Aquí dado un rango de menor a mayor y dado un número k. Tienes que encontrar el número de conteo en el que un número tiene el mismo dígito que k
Ejemplos:
Input: low = 2, high = 35, k = 2 Output: 4 Numbers are 2, 12, 22, 32 Input: low = 3, high = 30, k = 3 Output: 3 Numbers are 3, 13, 23
Un enfoque ingenuo es atravesar todos los números en un rango dado y verificar el último dígito de cada número e incrementar el resultado si el último dígito es igual a k.
C++
// Simple CPP program to count numbers with
// last digit as k in given range.
#include <bits/stdc++.h>
using namespace std;
// Returns count of numbers with k as last
// digit.
int countLastDigitK(int low, int high, int k)
{
int count = 0;
for (int i = low; i <= high; i++)
if (i % 10 == k)
count++;
return count;
}
// Driver Program
int main()
{
int low = 3, high = 35, k = 3;
cout << countLastDigitK(low, high, k);
return 0;
}
Java
// Simple Java program to count numbers with
// last digit as k in given range.
import java.util.*;
import java.lang.*;
public class GfG
{
// Returns count of numbers with
// k as last digit.
public static int countLastDigitK(int low,
int high, int k)
{
int count = 0;
for (int i = low; i <= high; i++)
if (i % 10 == k)
count++;
return count;
}
// driver function
public static void main(String args[])
{
int low = 3, high = 35, k = 3;
System.out.println(countLastDigitK(low, high, k));
}
}
// This code is contributed by Sagar Shukla
Python3
# Simple python program to count numbers with # last digit as k in given range. # Returns count of numbers with k as last # digit. def countLastDigitK(low, high, k): count = 0 for i in range(low, high+1): if (i % 10 == k): count+=1 return count # Driver Program low = 3 high = 35 k = 3 print(countLastDigitK(low, high, k)) # This code is contributed by # Smitha Dinesh Semwal
C#
// Simple C# program to count numbers with
// last digit as k in given range.
using System;
public class GfG
{
// Returns count of numbers with
// k as last digit.
public static int countLastDigitK(int low,
int high, int k)
{
int count = 0;
for (int i = low; i <= high; i++)
if (i % 10 == k)
count++;
return count;
}
// Driver function
public static void Main()
{
int low = 3, high = 35, k = 3;
Console.WriteLine(countLastDigitK(low, high, k));
}
}
// This code is contributed by vt_m
PHP
<?php
// Simple PHP program to count numbers with
// last digit as k in given range.
// Returns count of numbers with
// k as last digit.
function countLastDigitK($low, $high, $k)
{
$count = 0;
for ($i = $low; $i <= $high; $i++)
if ($i % 10 == $k)
$count++;
return $count;
}
// Driver Code
$low = 3;
$high = 35;
$k = 3;
echo countLastDigitK($low, $high, $k);
// This code is contributed by ajit
?>
Javascript
<script>
// java script PHP program to count numbers with
// last digit as k in given range.
// Returns count of numbers with
// k as last digit.
function countLastDigitK(low, high, k)
{
let count = 0;
for (let i = low; i <= high; i++)
if (i % 10 == k)
count++;
return count;
}
// Driver Code
let low = 3;
let high = 35;
let k = 3;
document.write( countLastDigitK(low, high, k));
// This code is contributed
// by sravan kumar
</script>
Producción:
4
Complejidad de tiempo: O (alta – baja)
Espacio Auxiliar: O(1)
Una solución eficiente se basa en que cada dígito aparece una vez como el último dígito de cada 10 números consecutivos.
C++
// Efficient CPP program to count numbers
// with last digit as k in given range.
#include <bits/stdc++.h>
using namespace std;
// Returns count of numbers with k as last
// digit.
int countLastDigitK(long long low,
long long high, long long K)
{
long long mlow = 10 * ceil(low/10.0);
long long mhigh = 10 * floor(high/10.0);
int count = (mhigh - mlow)/10;
if (high % 10 >= K)
count++;
if (low % 10 <=K && (low%10))
count++;
return count;
}
// Driver Code
int main()
{
int low = 3, high = 35, k = 3;
cout << countLastDigitK(low, high, k);
return 0;
}
Java
// Efficient Java program to count numbers
// with last digit as k in given range.
import java.util.*;
import java.lang.*;
public class GfG
{
// Returns count of numbers with
// k as last digit.
public static int counLastDigitK(int low,
int high, int k)
{
int mlow = 10 * (int)
Math.ceil(low/10.0);
int mhigh = 10 * (int)
Math.floor(high/10.0);
int count = (mhigh - mlow)/10;
if (high % 10 >= k)
count++;
if (low % 10 <= k && (low%10) > 0)
count++;
return count;
}
// driver function
public static void main(String argc[])
{
int low = 3, high = 35, k = 3;
System.out.println(counLastDigitK(low, high, k));
}
}
// This code is contributed by Sagar Shukla
Python3
import math # Efficient python program to count numbers # with last digit as k in given range. # Returns count of numbers with k as last # digit. def counLastDigitK(low, high, k): mlow = 10 * math.ceil(low/10.0) mhigh = 10 * int(high/10.0) count = (mhigh - mlow)/10 if (high % 10 >= k): count += 1 if (low % 10 <= k and \ (low%10) > 0): count += 1 return int(count) # Driver Code low = 3 high = 35 k = 3 print(counLastDigitK(low, high, k)) # This code is contributed by # Smitha Dinesh Semwal
C#
// Efficient Java program to count numbers
// with last digit as k in given range.
using System;
public class GfG
{
// Returns count of numbers with
// k as last digit.
public static int counLastDigitK(int low,
int high, int k)
{
int mlow = 10 * Convert.ToInt32(
Math.Ceiling(low/10.0));
int mhigh = 10 * Convert.ToInt32(
Math.Floor(high/10.0));
int count = (mhigh - mlow) / 10;
if (high % 10 >= k)
count++;
if (low % 10 <= k && (low%10) > 0)
count++;
return count;
}
// Driver function
public static void Main()
{
int low = 3, high = 35, k = 3;
Console.WriteLine(
counLastDigitK(low, high, k));
}
}
// This code is contributed by vt_m
Javascript
<script>
// Efficient Javascript program to count numbers
// with last digit as k in given range.
// Returns count of numbers with
// k as last digit.
function counLastDigitK(low, high, k)
{
let mlow = 10 * (Math.ceil(low/10.0));
let mhigh = 10 * (Math.floor(high/10.0));
let count = (mhigh - mlow) / 10;
if (high % 10 >= k)
count++;
if (low % 10 <= k && (low%10) > 0)
count++;
return count;
}
let low = 3, high = 35, k = 3;
document.write(counLastDigitK(low, high, k));
</script>
Producción
4
Complejidad de tiempo : O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por shrikanth13 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA