Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
Chef Likes Good Sequences CodeChef Solution
Problem -Chef Likes Good Sequences CodeChef Solution
“This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.
Chef Likes Good Sequences CodeChef Solution in C++17
“#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> //find_by_order,order_of_key
#define int long long
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define fr first
#define sc second
#define all(x) (x).begin(),(x).end()
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define sz(x) (int)((x).size())
#define pii pair<int,int>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define mem1(a) memset(a,-1,sizeof(a))
#define mem0(a) memset(a,0,sizeof(a))
#define ppc __builtin_popcount
#define ppcll __builtin_popcountll
template<typename T> istream &operator>>(istream &input, vector<T> &v) {for (auto &it : v) input >> it; return input;}
template<typename T> ostream &operator<<(ostream &output, vector<T> &v) {for (auto &it : v) output << it << " "; return output;}
template<typename T, typename T1>T amax(T &a, T1 b) {if (b > a)a = b; return a;}
template<typename T, typename T1>T amin(T &a, T1 b) {if (b < a)a = b; return a;}
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 1e6 + 7;
const int INF = 1e17;
const int dx[4] {1, 0, -1, 0}, dy[4] {0, 1, 0, -1};
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define debug(x)
#endif
void Solve() {
int n,q;
cin >> n >> q;
vector<int> v(n);
cin >> v;
int tot_cnt = 1;
for(int i = 1;i<n;i++){
if(v[i] != v[i-1]) tot_cnt++;
}
for(int i = 0;i < q;i++){
int idx,x;
cin >> idx >> x;
--idx;
int cnt = 0;
if(n == 1){
cout << 1 <<'\n';
continue;
}
if(idx==0){
if(v[idx] != v[idx + 1]) cnt++;
}
else if(idx != 0 and idx == n-1){
if(v[idx] != v[idx - 1]) cnt++;
}
else{
if(v[idx] != v[idx + 1]) cnt++;
if(v[idx] != v[idx - 1]) cnt++;
}
tot_cnt-=cnt;
cnt = 0;
v[idx] = x;
if(idx==0){
if(v[idx] != v[idx + 1]) cnt++;
}
else if(idx != 0 and idx == n-1){
if(v[idx] != v[idx - 1]) cnt++;
}
else{
if(v[idx] != v[idx + 1]) cnt++;
if(v[idx] != v[idx - 1]) cnt++;
}
tot_cnt+=cnt;
cout << tot_cnt <<'\n';
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("Error.txt", "w", stderr);
#endif
int t = 1;
cin >> t;
while (t--) {
Solve();
}
return 0;
}Chef Likes Good Sequences CodeChef Solution in C++14
“#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
// /freopen("alimagde.in","r",stdin);
ll t;
cin>>t;
while(t--)
{
ll n,m;
cin>>n>>m;
vector<ll>v;
ll a,x,y;
for(ll i=1; i<=n; i++) cin>>a,v.push_back(a);
ll ans=1;
for(ll i = 1 ; i<n; i++)
{
if(v[i]!=v[i-1])
ans++;
}
for(ll i=0; i<m; i++)
{
cin>>x>>y;
x--;
ll cur=1;
if(x+1<n and v[x]!=v[x+1])
cur++;
if(x-1>=0 and v[x]!=v[x-1])
cur++;
ll now=1;
v[x]=y;
if(x+1<n and v[x]!=v[x+1])
now++;
if(x-1>=0 and v[x]!=v[x-1])
now++;
ans-=cur;
ans+=now;
cout<<ans<<'\n';
}
}
}
Chef Likes Good Sequences CodeChef Solution in PYTH 3
“# cook your dish here
tc = int(input())
while tc > 0:
n,q = map(int,input().split())
arr = [int(i) for i in input().split()]
currlen = 1
i = 1
while i<n:
if arr[i] != arr[i-1]:
currlen += 1
else:
while i<n and arr[i] == arr[i-1]:
i += 1
if i<n:
currlen += 1
i += 1
while q>0:
x,y = map(int,input().split())
if arr[x-1] == y:
print(currlen)
else:
p = ne = 0
p += 1 if (x-1)>0 and arr[x-1] != arr[x-2] else 0
p +=1 if (x-1)<(n-1) and arr[x-1] != arr[x] else 0
arr[x-1] = y
ne += 1 if (x-1)>0 and arr[x-1] != arr[x-2] else 0
ne += 1 if (x-1)<(n-1) and arr[x-1] != arr[x] else 0
currlen = currlen - p + ne
print(currlen)
q -= 1
tc -= 1Chef Likes Good Sequences CodeChef Solution in C
“#include<stdio.h>
#include<stdlib.h>
typedef long long int ll;
int main()
{
int t,i;
scanf("%d",&t);
for(i=0;i<t;i++)
{
ll n,q,*arr,j,x,y,count=1;
scanf("%lld%lld",&n,&q);
arr=(ll *)malloc(n*sizeof(ll));
for(j=0;j<n;j++)
{
scanf("%lld",&arr[j]);
if(j!=0)
{
if(arr[j]!=arr[j-1])
{
count++; //count initial length of subsequence;
}
}
}
for(j=0;j<q;j++)
{
scanf("%lld%lld",&x,&y);
if(y==arr[x-1]) //there is no change in the array after introducing y at x;
{
printf("%lld\n",count);
continue;
}
else
{
if(arr[x-1]==arr[x-2] && arr[x-1]==arr[x]) //both neighbours are equal to arr[x] ,
{ // comes in same segment, then,
if(y!=arr[x-1]) //middle is different and divide one segment
{ // into three of different elements
count+=2;
}
}
else if(arr[x-1]==arr[x-2] && arr[x-1]!=arr[x]) // arr[x] is identical to its left neighbour
{ // means two segment , divide it into three.
if(y!=arr[x-1] && y!=arr[x])
{
count++;
}
}
else if(arr[x-1]!=arr[x-2] && arr[x-1]==arr[x]) // arr[x] is identical to its right neighbour.
{
if(y!=arr[x-1] && y!=arr[x-2])
{
count++;
}
}
else if(arr[x-1]!=arr[x-2] && arr[x-1]!=arr[x]) // when arr[x] is different from its neighbour
{
if(y==arr[x-2] && y==arr[x])
{
count-=2;
}
else if(y==arr[x-2] && y!=arr[x])
{
count--;
}
else if(y==arr[x] && y!=arr[x-2])
{
count--;
}
}
arr[x-1]=y; // change the array after every query
printf("%lld\n",count);
}
}
}
return 0;
}Chef Likes Good Sequences 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
{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Reader sc =new Reader();
int t = sc.nextInt();
while(t-->0)
{
int n = sc.nextInt();
int q = sc.nextInt();
int arr[] = new int[n+2];
int ans=0;
for(int i=1;i<=n;i++){
arr[i] =sc.nextInt();
ans = arr[i]==arr[i-1] ? ans : ans+1;
}
StringBuilder sb = new StringBuilder();
while( q-- > 0){
int x= sc.nextInt();
int y = sc.nextInt();
if(arr[x] != y){
// 2 2 3 -> 2 3 3 same, 2 4 3 +1
if(arr[x]!=arr[x-1])
ans--;
if(arr[x]!=arr[x+1])
ans--;
if(y!=arr[x-1])
ans++;
if(y!=arr[x+1])
ans++;
arr[x] = y;
}
sb.append(ans+"\n");
}
System.out.print(sb.toString());
}
}
}Chef Likes Good Sequences CodeChef Solution in PYPY 3
“import sys
#from collections import deque
#from functools import *
#from fractions import Fraction as f
from copy import *
from bisect import *
#from heapq import *
from math import gcd,ceil,sqrt
from itertools import permutations as prm,product
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def string(s):
return "".join(s)
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def bo(i):
return ord(i)-ord('A')
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for xx in input().split()]
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=fi()
while t>0:
t-=1
n,q=mi()
a=li()
ans=0
for i in range(n):
if i==0:
ans+=1
elif a[i]!=a[i-1]:
ans+=1
for i in range(q):
x,val=mi()
x-=1
if a[x]==val:
print(ans)
continue
else:
if x>0 and a[x-1]==val:
ans-=1
if x<n-1 and a[x+1]==val:
ans-=1
if x>0 and a[x-1]==a[x]:
ans+=1
if x<n-1 and a[x+1]==a[x]:
ans+=1
print(ans)
a[x]=val Chef Likes Good Sequences CodeChef Solution in PYTH
“t = int(raw_input())
for i in range(t):
st = raw_input().split()
N = int(st[0])
Q = int(st[1])
st = raw_input().split()
A = [-1]
for x in st:
n = int(x)
A.append(n)
# endfor x
tot = 0
for k in range(1,N+1):
if A[k] <> A[k-1]:
tot += 1
# endif
# endfor k
A.append(-1)
for k in range(Q):
st = raw_input().split()
x = int(st[0])
y = int(st[1])
if A[x] <> y:
if A[x] <> A[x-1]:
tot -= 1
# endif
if A[x] <> A[x+1]:
tot -= 1
# endif
A[x] = y
if A[x] <> A[x-1]:
tot += 1
# endif
if A[x] <> A[x+1]:
tot += 1
# endif
# endif
print tot
# endfor k
# endfor i
Chef Likes Good Sequences CodeChef Solution in C#
“using System;
using System.Text;
public class Test
{
public static void Main()
{
int test = int.Parse(Console.ReadLine());
StringBuilder sb = new StringBuilder();
for(int i=0;i<test;++i){
string[] values = Console.ReadLine().Split(' ');
int n = int.Parse(values[0]);
int q = int.Parse(values[1]);
int[] arr = new int[n+2];
arr[0] = -1;
arr[n+1] = -1;
values = Console.ReadLine().Split(' ');
for(int j=1;j<=n;++j){
arr[j] = int.Parse(values[j-1]);
}
long totalCount = 2;
for(int j=1;j<=n;++j){
if(arr[j-1] != arr[j]){
totalCount++;
}
}
for(int j=0;j<q;++j){
values = Console.ReadLine().Split(' ');
int index = int.Parse(values[0]);
int valueAtIndex = int.Parse(values[1]);
if(arr[index] == valueAtIndex){
sb.Append((totalCount-2)+"\n");
}
else{
if(arr[index-1] == arr[index+1] && arr[index] == arr[index-1]
&& arr[index] == arr[index+1]){
totalCount+=2;
}
else if(arr[index] == arr[index+1] && valueAtIndex != arr[index-1]){
totalCount+=1;
}
else if(arr[index] == arr[index-1] && valueAtIndex != arr[index+1]){
totalCount+=1;
}
if(valueAtIndex == arr[index-1] && arr[index] != arr[index+1]){
totalCount -= 1;
}
if(valueAtIndex == arr[index+1] && arr[index] != arr[index-1]){
totalCount -= 1;
}
arr[index] = valueAtIndex;
sb.Append((totalCount-2)+"\n");
}
}
}
Console.WriteLine(sb.ToString());
}
}Chef Likes Good Sequences CodeChef Solution in GO
“package main
import (
"bufio"
"bytes"
"fmt"
"os"
)
const N = 100010
var arr [N]int
var bak [N]int
func main() {
reader := bufio.NewReader(os.Stdin)
tc := readNum(reader)
var buf bytes.Buffer
for tc > 0 {
tc--
n, q := readTwoNums(reader)
A := readNNums(reader, n)
solver := NewSolver(n, A)
for q > 0 {
q--
x, y := readTwoNums(reader)
res := solver.Update(x, y)
buf.WriteString(fmt.Sprintf("%d\n", res))
}
}
fmt.Print(buf.String())
}
func readInt(bytes []byte, from int, val *int) int {
i := from
sign := 1
if bytes[i] == '-' {
sign = -1
i++
}
tmp := 0
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + int(bytes[i]-'0')
i++
}
*val = tmp * sign
return i
}
func readNum(reader *bufio.Reader) (a int) {
bs, _ := reader.ReadBytes('\n')
readInt(bs, 0, &a)
return
}
func readTwoNums(reader *bufio.Reader) (a int, b int) {
res := readNNums(reader, 2)
a, b = res[0], res[1]
return
}
func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
res := readNNums(reader, 3)
a, b, c = res[0], res[1], res[2]
return
}
func readNNums(reader *bufio.Reader, n int) []int {
res := bak[:n]
x := 0
bs, _ := reader.ReadBytes('\n')
for i := 0; i < n; i++ {
for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
x++
}
x = readInt(bs, x, &res[i])
}
return res
}
func readUint64(bytes []byte, from int, val *uint64) int {
i := from
var tmp uint64
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + uint64(bytes[i]-'0')
i++
}
*val = tmp
return i
}
type Solver struct {
arr []int
cur int
n int
}
func NewSolver(n int, A []int) Solver {
var ans = 2
arr := make([]int, n+2)
copy(arr[1:], A)
arr[0] = -1
arr[n+1] = -1
for i := 1; i <= n; i++ {
if arr[i] != arr[i-1] {
ans++
}
}
return Solver{arr, ans, n}
}
func max(a, b int) int {
if a >= b {
return a
}
return b
}
func (solver *Solver) Update(x int, y int) int {
arr := solver.arr
cur := solver.cur
if arr[x] == y {
// no change
return cur - 2
}
if arr[x] == arr[x-1] && arr[x] == arr[x+1] {
cur += 2
} else if (arr[x] != arr[x-1]) && (arr[x] == arr[x+1]) {
cur++
if y == arr[x-1] {
cur--
}
} else if arr[x] == arr[x-1] && arr[x] != arr[x+1] {
cur++
if y == arr[x+1] {
cur--
}
} else {
// arr[x] != arr[x-1] && arr[x] != arr[x+1]
if y == arr[x-1] {
cur--
}
if y == arr[x+1] {
cur--
}
}
arr[x] = y
solver.cur = cur
return cur - 2
}Chef Likes Good Sequences CodeChef Solution Review:
In our experience, we suggest you solve this Chef Likes Good Sequences CodeChef Solution and gain some new skills from Professionals completely free and we assure you will be worth it.
If you are stuck anywhere between any coding problem, just visit Queslers to get the Chef Likes Good Sequences CodeChef Solution.
“Conclusion:
I hope this Chef Likes Good Sequences CodeChef Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.
This Problem is intended for audiences of all experiences who are interested in learning about Programming Language in a business context; there are no prerequisites.
Keep Learning!
