Dado un arreglo de N enteros, la tarea es encontrar la longitud del subarreglo más largo tal que los elementos adyacentes del subarreglo tengan al menos un dígito en común.
Ejemplos:
Input : arr[] = {12, 23, 45, 43, 36, 97}
Output : 3
Explanation: The subarray is 45 43 36 which has
4 common in 45, 43 and 3 common in 43, 36.
Input : arr[] = {11, 22, 33, 44, 54, 56, 63}
Output : 4
Explanation: Subarray is 44, 54, 56, 63
La solución discutida en la publicación anterior usa O (N) espacio adicional. El problema se puede resolver utilizando el espacio constante. Se utiliza un mapa hash de tamaño constante para almacenar si un dígito está presente en un elemento de array dado o no. Para verificar si los elementos adyacentes tienen un dígito común, solo se requiere contar los dígitos de dos elementos adyacentes. Entonces, la cantidad de filas requeridas en hashmap se puede reducir a 2. La variable currRow representa la fila actual y 1: currRow representa la fila anterior en hashmap. Si los elementos adyacentes tienen un dígito común, aumente la longitud actual en 1 y compárela con la longitud máxima. De lo contrario, establezca la longitud actual en 1.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
#include <bits/stdc++.h>
using namespace std;
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
int longestSubarray(int arr[], int n)
{
int i, d;
// To mark presence of digit in current
// element.
int hash[2][10];
memset(hash, 0, sizeof(hash));
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0) {
hash[0][tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++) {
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow][d] = 0;
// Find all digits in element.
while (tmp > 0) {
hash[currRow][tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++) {
if (hash[currRow][d] && hash[1 - currRow][d]) {
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10) {
len = 1;
}
maxLen = max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
int main()
{
int arr[] = { 11, 22, 33, 44, 54, 56, 63 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << longestSubarray(arr, n);
return 0;
}
Java
// Java program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
class GFG
{
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
static int longestSubarray(int arr[], int n)
{
int i, d;
// To mark presence of digit in current
// element.
int hash[][] = new int[2][10];
for( i = 0; i < 2; i++)
for(int j = 0; j < 10; j++)
hash[i][j] = 0;
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0)
{
hash[0][tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++)
{
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow][d] = 0;
// Find all digits in element.
while (tmp > 0)
{
hash[currRow][tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++)
{
if (hash[currRow][d] != 0 && hash[1 - currRow][d] != 0)
{
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10)
{
len = 1;
}
maxLen = Math.max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 11, 22, 33, 44, 54, 56, 63 };
int n = arr.length;
System.out.println( longestSubarray(arr, n));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 program to print the length of the # longest subarray such that adjacent elements # of the subarray have at least one digit in # common import math # Function to print the longest subarray # such that adjacent elements have at least # one digit in common def longestSubarray(arr, n): i = d = 0; # To mark presence of digit in current # element. HASH1 = [[0 for x in range(10)] for y in range(2)]; # To store current row. currRow = 0; # To store maximum length subarray length. maxLen = 1; # To store current subarray length. len1 = 0; # To store current array element. tmp = 0; # Mark the presence of digits # of first element. tmp = arr[0]; while (tmp > 0): HASH1[0][tmp % 10] = 1; tmp = tmp // 10; currRow = 1; # Find digits of each element and check # if adjacent elements have common digit # and update len. for i in range(0, n): tmp = arr[i]; for d in range(0, 10): HASH1[currRow][d] = 0; # Find all digits in element. while (tmp > 0): HASH1[currRow][tmp % 10] = 1; tmp = tmp // 10; # Find common digit in adjacent element. for d in range(0, 10): if (HASH1[currRow][d] and HASH1[1 - currRow][d]): len1 += 1; break; # If no common digit is found a new subarray # has to start from current element. if (d == 10): len1 = 1; maxLen = max(maxLen, len1); currRow = 1 - currRow; return maxLen; # Driver Code arr = [ 11, 22, 33, 44, 54, 56, 63 ]; n = len(arr); print(longestSubarray(arr, n)); # This code is contributed by chandan_jnu
C#
// C# program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
using System;
class GFG
{
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
static int longestSubarray(int []arr, int n)
{
int i, d;
// To mark presence of digit in current
// element.
int[,] hash = new int[2,10];
for( i = 0; i < 2; i++)
for(int j = 0; j < 10; j++)
hash[i,j] = 0;
// To store current row.
int currRow;
// To store maximum length subarray length.
int maxLen = 1;
// To store current subarray length.
int len = 0;
// To store current array element.
int tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0)
{
hash[0,tmp % 10] = 1;
tmp /= 10;
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++)
{
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow,d] = 0;
// Find all digits in element.
while (tmp > 0)
{
hash[currRow,tmp % 10] = 1;
tmp /= 10;
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++)
{
if (hash[currRow,d] != 0 &&
hash[1 - currRow,d] != 0)
{
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10)
{
len = 1;
}
maxLen = Math.Max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
static void Main()
{
int []arr = { 11, 22, 33, 44, 54, 56, 63 };
int n = arr.Length;
Console.WriteLine( longestSubarray(arr, n));
}
}
// This code is contributed by chandan_jnu
PHP
<?php
// PHP program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
function longestSubarray($arr, $n)
{
$i = $d = 0;
// To mark presence of digit in current
// element.
$hash = array_fill(0, 2, array_fill(0, 10, 0));
// To store current row.
$currRow = 0;
// To store maximum length subarray length.
$maxLen = 1;
// To store current subarray length.
$len = 0;
// To store current array element.
$tmp = 0;
// Mark the presence of digits
// of first element.
$tmp = $arr[0];
while ($tmp > 0)
{
$hash[0][$tmp % 10] = 1;
$tmp = (int)($tmp / 10);
}
$currRow = 1;
// Find digits of each element and check
// if adjacent elements have common digit
// and update len.
for ($i = 1; $i < $n; $i++)
{
$tmp = $arr[$i];
for ($d = 0; $d <= 9; $d++)
$hash[$currRow][$d] = 0;
// Find all digits in element.
while ($tmp > 0)
{
$hash[$currRow][$tmp % 10] = 1;
$tmp =(int)($tmp/10);
}
// Find common digit in adjacent element.
for ($d = 0; $d <= 9; $d++)
{
if ($hash[$currRow][$d] &&
$hash[1 - $currRow][$d])
{
$len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if ($d == 10)
{
$len = 1;
}
$maxLen = max($maxLen, $len);
$currRow = 1 - $currRow;
}
return $maxLen;
}
// Driver Code
$arr = array( 11, 22, 33, 44, 54, 56, 63 );
$n = count($arr);
echo longestSubarray($arr, $n);
// This code is contributed by chandan_jnu
?>
Javascript
<script>
// Javascript program to print the length of the
// longest subarray such that adjacent elements
// of the subarray have at least one digit in
// common
// Function to print the longest subarray
// such that adjacent elements have at least
// one digit in common
function longestSubarray(arr, n)
{
var i, d;
// To mark presence of digit in current
// element.
var hash = Array.from(Array(2), () => Array(10));
// To store current row.
var currRow;
// To store maximum length subarray length.
var maxLen = 1;
// To store current subarray length.
var len = 0;
// To store current array element.
var tmp;
// Mark the presence of digits of first element.
tmp = arr[0];
while (tmp > 0) {
hash[0][tmp % 10] = 1;
tmp = parseInt(tmp/10);
}
currRow = 1;
// Find digits of each element and check if adjacent
// elements have common digit and update len.
for (i = 1; i < n; i++) {
tmp = arr[i];
for (d = 0; d <= 9; d++)
hash[currRow][d] = 0;
// Find all digits in element.
while (tmp > 0) {
hash[currRow][tmp % 10] = 1;
tmp = parseInt(tmp/10);
}
// Find common digit in adjacent element.
for (d = 0; d <= 9; d++) {
if (hash[currRow][d] && hash[1 - currRow][d]) {
len++;
break;
}
}
// If no common digit is found a new subarray
// has to start from current element.
if (d == 10) {
len = 1;
}
maxLen = Math.max(maxLen, len);
currRow = 1 - currRow;
}
return maxLen;
}
// Driver Code
var arr = [ 11, 22, 33, 44, 54, 56, 63 ];
var n = arr.length;
document.write( longestSubarray(arr, n));
// This code is contributed by noob2000.
</script>
Producción:
4
Complejidad temporal: O(N)
Espacio auxiliar: O(1)