Check if mirror image of a number is same if displayed in seven segment display
Given a positive number n. The task is to check if the mirror image of the number is equal to given number or not if displayed on Seven Line Segment. A mirror image of a number is a reflected duplication of the number that appears almost identical but is reversed in the direction perpendicular to the mirror surface.
Examples:
Input : n = 101
Output: Yes
Mirror image of 101 is 101 on seven line segment. So, print "Yes".
Input : n = 020
Output : No
Observe each digit when displayed on seven line segment, only digit 0, 1, 8 remain same in their mirror image. So, for a number to be equal to its mirror image, it should contain only 0, 1, 8 digits. Also, observe, for two number to be equal their corresponding position digit should be same. Therefore, mirror image should also contain the same digit on their corresponding digit position. So, the number should also be a palindrome.
Below is implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
string checkEqual(string S)
{
for ( int i = 0; i < S.size(); i++) {
if (S[i] != '1' && S[i] != '0' && S[i] != '8' ) {
return "No" ;
}
}
int start = 0, end = S.size() - 1;
while (start < end) {
if (S[start] != S[end]) {
return "No" ;
}
start++;
end--;
}
return "Yes" ;
}
int main()
{
string S = "101" ;
cout << checkEqual(S) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static String checkEqual(String S)
{
for ( int i = 0 ;
i < S.length(); i++)
{
if (S.charAt(i) != '1' &&
S.charAt(i) != '0' &&
S.charAt(i) != '8' )
{
return "No" ;
}
}
int start = 0 ,
end = S.length() - 1 ;
while (start < end)
{
if (S.charAt(start) !=
S.charAt(end))
{
return "No" ;
}
start++;
end--;
}
return "Yes" ;
}
public static void main (String[] args)
{
String S = "101" ;
System.out.println(checkEqual(S));
}
}
|
Python3
def checkEqual(S):
for i in range ( len (S)):
if (S[i] ! = '1' and S[i] ! = '0'
and S[i] ! = '8' ):
return "No" ;
start = 0 ;
end = len (S) - 1 ;
while (start < end):
if (S[start] ! = S[end]):
return "No" ;
start + = 1 ;
end - = 1 ;
return "Yes" ;
S = "101" ;
print (checkEqual(S));
|
C#
using System;
class GFG
{
static string checkEqual( string S)
{
for ( int i = 0; i < S.Length; i++)
{
if (S[i] != '1' &&
S[i] != '0' &&
S[i] != '8' )
{
return "No" ;
}
}
int start = 0, end = S.Length - 1;
while (start < end)
{
if (S[start] !=
S[end])
{
return "No" ;
}
start++;
end--;
}
return "Yes" ;
}
public static void Main()
{
string S = "101" ;
Console.WriteLine(checkEqual(S));
}
}
|
PHP
<?php
function checkEqual( $S )
{
for ( $i = 0; $i < strlen ( $S ); $i ++)
{
if ( $S [ $i ] != '1' && $S [ $i ] != '0' &&
$S [ $i ] != '8' )
{
return "No" ;
}
}
$start = 0;
$end = strlen ( $S ) - 1;
while ( $start < $end )
{
if ( $S [ $start ] != $S [ $end ])
{
return "No" ;
}
$start ++;
$end --;
}
return "Yes" ;
}
$S = "101" ;
echo checkEqual( $S );
?>
|
Javascript
<script>
function checkEqual(S)
{
for ( var i = 0; i < S.length; i++) {
if (S[i] != '1' && S[i] != '0' && S[i] != '8' ) {
return "No" ;
}
}
var start = 0, end = S.length - 1;
while (start < end) {
if (S[start] != S[end]) {
return "No" ;
}
start++;
end--;
}
return "Yes" ;
}
var S = "101" ;
document.write( checkEqual(S));
</script>
|
Output
Yes
Time Complexity: O(n), where n is the size of the given string
Auxiliary Space: O(1), as no extra space is required
Last Updated :
23 Jun, 2022
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