printing the elements by removing middle element of stack











up vote
0
down vote

favorite












Input: First line of input contains a single integer T which denotes the number of test cases. T test cases follows, first line of each test case contains a integer n. Second line consists of n spaced integers.



Output: Print the elements of the stack after deleting the middle element in reverse order.



Input:1



7



1 2 3 4 5 6 7



output is:




7 6 5 3 2 1




actually i am able to do printing in reverse fashion but i don't know how to remove middle element from the stack.please help



import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String args)
{
Scanner s=new Scanner(System.in);
int test=s.nextInt();
for(int t=0;t<test;t++)
{
int n=s.nextInt();
int a=new int[n];
for(int i=0;i<n;i++)
a[i]=s.nextInt();
Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++)
{
stack.push(a[i]);
}
ListIterator<Integer> lstIterator=stack.listIterator(stack.size());
while(lstIterator.hasPrevious())
{
Integer res=lstIterator.previous();
//what condition should i give so that it would print all the
elements except middle one.
System.out.print(res+" ");
}
System.out.println();
}
}
}









share|improve this question




















  • 1




    How do you deal with middle element in case there are even number of elements?
    – Pushpesh Kumar Rajwanshi
    Nov 9 at 9:37












  • for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52










  • kindly tell me the logic to delete the middle element for the odd no.of elements
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52















up vote
0
down vote

favorite












Input: First line of input contains a single integer T which denotes the number of test cases. T test cases follows, first line of each test case contains a integer n. Second line consists of n spaced integers.



Output: Print the elements of the stack after deleting the middle element in reverse order.



Input:1



7



1 2 3 4 5 6 7



output is:




7 6 5 3 2 1




actually i am able to do printing in reverse fashion but i don't know how to remove middle element from the stack.please help



import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String args)
{
Scanner s=new Scanner(System.in);
int test=s.nextInt();
for(int t=0;t<test;t++)
{
int n=s.nextInt();
int a=new int[n];
for(int i=0;i<n;i++)
a[i]=s.nextInt();
Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++)
{
stack.push(a[i]);
}
ListIterator<Integer> lstIterator=stack.listIterator(stack.size());
while(lstIterator.hasPrevious())
{
Integer res=lstIterator.previous();
//what condition should i give so that it would print all the
elements except middle one.
System.out.print(res+" ");
}
System.out.println();
}
}
}









share|improve this question




















  • 1




    How do you deal with middle element in case there are even number of elements?
    – Pushpesh Kumar Rajwanshi
    Nov 9 at 9:37












  • for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52










  • kindly tell me the logic to delete the middle element for the odd no.of elements
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Input: First line of input contains a single integer T which denotes the number of test cases. T test cases follows, first line of each test case contains a integer n. Second line consists of n spaced integers.



Output: Print the elements of the stack after deleting the middle element in reverse order.



Input:1



7



1 2 3 4 5 6 7



output is:




7 6 5 3 2 1




actually i am able to do printing in reverse fashion but i don't know how to remove middle element from the stack.please help



import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String args)
{
Scanner s=new Scanner(System.in);
int test=s.nextInt();
for(int t=0;t<test;t++)
{
int n=s.nextInt();
int a=new int[n];
for(int i=0;i<n;i++)
a[i]=s.nextInt();
Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++)
{
stack.push(a[i]);
}
ListIterator<Integer> lstIterator=stack.listIterator(stack.size());
while(lstIterator.hasPrevious())
{
Integer res=lstIterator.previous();
//what condition should i give so that it would print all the
elements except middle one.
System.out.print(res+" ");
}
System.out.println();
}
}
}









share|improve this question















Input: First line of input contains a single integer T which denotes the number of test cases. T test cases follows, first line of each test case contains a integer n. Second line consists of n spaced integers.



Output: Print the elements of the stack after deleting the middle element in reverse order.



Input:1



7



1 2 3 4 5 6 7



output is:




7 6 5 3 2 1




actually i am able to do printing in reverse fashion but i don't know how to remove middle element from the stack.please help



import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String args)
{
Scanner s=new Scanner(System.in);
int test=s.nextInt();
for(int t=0;t<test;t++)
{
int n=s.nextInt();
int a=new int[n];
for(int i=0;i<n;i++)
a[i]=s.nextInt();
Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++)
{
stack.push(a[i]);
}
ListIterator<Integer> lstIterator=stack.listIterator(stack.size());
while(lstIterator.hasPrevious())
{
Integer res=lstIterator.previous();
//what condition should i give so that it would print all the
elements except middle one.
System.out.print(res+" ");
}
System.out.println();
}
}
}






java stack






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 9 at 9:54









Bas de Groot

532115




532115










asked Nov 9 at 9:28









RAVI KUMAR SHARMA CSE16

215




215








  • 1




    How do you deal with middle element in case there are even number of elements?
    – Pushpesh Kumar Rajwanshi
    Nov 9 at 9:37












  • for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52










  • kindly tell me the logic to delete the middle element for the odd no.of elements
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52














  • 1




    How do you deal with middle element in case there are even number of elements?
    – Pushpesh Kumar Rajwanshi
    Nov 9 at 9:37












  • for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52










  • kindly tell me the logic to delete the middle element for the odd no.of elements
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:52








1




1




How do you deal with middle element in case there are even number of elements?
– Pushpesh Kumar Rajwanshi
Nov 9 at 9:37






How do you deal with middle element in case there are even number of elements?
– Pushpesh Kumar Rajwanshi
Nov 9 at 9:37














for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:52




for that situation i will check for if n is even or odd and if it is even then i will delete the (n/2)-1 position as per the test case.for eg-n=6 and element is 1 2 3 4 5 6 then my output will be 6 5 4 2 1
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:52












kindly tell me the logic to delete the middle element for the odd no.of elements
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:52




kindly tell me the logic to delete the middle element for the odd no.of elements
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:52












2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










Do not insert the middle element of input array.



get the middle element index :



 int middle = a.length/2;


do not push the middle element in the stack:



Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++){
if(i != middle)
stack.push(a[i]);
}


Everything else looks okay. Just make sure to provide meaningful names to variables.






share|improve this answer





















  • it worked for me..thanks buddy
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:59


















up vote
1
down vote













You can do this with using pop() method popping returns and removes the top element of the stack so that way you can create and fill a new stack with reverse order, don't need to reverse the iterator, have a look at the code below.



import java.util.ListIterator;
import java.util.Scanner;
import java.util.Stack;

class GFG {

public static void main(String args) {
Scanner s = new Scanner(System.in);
//Define stacks here
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> new_stack = new Stack<Integer>();
int test = s.nextInt();
for (int t = 0; t < test; t++) {
int n = s.nextInt();
int a = new int[n];
double middle = Math.ceil((double) n / 2);
System.out.println("Middle is : " + middle);
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
}

// add elements to stack

for (int i = 0; i < n; i++) {
stack.push(a[i]);
}

//popping the elements of stack

for (int j = 0; j < n; j++) {
Integer element = stack.pop();
if (j != middle -1) {
new_stack.push(element);
}
}

ListIterator<Integer> lstIterator = new_stack.listIterator(stack.size());
while (lstIterator.hasNext()) {
Integer res = lstIterator.next();
//what condition should i give so that it would print all the elements except middle one.
System.out.print(res + " ");
}
System.out.println();
}
}
}





share|improve this answer





















  • thanks buddy for the alternate answer.
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 10:08










  • you're welcome.
    – Alican Beydemir
    Nov 9 at 10:15











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Do not insert the middle element of input array.



get the middle element index :



 int middle = a.length/2;


do not push the middle element in the stack:



Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++){
if(i != middle)
stack.push(a[i]);
}


Everything else looks okay. Just make sure to provide meaningful names to variables.






share|improve this answer





















  • it worked for me..thanks buddy
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:59















up vote
0
down vote



accepted










Do not insert the middle element of input array.



get the middle element index :



 int middle = a.length/2;


do not push the middle element in the stack:



Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++){
if(i != middle)
stack.push(a[i]);
}


Everything else looks okay. Just make sure to provide meaningful names to variables.






share|improve this answer





















  • it worked for me..thanks buddy
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:59













up vote
0
down vote



accepted







up vote
0
down vote



accepted






Do not insert the middle element of input array.



get the middle element index :



 int middle = a.length/2;


do not push the middle element in the stack:



Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++){
if(i != middle)
stack.push(a[i]);
}


Everything else looks okay. Just make sure to provide meaningful names to variables.






share|improve this answer












Do not insert the middle element of input array.



get the middle element index :



 int middle = a.length/2;


do not push the middle element in the stack:



Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<n;i++){
if(i != middle)
stack.push(a[i]);
}


Everything else looks okay. Just make sure to provide meaningful names to variables.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 9 at 9:36









janardhan sharma

2507




2507












  • it worked for me..thanks buddy
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:59


















  • it worked for me..thanks buddy
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 9:59
















it worked for me..thanks buddy
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:59




it worked for me..thanks buddy
– RAVI KUMAR SHARMA CSE16
Nov 9 at 9:59












up vote
1
down vote













You can do this with using pop() method popping returns and removes the top element of the stack so that way you can create and fill a new stack with reverse order, don't need to reverse the iterator, have a look at the code below.



import java.util.ListIterator;
import java.util.Scanner;
import java.util.Stack;

class GFG {

public static void main(String args) {
Scanner s = new Scanner(System.in);
//Define stacks here
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> new_stack = new Stack<Integer>();
int test = s.nextInt();
for (int t = 0; t < test; t++) {
int n = s.nextInt();
int a = new int[n];
double middle = Math.ceil((double) n / 2);
System.out.println("Middle is : " + middle);
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
}

// add elements to stack

for (int i = 0; i < n; i++) {
stack.push(a[i]);
}

//popping the elements of stack

for (int j = 0; j < n; j++) {
Integer element = stack.pop();
if (j != middle -1) {
new_stack.push(element);
}
}

ListIterator<Integer> lstIterator = new_stack.listIterator(stack.size());
while (lstIterator.hasNext()) {
Integer res = lstIterator.next();
//what condition should i give so that it would print all the elements except middle one.
System.out.print(res + " ");
}
System.out.println();
}
}
}





share|improve this answer





















  • thanks buddy for the alternate answer.
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 10:08










  • you're welcome.
    – Alican Beydemir
    Nov 9 at 10:15















up vote
1
down vote













You can do this with using pop() method popping returns and removes the top element of the stack so that way you can create and fill a new stack with reverse order, don't need to reverse the iterator, have a look at the code below.



import java.util.ListIterator;
import java.util.Scanner;
import java.util.Stack;

class GFG {

public static void main(String args) {
Scanner s = new Scanner(System.in);
//Define stacks here
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> new_stack = new Stack<Integer>();
int test = s.nextInt();
for (int t = 0; t < test; t++) {
int n = s.nextInt();
int a = new int[n];
double middle = Math.ceil((double) n / 2);
System.out.println("Middle is : " + middle);
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
}

// add elements to stack

for (int i = 0; i < n; i++) {
stack.push(a[i]);
}

//popping the elements of stack

for (int j = 0; j < n; j++) {
Integer element = stack.pop();
if (j != middle -1) {
new_stack.push(element);
}
}

ListIterator<Integer> lstIterator = new_stack.listIterator(stack.size());
while (lstIterator.hasNext()) {
Integer res = lstIterator.next();
//what condition should i give so that it would print all the elements except middle one.
System.out.print(res + " ");
}
System.out.println();
}
}
}





share|improve this answer





















  • thanks buddy for the alternate answer.
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 10:08










  • you're welcome.
    – Alican Beydemir
    Nov 9 at 10:15













up vote
1
down vote










up vote
1
down vote









You can do this with using pop() method popping returns and removes the top element of the stack so that way you can create and fill a new stack with reverse order, don't need to reverse the iterator, have a look at the code below.



import java.util.ListIterator;
import java.util.Scanner;
import java.util.Stack;

class GFG {

public static void main(String args) {
Scanner s = new Scanner(System.in);
//Define stacks here
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> new_stack = new Stack<Integer>();
int test = s.nextInt();
for (int t = 0; t < test; t++) {
int n = s.nextInt();
int a = new int[n];
double middle = Math.ceil((double) n / 2);
System.out.println("Middle is : " + middle);
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
}

// add elements to stack

for (int i = 0; i < n; i++) {
stack.push(a[i]);
}

//popping the elements of stack

for (int j = 0; j < n; j++) {
Integer element = stack.pop();
if (j != middle -1) {
new_stack.push(element);
}
}

ListIterator<Integer> lstIterator = new_stack.listIterator(stack.size());
while (lstIterator.hasNext()) {
Integer res = lstIterator.next();
//what condition should i give so that it would print all the elements except middle one.
System.out.print(res + " ");
}
System.out.println();
}
}
}





share|improve this answer












You can do this with using pop() method popping returns and removes the top element of the stack so that way you can create and fill a new stack with reverse order, don't need to reverse the iterator, have a look at the code below.



import java.util.ListIterator;
import java.util.Scanner;
import java.util.Stack;

class GFG {

public static void main(String args) {
Scanner s = new Scanner(System.in);
//Define stacks here
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> new_stack = new Stack<Integer>();
int test = s.nextInt();
for (int t = 0; t < test; t++) {
int n = s.nextInt();
int a = new int[n];
double middle = Math.ceil((double) n / 2);
System.out.println("Middle is : " + middle);
for (int i = 0; i < n; i++) {
a[i] = s.nextInt();
}

// add elements to stack

for (int i = 0; i < n; i++) {
stack.push(a[i]);
}

//popping the elements of stack

for (int j = 0; j < n; j++) {
Integer element = stack.pop();
if (j != middle -1) {
new_stack.push(element);
}
}

ListIterator<Integer> lstIterator = new_stack.listIterator(stack.size());
while (lstIterator.hasNext()) {
Integer res = lstIterator.next();
//what condition should i give so that it would print all the elements except middle one.
System.out.print(res + " ");
}
System.out.println();
}
}
}






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 9 at 10:01









Alican Beydemir

226312




226312












  • thanks buddy for the alternate answer.
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 10:08










  • you're welcome.
    – Alican Beydemir
    Nov 9 at 10:15


















  • thanks buddy for the alternate answer.
    – RAVI KUMAR SHARMA CSE16
    Nov 9 at 10:08










  • you're welcome.
    – Alican Beydemir
    Nov 9 at 10:15
















thanks buddy for the alternate answer.
– RAVI KUMAR SHARMA CSE16
Nov 9 at 10:08




thanks buddy for the alternate answer.
– RAVI KUMAR SHARMA CSE16
Nov 9 at 10:08












you're welcome.
– Alican Beydemir
Nov 9 at 10:15




you're welcome.
– Alican Beydemir
Nov 9 at 10:15


















 

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ꓛꓣだゔៀៅຸ໢ທຮ໕໒ ,ໂ'໥໓າ໼ឨឲ៵៭ៈゎゔit''䖳𥁄卿' ☨₤₨こゎもょの;ꜹꟚꞖꞵꟅꞛေၦေɯ,ɨɡ𛃵𛁹ޝ޳ޠ޾,ޤޒޯ޾𫝒𫠁သ𛅤チョ'サノބޘދ𛁐ᶿᶇᶀᶋᶠ㨑㽹⻮ꧬ꧹؍۩وَؠ㇕㇃㇪ ㇦㇋㇋ṜẰᵡᴠ 軌ᵕ搜۳ٰޗޮ޷ސޯ𫖾𫅀ल, ꙭ꙰ꚅꙁꚊꞻꝔ꟠Ꝭㄤﺟޱސꧨꧼ꧴ꧯꧽ꧲ꧯ'⽹⽭⾁⿞⼳⽋២៩ញណើꩯꩤ꩸ꩮᶻᶺᶧᶂ𫳲𫪭𬸄𫵰𬖩𬫣𬊉ၲ𛅬㕦䬺𫝌𫝼,,𫟖𫞽ហៅ஫㆔ాఆఅꙒꚞꙍ,Ꙟ꙱エ ,ポテ,フࢰࢯ𫟠𫞶 𫝤𫟠ﺕﹱﻜﻣ𪵕𪭸𪻆𪾩𫔷ġ,ŧآꞪ꟥,ꞔꝻ♚☹⛵𛀌ꬷꭞȄƁƪƬșƦǙǗdžƝǯǧⱦⱰꓕꓢႋ神 ဴ၀க௭எ௫ឫោ ' េㇷㇴㇼ神ㇸㇲㇽㇴㇼㇻㇸ'ㇸㇿㇸㇹㇰㆣꓚꓤ₡₧ ㄨㄟ㄂ㄖㄎ໗ツڒذ₶।ऩछएोञयूटक़कयँृी,冬'𛅢𛅥ㇱㇵㇶ𥄥𦒽𠣧𠊓𧢖𥞘𩔋цѰㄠſtʯʭɿʆʗʍʩɷɛ,əʏダヵㄐㄘR{gỚṖḺờṠṫảḙḭᴮᵏᴘᵀᵷᵕᴜᴏᵾq﮲ﲿﴽﭙ軌ﰬﶚﶧ﫲Ҝжюїкӈㇴffצּ﬘﭅﬈軌'ffistfflſtffतभफɳɰʊɲʎ𛁱𛁖𛁮𛀉 𛂯𛀞నఋŀŲ 𫟲𫠖𫞺ຆຆ ໹້໕໗ๆทԊꧢꧠ꧰ꓱ⿝⼑ŎḬẃẖỐẅ ,ờỰỈỗﮊDžȩꭏꭎꬻ꭮ꬿꭖꭥꭅ㇭神 ⾈ꓵꓑ⺄㄄ㄪㄙㄅㄇstA۵䞽ॶ𫞑𫝄㇉㇇゜軌𩜛𩳠Jﻺ‚Üမ႕ႌႊၐၸဓၞၞၡ៸wyvtᶎᶪᶹစဎ꣡꣰꣢꣤ٗ؋لㇳㇾㇻㇱ㆐㆔,,㆟Ⱶヤマފ޼ޝަݿݞݠݷݐ',ݘ,ݪݙݵ𬝉𬜁𫝨𫞘くせぉて¼óû×ó£…𛅑הㄙくԗԀ5606神45,神796'𪤻𫞧ꓐ㄁ㄘɥɺꓵꓲ3''7034׉ⱦⱠˆ“𫝋ȍ,ꩲ軌꩷ꩶꩧꩫఞ۔فڱێظペサ神ナᴦᵑ47 9238їﻂ䐊䔉㠸﬎ffiﬣ,לּᴷᴦᵛᵽ,ᴨᵤ ᵸᵥᴗᵈꚏꚉꚟ⻆rtǟƴ𬎎

Guess what letter conforming each word