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# Yet Another Palindrome Problem CodeChef Solution

## 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() {
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 {

while (t-- > 0) {
// String[] valuesA = br.readLine().split(" ");
// 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));
}
}

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 n = 0;
string[] input = null;
long[] diff = null;

while (t-- > 0) {
if (n == 1) {
Console.WriteLine(0);
continue;
}
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');

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();
});

return inputString[currentline++];
}

function main(){

while(t-- > 0){

.map((x)=> parseInt(x, 10));

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() {

var buf bytes.Buffer

for tc > 0 {
tc--
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
}

for i := 0; i < len(s); i++ {
if s[i] == '\n' || s[i] == '\r' {
return s[:i]
}
}
return s
}

return
}

a, b = res[0], res[1]
return
}

a, b, c = res[0], res[1], res[2]
return
}

res := make([]int, n)
x := 0
for i := 0; i < n; i++ {
for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
x++
}
}
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
}

for p >= 0 {
if cur > A[q] {
return -1
}
p--
q++
}

}``````
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