Dada una string ‘S’ (compuesta por dígitos) y un entero ‘X’, la tarea es contar todas las substrings de ‘S’ que cumplan las siguientes condiciones:
- La substring no debe comenzar con el dígito ‘0’.
- Y el número numérico que representa debe ser mayor que ‘X’.
Nota: dos formas de seleccionar una substring son diferentes si comienzan o terminan en índices diferentes.
Ejemplos:
Input: S = "471", X = 47 Output: 2 Only the sub-strings "471" and "71" satisfy the given conditions. Input: S = "2222", X = 97 Output: 3 Valid strings are "222", "222" and "2222".
Acercarse:
- Itere sobre cada dígito de la string ‘S’ y elija los dígitos que son mayores que ‘0’.
- Ahora, tome todas las substrings posibles a partir del carácter elegido en el paso anterior y convierta cada substring en un número entero.
- Compara el número entero del paso anterior con ‘X’. Si el número es mayor que ‘X’, incremente la variable de conteo.
- Finalmente, imprima el valor de la variable de conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that counts
// valid sub-strings
int count(string S, int X)
{
int count = 0;
const int N = S.length();
for (int i = 0; i < N; ++i) {
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0') {
for (int len = 1; (i + len) <= N; ++len) {
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
if (stoi(S.substr(i, len)) > X)
count++;
}
}
}
return count;
}
// Driver code
int main()
{
string S = "2222";
int X = 97;
cout << count(S, X);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that counts
// valid sub-strings
static int count(String S, int X)
{
int count = 0;
int N = S.length();
for (int i = 0; i < N; ++i)
{
// Only take those numbers
// that do not start with '0'.
if (S.charAt(i) != '0')
{
for (int len = 1;
(i + len) <= N; ++len)
{
// converting the sub-string
// starting from index 'i'
// and having length 'len' to
// int and checking if it is
// greater than X or not
int num = Integer.parseInt(S.substring(i, i + len));
if (num > X)
count++;
}
}
}
return count;
}
// Driver code
public static void main(String []args)
{
String S = "2222";
int X = 97;
System.out.println(count(S, X));
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of # the approach # Function that counts # valid sub-strings def countSubStr(S, X): cnt = 0 N = len(S) for i in range(0, N): # Only take those numbers # that do not start with '0'. if (S[i] != '0'): j = 1 while((j + i) <= N): # converting the sub-string # starting from index 'i' # and having length 'len' to # int and checking if it is # greater than X or not num = int(S[i : i + j]) if (num > X): cnt = cnt + 1 j = j + 1 return cnt; # Driver code S = "2222"; X = 97; print(countSubStr(S, X)) # This code is contributed by ihritik
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that counts
// valid sub-strings
static int count(string S, int X)
{
int count = 0;
int N = S.Length;
for (int i = 0; i < N; ++i)
{
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0')
{
for (int len = 1;
(i + len) <= N; ++len)
{
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
int num = Int32.Parse(S.Substring(i, len));
if (num > X)
count++;
}
}
}
return count;
}
// Driver code
public static void Main(String []args)
{
string S = "2222";
int X = 97;
Console.WriteLine(count(S, X));
}
}
// This code is contributed by ihritik
PHP
<?php
// PHP implementation of the approach
// Function that counts
// valid sub-strings
function countSubStr($S, $X)
{
$cnt = 0;
$N = strlen($S);
for ($i = 0; $i < $N; ++$i)
{
// Only take those numbers
// that do not start w$ith '0'.
if ($S[$i] != '0')
{
for ($len = 1;
($i + $len) <= $N; ++$len)
{
// converting the sub-str$ing
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
$num = intval(substr($S, $i, $len));
if ($num > $X)
$cnt++;
}
}
}
return $cnt;
}
// Driver code
$S = "2222";
$X = 97;
echo countSubStr($S, $X);
// This code is contributed by ihritik
?>
Javascript
<script>
// Javascript implementation of the approach
// Function that counts
// valid sub-strings
function count(S, X)
{
var count = 0;
var N = S.length;
for (var i = 0; i < N; ++i) {
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0') {
for (var len = 1; (i + len) <= N; ++len) {
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
if (parseInt(S.substring(i, i+len)) > X)
count++;
}
}
}
return count;
}
// Driver code
var S = "2222";
var X = 97;
document.write( count(S, X));
</script>
Producción:
3