Check if a number has digits in the given Order
Last Updated :
24 Jun, 2024
Given a number N. The task is to check if the digits of the number follow any of the below order:
- The digits are in strictly increasing order.
- Or, the digits are in strictly decreasing order.
- Or, the digits follow strictly increasing order first and then strictly decreasing.
If the number follows any of the above order then print YES otherwise print NO.
Examples:
Input : N = 415
Output : NO
Input : N = 123454321
Output : YES
Traverse the number from right to left by extracting each digit one by one. Keep a pointer to tell that whether the current sequence is descending or ascending sequence, -1 denotes strictly ascending and 1 denotes strictly descending sequence. At first the sequence should be strictly increasing as we are going from right to left. As we encounter a digit which is lesser than the previous digit, change the flag to decreasing(i.e -1) and while in increasing order we get any digit which is equal to the previous digit we directly print NO.
Below is the implementation of the above approach:
C++
// CPP program to check if the digits are in
// the given order
#include<bits/stdc++.h>
using namespace std;
// Check if the digits follow the correct order
bool isCorrectOrder(int n)
{
bool flag = true;
// to store the previous digit
int prev = -1;
// pointer to tell what type of sequence
// are we dealing with
int type = -1;
while(n != 0)
{
if(type ==-1)
{
if(prev ==-1)
{
prev = n % 10;
n = n/10;
continue;
}
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// checking the peak point of the number
if(prev > n % 10)
{
type = 1;
prev = n % 10;
n = n/10;
continue;
}
prev = n % 10;
n = n / 10;
}
else
{
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// check if the digit is greater than
// the previous one
// If true, then break from the loop as
// we are in descending order part
if(prev < n % 10)
{
flag = false;
break;
}
prev = n % 10;
n = n / 10;
}
}
return flag;
}
// Driver code
int main()
{
int n = 123454321;
if(isCorrectOrder(n))
cout<<"YES";
else
cout<<"NO";
return 0;
}
// Java program to check if the digits are in
// the given order
import java.io.*;
class GFG {
// Check if the digits follow the correct order
static boolean isCorrectOrder(int n)
{
boolean flag = true;
// to store the previous digit
int prev = -1;
// pointer to tell what type of sequence
// are we dealing with
int type = -1;
while(n != 0)
{
if(type ==-1)
{
if(prev ==-1)
{
prev = n % 10;
n = n/10;
continue;
}
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// checking the peak point of the number
if(prev > n % 10)
{
type = 1;
prev = n % 10;
n = n/10;
continue;
}
prev = n % 10;
n = n / 10;
}
else
{
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// check if the digit is greater than
// the previous one
// If true, then break from the loop as
// we are in descending order part
if(prev < n % 10)
{
flag = false;
break;
}
prev = n % 10;
n = n / 10;
}
}
return flag;
}
// Driver code
public static void main (String[] args) {
int n = 123454321;
if(isCorrectOrder(n))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by ajit
# Python3 program to check if the
# digits are in the given order
# Check if the digits follow
# the correct order
def isCorrectOrder(n):
flag = True;
# to store the previous digit
prev = -1;
# pointer to tell what type of
# sequence are we dealing with
type = -1;
while(n != 0):
if(type ==-1):
if(prev ==-1):
prev = n % 10;
n = int(n / 10);
continue;
# check if we have same digit
# as the previous digit
if(prev == n % 10):
flag = False;
break;
# checking the peak point
# of the number
if(prev > n % 10):
type = 1;
prev = n % 10;
n = int(n / 10);
continue;
prev = n % 10;
n = int(n / 10);
else:
# check if we have same digit
# as the previous digit
if(prev == n % 10):
flag = False;
break;
# check if the digit is greater
# than the previous one
# If true, then break from the
# loop as we are in descending
# order part
if(prev < n % 10):
flag = False;
break;
prev = n % 10;
n = int(n / 10);
return flag;
# Driver code
n = 123454321;
if(isCorrectOrder(n)):
print("YES");
else:
print("NO");
# This Code is contributed by mits
// C# program to check if the
// digits are in the given order
using System;
class GFG
{
// Check if the digits follow
// the correct order
static bool isCorrectOrder(int n)
{
bool flag = true;
// to store the previous digit
int prev = -1;
// pointer to tell what type of
// sequence are we dealing with
int type = -1;
while(n != 0)
{
if(type == -1)
{
if(prev == -1)
{
prev = n % 10;
n = n / 10;
continue;
}
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// checking the peak point
// of the number
if(prev > n % 10)
{
type = 1;
prev = n % 10;
n = n / 10;
continue;
}
prev = n % 10;
n = n / 10;
}
else
{
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// check if the digit is greater
// than the previous one
// If true, then break from the
// loop as we are in descending
// order part
if(prev < n % 10)
{
flag = false;
break;
}
prev = n % 10;
n = n / 10;
}
}
return flag;
}
// Driver code
public static void Main ()
{
int n = 123454321;
if(isCorrectOrder(n))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by Shashank
<script>
// JavaScript program to check if the digits are in
// the given order
// Check if the digits follow the correct order
function isCorrectOrder(n)
{
let flag = true;
// to store the previous digit
let prev = -1;
// pointer to tell what type of sequence
// are we dealing with
let type = -1;
while(n != 0)
{
if(type ===-1)
{
if(prev ===-1)
{
prev = n % 10;
n = Math.floor(n/10);
continue;
}
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// checking the peak point of the number
if(prev > n % 10)
{
type = 1;
prev = n % 10;
n = Math.floor(n/10);
continue;
}
prev = n % 10;
n = Math.floor(n / 10);
}
else
{
// check if we have same digit
// as the previous digit
if(prev == n % 10)
{
flag = false;
break;
}
// check if the digit is greater than
// the previous one
// If true, then break from the loop as
// we are in descending order part
if(prev < n % 10)
{
flag = false;
break;
}
prev = n % 10;
n = Math.floor(n / 10);
}
}
return flag;
}
// Driver code
let n = 123454321;
if(isCorrectOrder(n))
document.write("YES");
else
document.write("NO");
// This code is contributed by Surbhi Tyagi
</script>
<?php
// PHP program to check if the
// digits are in the given order
// Check if the digits follow
// the correct order
function isCorrectOrder($n)
{
$flag = true;
// to store the previous digit
$prev = -1;
// pointer to tell what type of
// sequence are we dealing with
$type = -1;
while($n != 0)
{
if($type ==-1)
{
if($prev ==-1)
{
$prev = $n % 10;
$n = (int)$n / 10;
continue;
}
// check if we have same digit
// as the previous digit
if($prev == $n % 10)
{
$flag = false;
break;
}
// checking the peak point
// of the number
if($prev > $n % 10)
{
$type = 1;
$prev = $n % 10;
$n = (int)$n / 10;
continue;
}
$prev = $n % 10;
$n = (int)$n / 10;
}
else
{
// check if we have same digit
// as the previous digit
if($prev == $n % 10)
{
$flag = false;
break;
}
// check if the digit is greater
// than the previous one
// If true, then break from the
// loop as we are in descending
// order part
if($prev < $n % 10)
{
$flag = false;
break;
}
$prev = $n % 10;
$n = (int) $n / 10;
}
}
return $flag;
}
// Driver code
$n = 123454321;
if(isCorrectOrder($n))
echo "YES";
else
echo "NO";
// This Code is contributed by ajit
?>
Time Complexity: O(logN)
Auxiliary Space: O(1), since no extra space has been taken.
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