Dada una string de dígitos, determine si es una ‘string de suma’. Una string S se llama string de suma si la substring más a la derecha se puede escribir como la suma de dos substrings antes de ella y lo mismo es cierto recursivamente para las substrings anteriores.
Ejemplos:
“12243660” is a sum string. Explanation : 24 + 36 = 60, 12 + 24 = 36 “1111112223” is a sum string. Explanation: 111+112 = 223, 1+111 = 112 “2368” is not a sum string
En general, una string S se denomina string de suma si cumple las siguientes propiedades:
sub-string(i, x) + sub-string(x+1, j) = sub-string(j+1, l) and sub-string(x+1, j)+sub-string(j+1, l) = sub-string(l+1, m) and so on till end.
De los ejemplos, podemos ver que nuestra decisión depende de los dos primeros números elegidos. Así que elegimos todos los dos primeros números posibles para una string dada. Luego, para cada dos números elegidos, verificamos si es una string de suma o no. Así que el enfoque es muy simple. Generamos todos los primeros dos números posibles usando dos substrings s1 y s2 usando dos bucles. luego comprobamos si es posible hacer el número s3 = (s1 + s2) o no. Si podemos hacer s3 entonces buscamos recursivamente s2 + s3 y así sucesivamente.
Implementación:
C++
// C++ program to check if a given string
// is sum-string or not
#include <bits/stdc++.h>
using namespace std;
// this is function for finding sum of two
// numbers as string
string string_sum(string str1, string str2)
{
if (str1.size() < str2.size())
swap(str1, str2);
int m = str1.size();
int n = str2.size();
string ans = "";
// sum the str2 with str1
int carry = 0;
for (int i = 0; i < n; i++) {
// Sum of current digits
int ds = ((str1[m - 1 - i] - '0')
+ (str2[n - 1 - i] - '0') + carry)
% 10;
carry = ((str1[m - 1 - i] - '0')
+ (str2[n - 1 - i] - '0') + carry)
/ 10;
ans = char(ds + '0') + ans;
}
for (int i = n; i < m; i++) {
int ds = (str1[m - 1 - i] - '0' + carry) % 10;
carry = (str1[m - 1 - i] - '0' + carry) / 10;
ans = char(ds + '0') + ans;
}
if (carry)
ans = char(carry + '0') + ans;
return ans;
}
// Returns true if two substrings of given
// lengths of str[beg..] can cause a positive
// result.
bool checkSumStrUtil(string str, int beg, int len1,
int len2)
{
// Finding two substrings of given lengths
// and their sum
string s1 = str.substr(beg, len1);
string s2 = str.substr(beg + len1, len2);
string s3 = string_sum(s1, s2);
int s3_len = s3.size();
// if number of digits s3 is greater then
// the available string size
if (s3_len > str.size() - len1 - len2 - beg)
return false;
// we got s3 as next number in main string
if (s3 == str.substr(beg + len1 + len2, s3_len)) {
// if we reach at the end of the string
if (beg + len1 + len2 + s3_len == str.size())
return true;
// otherwise call recursively for n2, s3
return checkSumStrUtil(str, beg + len1, len2,
s3_len);
}
// we do not get s3 in main string
return false;
}
// Returns true if str is sum string, else false.
bool isSumStr(string str)
{
int n = str.size();
// choosing first two numbers and checking
// whether it is sum-string or not.
for (int i = 1; i < n; i++)
for (int j = 1; i + j < n; j++)
if (checkSumStrUtil(str, 0, i, j))
return true;
return false;
}
// Driver code
int main()
{
bool result;
result = isSumStr("1212243660");
cout << (result == 1 ? "True\n" : "False\n");
result = isSumStr("123456787");
cout << (result == 1 ? "True\n" : "False\n");
return 0;
}
Java
// Java program to check if a given string
// is sum-string or not
import java.util.*;
class GFG {
// this is function for finding sum of two
// numbers as string
public static String string_sum(String str1,
String str2)
{
if (str1.length() < str2.length()) {
String temp = str1;
str1 = str2;
str2 = temp;
}
int m = str1.length();
int n = str2.length();
String ans = "";
// sum the str2 with str1
int carry = 0;
for (int i = 0; i < n; i++) {
// Sum of current digits
int ds
= ((str1.charAt(m - 1 - i) - '0')
+ (str2.charAt(n - 1 - i) - '0') + carry)
% 10;
carry
= ((str1.charAt(m - 1 - i) - '0')
+ (str2.charAt(n - 1 - i) - '0') + carry)
/ 10;
ans = Character.toString((char)(ds + '0'))
+ ans;
}
for (int i = n; i < m; i++) {
int ds = (str1.charAt(m - 1 - i) - '0' + carry)
% 10;
carry = (str1.charAt(m - 1 - i) - '0' + carry)
/ 10;
ans = Character.toString((char)(ds + '0'))
+ ans;
}
if (carry != 0) {
ans = Character.toString((char)(carry + '0'))
+ ans;
}
return ans;
}
// Returns true if two substrings of given
// lengths of str[beg..] can cause a positive
// result.
public static boolean
checkSumStrUtil(String str, int beg, int len1, int len2)
{
// Finding two substrings of given lengths
// and their sum
String s1 = str.substring(beg, beg + len1);
String s2
= str.substring(beg + len1, beg + len1 + len2);
String s3 = string_sum(s1, s2);
int s3_len = s3.length();
// if number of digits s3 is greater then
// the available string size
if (s3_len > str.length() - len1 - len2 - beg)
return false;
// we got s3 as next number in main string
if (s3.equals(str.substring(beg + len1 + len2,
beg + len1 + len2
+ s3_len))) {
// if we reach at the end of the string
if (beg + len1 + len2 + s3_len
== str.length()) {
return true;
}
// otherwise call recursively for n2, s3
return checkSumStrUtil(str, beg + len1, len2,
s3_len);
}
// we do not get s3 in main string
return false;
}
// Returns true if str is sum string, else false.
public static boolean isSumStr(String str)
{
int n = str.length();
// choosing first two numbers and checking
// whether it is sum-string or not.
for (int i = 1; i < n; i++)
for (int j = 1; i + j < n; j++)
if (checkSumStrUtil(str, 0, i, j))
return true;
return false;
}
// Driver Code
public static void main(String[] args)
{
boolean result;
result = isSumStr("1212243660");
System.out.println(result == true ? "True"
: "False");
result = isSumStr("123456787");
System.out.println(result == true ? "True"
: "False");
}
}
// This code is contributed by Tapesh (tapeshdua420)
Python3
# Python code for the above approach
# this is function for finding sum of two
# numbers as string
def string_sum(str1, str2):
if (len(str1) < len(str2)):
str1, str2 = str2,str1
m = len(str1)
n = len(str2)
ans = ""
# sum the str2 with str1
carry = 0
for i in range(n):
# Sum of current digits
ds = ((ord(str1[m - 1 - i]) - ord('0')) +
(ord(str2[n - 1 - i]) - ord('0')) +
carry) % 10
carry = ((ord(str1[m - 1 - i]) - ord('0')) +
(ord(str2[n - 1 - i]) - ord('0')) +
carry) // 10
ans = str(ds) + ans
for i in range(n,m):
ds = (ord(str1[m - 1 - i]) - ord('0') +
carry) % 10
carry = (ord(str1[m - 1 - i]) - ord('0') +
carry) // 10
ans = str(ds) + ans
if (carry):
ans = str(carry) + ans
return ans
# Returns True if two substrings of given
# lengths of str[beg..] can cause a positive
# result.
def checkSumStrUtil(Str, beg,len1, len2):
# Finding two substrings of given lengths
# and their sum
s1 = Str[beg: beg+len1]
s2 = Str[beg + len1: beg + len1 +len2]
s3 = string_sum(s1, s2)
s3_len = len(s3)
# if number of digits s3 is greater then
# the available string size
if (s3_len > len(Str) - len1 - len2 - beg):
return False
# we got s3 as next number in main string
if (s3 == Str[beg + len1 + len2: beg + len1 + len2 +s3_len]):
# if we reach at the end of the string
if (beg + len1 + len2 + s3_len == len(Str)):
return True
# otherwise call recursively for n2, s3
return checkSumStrUtil(Str, beg + len1, len2,s3_len)
# we do not get s3 in main string
return False
# Returns True if str is sum string, else False.
def isSumStr(Str):
n = len(Str)
# choosing first two numbers and checking
# whether it is sum-string or not.
for i in range(1,n):
for j in range(1,n-i):
if (checkSumStrUtil(Str, 0, i, j)):
return True
return False
# Driver code
print(isSumStr("1212243660"))
print(isSumStr("123456787"))
# This code is contributed by shinjanpatra
C#
// C# program to check if a given string
// is sum-string or not
using System;
class sub_string {
// this is function for finding sum of two
// numbers as string
static String string_sum(String str1, String str2)
{
if (str1.Length < str2.Length) {
String temp = str1;
str1 = str2;
str2 = temp;
}
int m = str1.Length;
int n = str2.Length;
String ans = "";
// sum the str2 with str1
int carry = 0;
for (int i = 0; i < n; i++) {
// Sum of current digits
int ds = ((str1[m - 1 - i] - '0')
+ (str2[n - 1 - i] - '0') + carry)
% 10;
carry = ((str1[m - 1 - i] - '0')
+ (str2[n - 1 - i] - '0') + carry)
/ 10;
ans = (char)(ds + '0') + ans;
}
for (int i = n; i < m; i++) {
int ds = (str1[m - 1 - i] - '0' + carry) % 10;
carry = (str1[m - 1 - i] - '0' + carry) / 10;
ans = (char)(ds + '0') + ans;
}
if (carry > 0)
ans = (char)(carry + '0') + ans;
return ans;
}
// Returns true if two substrings of given
// lengths of str[beg..] can cause a positive
// result.
static bool checkSumStrUtil(String str, int beg,
int len1, int len2)
{
// Finding two substrings of given lengths
// and their sum
String s1 = str.Substring(beg, len1);
String s2 = str.Substring(beg + len1, len2);
String s3 = string_sum(s1, s2);
int s3_len = s3.Length;
// if number of digits s3 is greater then
// the available string size
if (s3_len > str.Length - len1 - len2 - beg)
return false;
// we got s3 as next number in main string
if (s3
== str.Substring(beg + len1 + len2, s3_len)) {
// if we reach at the end of the string
if (beg + len1 + len2 + s3_len == str.Length)
return true;
// otherwise call recursively for n2, s3
return checkSumStrUtil(str, beg + len1, len2,
s3_len);
}
// we do not get s3 in main string
return false;
}
// Returns true if str is sum string, else false.
static bool isSumStr(String str)
{
int n = str.Length;
// choosing first two numbers and checking
// whether it is sum-string or not.
for (int i = 1; i < n; i++)
for (int j = 1; i + j < n; j++)
if (checkSumStrUtil(str, 0, i, j))
return true;
return false;
}
// Driver code
public static void Main()
{
Console.WriteLine(isSumStr("1212243660"));
Console.WriteLine(isSumStr("123456787"));
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Javascript
<script>
// JavaScript code to implement the approach
// this is function for finding sum of two
// numbers as string
function string_sum(str1, str2){
if (str1.length < str2.length){
let temp = str1
str1 = str2
str2 = temp
}
let m = str1.length
let n = str2.length
let ans = ""
// sum the str2 with str1
let carry = 0
for(let i=0;i<n;i++){
// Sum of current digits
let ds = ((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
(str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
carry) % 10
carry = Math.floor(((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
(str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
carry) / 10)
ans = ds.toString() + ans
}
for(let i=n;i<m;i++){
let ds = ((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
(str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
carry) % 10
carry = Math.floor(((str1.charCodeAt(m - 1 - i) - '0'.charCodeAt(0)) +
(str2.charCodeAt(n - 1 - i) - '0'.charCodeAt(0)) +
carry) / 10)
ans = ds.toString() + ans
}
if (carry)
ans = carry.toString() + ans
return ans
}
// Returns true if two substrings of given
// lengths of str[beg..] can cause a positive
// result.
function checkSumStrUtil(Str, beg,len1, len2){
// Finding two substrings of given lengths
// and their sum
let s1 = Str.substring(beg,beg+len1)
let s2 = Str.substring(beg + len1, beg + len1 +len2)
let s3 = string_sum(s1, s2)
let s3_len = s3.length
// if number of digits s3 is greater then
// the available string size
if (s3_len > Str.length - len1 - len2 - beg)
return false
// we got s3 as next number in main string
if (s3 == Str.substring(beg + len1 + len2, beg + len1 + len2 +s3_len)){
// if we reach at the end of the string
if (beg + len1 + len2 + s3_len == Str.length)
return true
// otherwise call recursively for n2, s3
return checkSumStrUtil(Str, beg + len1, len2,s3_len)
}
// we do not get s3 in main string
return false
}
// Returns true if str is sum string, else false.
function isSumStr(Str){
let n = Str.length
// choosing first two numbers and checking
// whether it is sum-string or not.
for(let i=1;i<n;i++){
for(let j=1;j<n-i;j++){
if (checkSumStrUtil(Str, 0, i, j))
return true
}
}
return false
}
// Driver code
document.write(isSumStr("1212243660"))
document.write(isSumStr("123456787"))
// This code is contributed by shinjanpatra
</script>
Producción
True False
Complejidad de tiempo: O(n*n*n) , donde n es la longitud de la string.
Espacio auxiliar: O(n) , donde n es la longitud de la string.
Este artículo es una contribución de Jay Prakash Gupta . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA