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Yet Another Palindrome Problem CodeChef Solution
Problem -Yet Another Palindrome Problem 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.
Yet Another Palindrome Problem CodeChef Solution in C++17
“#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main() {
//ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int t; cin >> t;
while (t--) {
int n; cin >> n;
ll a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
bool flag = false;
ll ans[n];
for (int i = 0; i < n/2; i++) {
ans[i] = (a[n-i-1] - a[i]);
if (ans[i] < 0) {
flag = true;
break;
}
}
// bool pa = false;
for (int i = 0; i < n/2 - 1; i++) {
//cout << ans[i] << ans[i+1] << endl;
if (ans[i] < ans[i + 1]) {
flag = true;
break;
}
}
if (flag) {
cout << -1 << endl;
} else cout << a[n-1] - a[0] << endl;
}
}Yet Another Palindrome Problem CodeChef Solution in C++14
“#include <iostream>
using namespace std;
#define lli long long int
int main() {
// your code goes here
int t;
cin >> t;
while(t--){
int n;
cin >> n;
lli arr[n];
for(int i=0;i<n;i++){
cin >> arr[i];
}
if(n==1) cout << 0 << endl;
else{
int i,j;
i = n/2-1;
if(n%2==0){
j=n/2;
}
else{
j = n/2+1;
}
lli ans=0,curr_val;
bool flag=true;
while(i>=0){
curr_val=arr[i]+ans;
if(curr_val>arr[j]){
flag = false;
break;
}
else {
ans += (arr[j]-curr_val);
}
i--;j++;
}
cout << (flag?ans:-1) << endl;
}
}
return 0;
}Yet Another Palindrome Problem CodeChef Solution in PYTH 3
“for i in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
left=[]
k=True
for i in range(len(a)//2):
if(a[i]<=a[-i-1]):
left.append(a[-i-1]-a[i])
else:
k=False
print("-1")
break
if(k):
m=[]
m.extend(left)
m.reverse()
left.sort()
if(m==left):
print(m[-1])
else:
print("-1")
Yet Another Palindrome Problem CodeChef Solution in C
“#include <stdio.h>
int main(void) {
int T,N;
scanf("%d",&T);
while(T--)
{
scanf("%d",&N);
int A[N],D[N/2],i,j=0,k=0;
for(i=0;i<N;i++)
scanf("%d",&A[i]);
for(i=0;i<=N/2;i++)
{
D[i]=A[N-i-1]-A[i];
}
for(i=0;i<N/2;i++)
{
if(D[i+1]>D[i]) j++;
if (D[i]<0) k++;
}
if(j==0&&k==0) printf("%d\n",D[0]);
else printf("-1\n");
}
}
Yet Another Palindrome Problem CodeChef Solution in JAVA
“
// package Practice;
import java.io.*;
import java.util.*;
class Codechef {
private static long[] fact = new long[(int) 1e6 + 1];
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int N = Integer.parseInt(br.readLine().trim());
String[] values = br.readLine().split(" ");
// String[] valuesA = br.readLine().split(" ");
// String str = br.readLine();
// String str2 = br.readLine();
// String[] arrValues = br.readLine().split(" ");
// int x = Integer.parseInt(values[0]);
// int y = Integer.parseInt(values[1]);
// int z = Integer.parseInt(values[2]);
// int w = Integer.parseInt(values[3]);
// int[] arr = new int[values.length];
// int [] freq = new int[11];
// int maxValue = 0, value = 0;
// for (int i = 0; i < values.length; i++) {
// arr[i] = Integer.parseInt(values[i]);
// freq[arr[i]]++;
// }
// long x = Long.parseLong(values[0]);
// long y = Long.parseLong(values[1]);
System.out.println(solve(values));
}
}
// https://www.youtube.com/watch?v=MOStCA9W0PM
private static int solve(String [] values) {
int maxValue = (int)1e9, n = values.length, traceValue = (int)1e9;
for(int i = 0; i < n/2; i++){
int a = Integer.parseInt(values[i]);
int b = Integer.parseInt(values[n-1-i]);
if(a > b) return -1;
else {
int temp = b - a;
if(temp > traceValue) return -1;
else if(i == 0){
maxValue = temp;
traceValue = temp;
} else{
traceValue = temp;
}
}
}
return maxValue;
}
private static void display(long[] sumB) {
for (Long val : sumB) {
System.out.print(val + " ");
}
System.out.println();
}
private static boolean isPrime(int p) {
for (int i = 2; i * i <= p; i++) {
if (p % i == 0)
return false;
}
return true;
}
private static int lcm(int a, int b, int GCD) {
// gcd * lcm = a * b;
// lcm = a * b / gcd;
return a * b / GCD;
}
private static int gcd(int a, int b) {
int x = Math.max(a, b), y = Math.min(a, b);
while (x % y != 0) {
int rem = x % y;
x = y;
y = rem;
}
return y;
}
private static boolean isPerfectSquare(int i) {
double ans = Math.sqrt(i);
return ans == (int) ans;
}
private static void display(int[] arr) {
for (int ele : arr) {
System.out.print(ele + " ");
}
System.out.println();
}
private static void display(int[][] arr) {
for (int[] d : arr) {
for (int ele : d)
System.out.print(ele + " ");
System.out.println();
}
}
}Yet Another Palindrome Problem CodeChef Solution in PYPY 3
“for _ in range(int(input())):
n = int(input())
ls = list(map(int, input().split()))
i, j = 0, n - 1
ans = []
while i < j:
ans.append(ls[j] - ls[i])
i += 1
j -= 1
flag = True
for i in range(1, len(ans)):
if ans[i] > ans[i - 1]:
flag = False
break
if not flag:
print(-1)
else:
if min(ans) < 0:
print(-1)
else:
print(max(ans))
Yet Another Palindrome Problem CodeChef Solution in PYTH
“t = int(raw_input())
for i in range(t):
N = int(raw_input())
st = raw_input().split()
A = []
for x in st:
A.append(int(x))
# endfor x
r = 0
p = N/2 -1
q = p+1+N%2
while (r > -1) and (p >= 0):
if A[p]+r > A[q]:
r = -1
else:
r = A[q] - A[p]
# endif
p -= 1
q += 1
# endwhile
print r
# endfor i
Yet Another Palindrome Problem CodeChef Solution in C#
“using System;
public class Test
{
public static void Main()
{
var t = int.Parse(Console.ReadLine());
var n = 0;
string[] input = null;
long[] diff = null;
while (t-- > 0) {
n = int.Parse(Console.ReadLine());
if (n == 1) {
Console.WriteLine(0);
continue;
}
input = Console.ReadLine().Split(' ');
diff = new long[n / 2];
for(var i = 0; i < n / 2; i++) {
diff[i] = long.Parse(input[n - 1 - i]) - long.Parse(input[i]);
if (diff[i] < 0 || i > 0 && diff[i] > diff[i - 1]) {
Console.WriteLine(-1);
break;
}
else if (i == n / 2 - 1) {
Console.WriteLine(diff[0]);
}
}
}
}
}Yet Another Palindrome Problem CodeChef Solution in NODEJS
“process.stdin.resume();
process.stdin.setEncoding('utf8');
// your code goes here
let inputString = "";
let currentline = 0;
process.stdin.on("data", (inputStdin)=>{
inputString += inputStdin;
});
process.stdin.on("end", (inputStdin)=>{
inputString = inputString.trim()
.split("\n")
.map((string)=> string.trim());
main();
});
function readline(){
return inputString[currentline++];
}
function main(){
let t = parseInt(readline(), 10);
while(t-- > 0){
let n = parseInt(readline(), 10);
let arr = readline().split(" ")
.map((x)=> parseInt(x, 10));
// let arr = readline();
let ans = find(arr,n);
console.log(ans);
}
}
function find(arr, n){
let dd = [], max = 0;
for(let i =0 ; i < n/2; i++){
dd[i] = arr[n-1-i] - arr[i];
if(dd[i] < 0) return -1;
max = Math.max(max, dd[i]);
}
for(let i =0 ; i < dd.length-1; i++){
if(dd[i] < dd[i+1]) return -1;
}
return max;
}
Yet Another Palindrome Problem CodeChef Solution in GO
“package main
import (
"bufio"
"bytes"
"fmt"
"os"
)
func main() {
reader := bufio.NewReader(os.Stdin)
var buf bytes.Buffer
tc := readNum(reader)
for tc > 0 {
tc--
n := readNum(reader)
A := readNNums(reader, n)
res := solve(A)
buf.WriteString(fmt.Sprintf("%d\n", res))
}
fmt.Print(buf.String())
}
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
}
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 readString(reader *bufio.Reader) string {
s, _ := reader.ReadString('\n')
for i := 0; i < len(s); i++ {
if s[i] == '\n' || s[i] == '\r' {
return s[:i]
}
}
return s
}
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 := make([]int, 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 solve(A []int) int {
n := len(A)
var p, q int
if n&1 == 1 {
p = n/2 - 1
q = p + 2
} else {
p = n/2 - 1
q = p + 1
}
var add int
for p >= 0 {
cur := A[p] + add
if cur > A[q] {
return -1
}
add += A[q] - cur
p--
q++
}
return add
}Yet Another Palindrome Problem CodeChef Solution Review:
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“Conclusion:
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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.
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