Dada una string S , la tarea es imprimir el recuento de substrings de una string dada cuyo primer y último carácter son diferentes.
Ejemplos:
Entrada: S = “abcab”
Salida: 8
Explicación:
Hay 8 substrings que tienen el primer y el último carácter diferentes {ab, abc, abcab, bc, bca, ca, cab, ab}.Entrada: S = “aba”
Salida: 2
Explicación:
Hay 2 substrings que tienen el primer y el último carácter diferentes {ab, ba}.
Enfoque ingenuo: la idea es generar todas las substrings posibles de una string determinada y, para cada substring, verificar si el primer y el último carácter son diferentes o no. Si se encuentra que es cierto, incremente el conteo en 1 y verifique la siguiente substring. Imprime el conteo después del recorrido de toda la substring.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the substrings
// having different first and last
// characters
int countSubstring(string s, int n)
{
// Store the final count
int ans = 0;
// Loop to traverse the string
for (int i = 0; i < n; i++) {
// Counter for each iteration
int cnt = 0;
// Iterate over substrings
for (int j = i + 1; j < n; j++) {
// Compare the characters
if (s[j] != s[i])
// Increase count
cnt++;
}
// Adding count of substrings
// to the final count
ans += cnt;
}
// Print the final count
cout << ans;
}
// Driver Code
int main()
{
// Given string
string S = "abcab";
// Length of the string
int N = 5;
// Function Call
countSubstring(S, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to count the substrings
// having different first and last
// characters
static void countSubstring(String s, int n)
{
// Store the final count
int ans = 0;
// Loop to traverse the string
for(int i = 0; i < n; i++)
{
// Counter for each iteration
int cnt = 0;
// Iterate over substrings
for(int j = i + 1; j < n; j++)
{
// Compare the characters
if (s.charAt(j) != s.charAt(i))
// Increase count
cnt++;
}
// Adding count of substrings
// to the final count
ans += cnt;
}
// Print the final count
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
// Given string
String S = "abcab";
// Length of the string
int N = 5;
// Function call
countSubstring(S, N);
}
}
// This code is contributed by code_hunt
Python3
# Python3 program for the above approach # Function to count the substrings # having different first and last # characters def countSubstring(s, n): # Store the final count ans = 0 # Loop to traverse the string for i in range(n): # Counter for each iteration cnt = 0 # Iterate over substrings for j in range(i + 1, n): # Compare the characters if (s[j] != s[i]): # Increase count cnt += 1 # Adding count of substrings # to the final count ans += cnt # Print the final count print(ans) # Driver Code # Given string S = "abcab" # Length of the string N = 5 # Function call countSubstring(S, N) # This code is contributed by code_hunt
C#
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
class GFG{
// Function to count the substrings
// having different first and last
// characters
static void countSubstring(string s, int n)
{
// Store the final count
int ans = 0;
// Loop to traverse the string
for(int i = 0; i < n; i++)
{
// Counter for each iteration
int cnt = 0;
// Iterate over substrings
for(int j = i + 1; j < n; j++)
{
// Compare the characters
if (s[j] != s[i])
// Increase count
cnt++;
}
// Adding count of substrings
// to the final count
ans += cnt;
}
// Print the final count
Console.Write(ans);
}
// Driver Code
public static void Main(string[] args)
{
// Given string
string S = "abcab";
// Length of the string
int N = 5;
// Function call
countSubstring(S, N);
}
}
// This code is contributed by rutvik_56
Javascript
<script>
// JavaScript program for
// the above approach
// Function to count the substrings
// having different first and last
// characters
function countSubstring(s, n)
{
// Store the final count
let ans = 0;
// Loop to traverse the string
for(let i = 0; i < n; i++)
{
// Counter for each iteration
let cnt = 0;
// Iterate over substrings
for(let j = i + 1; j < n; j++)
{
// Compare the characters
if (s[j] != s[i])
// Increase count
cnt++;
}
// Adding count of substrings
// to the final count
ans += cnt;
}
// Print the final count
document.write(ans);
}
// Driver code
// Given string
let S = "abcab";
// Length of the string
let N = 5;
// Function call
countSubstring(S, N);
// This code is contributed by splevel62.
</script>
Producción:
8
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Enfoque eficiente: el enfoque anterior se puede optimizar utilizando Map by para almacenar la frecuencia de los caracteres de la string. Siga los pasos a continuación para resolver el problema:
- Inicialice dos variables, una para contar caracteres diferentes para cada iteración (digamos cur ) y otra para almacenar el recuento final de substrings (digamos ans ).
- Inicialice un mapa M para almacenar la frecuencia de todos los caracteres en él.
- Recorra la string dada y para cada carácter, siga los pasos a continuación:
- Iterar el mapa M .
- Si el primer elemento, es decir, la clave del mapa no es la misma que el carácter actual, entonces proceda.
- De lo contrario, agregue el valor correspondiente al carácter actual.
- Después de recorrer el Mapa , agregue cur al resultado final, es decir, ans += cur .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the substrings
// having different first & last character
int countSubstring(string s, int n)
{
// Stores frequency of each char
map<char, int> m;
// Loop to store frequency of
// the characters in a Map
for (int i = 0; i < n; i++)
m[s[i]]++;
// To store final result
int ans = 0;
// Traversal of string
for (int i = 0; i < n; i++) {
// Store answer for every
// iteration
int cnt = 0;
m[s[i]]--;
// Map traversal
for (auto value : m) {
// Compare current char
if (value.first == s[i]) {
continue;
}
else {
cnt += value.second;
}
}
ans += cnt;
}
// Print the final count
cout << ans;
}
// Driver Code
int main()
{
// Given string
string S = "abcab";
// Length of the string
int N = 5;
// Function Call
countSubstring(S, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{
// Stores frequency of each char
HashMap<Character,
Integer> mp = new HashMap<Character,
Integer>();
// Loop to store frequency of
// the characters in a Map
for(int i = 0; i < n; i++)
if (mp.containsKey(s[i]))
{
mp.put(s[i], mp.get(s[i]) + 1);
}
else
{
mp.put(s[i], 1);
}
// To store final result
int ans = 0;
// Traversal of String
for(int i = 0; i < n; i++)
{
// Store answer for every
// iteration
int cnt = 0;
if (mp.containsKey(s[i]))
{
mp.put(s[i], mp.get(s[i]) - 1);
// Map traversal
for(Map.Entry<Character,
Integer> value : mp.entrySet())
{
// Compare current char
if (value.getKey() == s[i])
{
continue;
}
else
{
cnt += value.getValue();
}
}
ans += cnt;
}
}
// Print the final count
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
// Given String
String S = "abcab";
// Length of the String
int N = 5;
// Function call
countSubString(S.toCharArray(), N);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach
# Function to count the substrings
# having different first & last character
def countSubstring(s, n):
# Stores frequency of each char
m = {}
# Loop to store frequency of
# the characters in a Map
for i in range(n):
if s[i] in m:
m[s[i]] += 1
else:
m[s[i]] = 1
# To store final result
ans = 0
# Traversal of string
for i in range(n):
# Store answer for every
# iteration
cnt = 0
if s[i] in m:
m[s[i]] -= 1
else:
m[s[i]] = -1
# Map traversal
for value in m:
# Compare current char
if (value == s[i]):
continue
else:
cnt += m[value]
ans += cnt
# Print the final count
print(ans)
# Driver code
# Given string
S = "abcab"
# Length of the string
N = 5
# Function Call
countSubstring(S, N)
# This code is contributed by divyeshrabadiya07
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{
// Stores frequency of each char
Dictionary<char,
int> mp = new Dictionary<char,
int>();
// Loop to store frequency of
// the characters in a Map
for(int i = 0; i < n; i++)
if (mp.ContainsKey(s[i]))
{
mp[s[i]] = mp[s[i]] + 1;
}
else
{
mp.Add(s[i], 1);
}
// To store readonly result
int ans = 0;
// Traversal of String
for(int i = 0; i < n; i++)
{
// Store answer for every
// iteration
int cnt = 0;
if (mp.ContainsKey(s[i]))
{
mp[s[i]] = mp[s[i]] - 1;
// Map traversal
foreach(KeyValuePair<char,
int> value in mp)
{
// Compare current char
if (value.Key == s[i])
{
continue;
}
else
{
cnt += value.Value;
}
}
ans += cnt;
}
}
// Print the readonly count
Console.Write(ans);
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String S = "abcab";
// Length of the String
int N = 5;
// Function call
countSubString(S.ToCharArray(), N);
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// Javascript program for the above approach
// Function to count the substrings
// having different first & last character
function countSubstring(s, n)
{
// Stores frequency of each char
var m = new Map();
// Loop to store frequency of
// the characters in a Map
for (var i = 0; i < n; i++)
{
if(m.has(s[i]))
m.set(s[i], m.get(s[i])+1)
else
m.set(s[i], 1)
}
// To store final result
var ans = 0;
// Traversal of string
for (var i = 0; i < n; i++) {
// Store answer for every
// iteration
var cnt = 0;
if(m.has(s[i]))
m.set(s[i], m.get(s[i])-1)
// Map traversal
m.forEach((value, key) => {
// Compare current char
if (key != s[i]) {
cnt += value;
}
});
ans += cnt;
}
// Print the final count
document.write( ans);
}
// Driver Code
// Given string
var S = "abcab";
// Length of the string
var N = 5;
// Function Call
countSubstring(S, N);
// This code is contributed by itsok.
</script>
Producción:
8
Complejidad temporal: O(N*26)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por ankitkumar774 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA