	defer writer.Flush()

	var nt int
	scanf("%d\n", &nt)
	for t := 0; t < nt; t++ {
		var n, k int
		scanf("%d %d\n", &n, &k)
		cakes := make([]int, n)
		for i := 0; i < n; i++ {
			scanf("%d", &cakes[i])
		}
		scanf("\n")
		ml := 0
		include := make(map[int]int)
		l, r := 0, 0
		count := 0
		for ; r < n; r++ {
			if include[cakes[r]] == 0 {
				count++
			}
			include[cakes[r]]++
			for ; count == k && l < r; l++ {
				if include[cakes[l]] == 1 {
					count--
				}
				include[cakes[l]]--
			}
			if ml < r-l+1 {
				ml = r - l + 1
			}
		}
		printf("%d\n", ml)
	}
}
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