先决条件： Digit-DP
给定的字符串str表示很大的数字，任务是找到给定字符串可以分割的最小段数，以使每个段都是1到10 9范围内的奇数。
例子：
Input: str = “123456789123456789123”
Output: 3
Explanation:
The number can be divided as {123456789, 123456789, 123}
Input: str = “123456”
Output: -1
Explanation:
We can’t split the given number such that all segments are odd.
方法：想法是在digit-dp概念的帮助下使用动态编程来解决此问题。因此，定义了splitDP []数组，其中splitDP [i]表示长度为’i’的前缀字符串中将其分成奇数细分所需的最小拆分数。
splitDP []数组按以下方式填充：
- 循环用于遍历给定字符串的所有索引。
- 对于上述循环中的每个索引’i’，将另一个循环从1迭代到9，以检查第(i + j)个索引的子字符串是否为奇数。
- 如果它形成一个奇数，则splitDP []上的值将更新为：
splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]); - 更新完数组的所有值后，最后一个索引处的值是整个字符串的最小拆分数。
下面是上述方法的实现：
C++
// C++ program to split the given string into odds
#include
using namespace std;
// Function to check whether a string
// is an odd number or not
bool checkOdd(string number)
{
int n = number.length();
int num = number[n - 1] - '0';
return (num & 1);
}
// A function to find the minimum
// number of segments the given string
// can be divided such that every
// segment is a odd number
int splitIntoOdds(string number)
{
int numLen = number.length();
// Declare a splitdp[] array
// and initialize to -1
int splitDP[numLen + 1];
memset(splitDP, -1, sizeof(splitDP));
// Build the DP table in
// a bottom-up manner
for (int i = 1; i <= numLen; i++) {
// Initially Check if the entire prefix is odd
if (i <= 9 && checkOdd(number.substr(0, i)))
splitDP[i] = 1;
// If the Given Prefix can be split into Odds
// then for the remaining string from i to j
// Check if Odd. If yes calculate
// the minimum split till j
if (splitDP[i] != -1) {
for (int j = 1; j <= 9
&& i + j <= numLen;
j++) {
// To check if the substring from i to j
// is a odd number or not
if (checkOdd(number.substr(i, j))) {
// If it is an odd number,
// then update the dp array
if (splitDP[i + j] == -1)
splitDP[i + j] = 1 + splitDP[i];
else
splitDP[i + j] = min(splitDP[i + j],
1 + splitDP[i]);
}
}
}
}
// Return the minimum number of splits
// for the entire string
return splitDP[numLen];
}
// Driver code
int main()
{
cout << splitIntoOdds("123456789123456789123") << "\n";
return 0;
} Java
// Java program to split the given string into odds
class GFG {
// Function to check whether a string
// is an odd number or not
static int checkOdd(String number)
{
int n = number.length();
int num = number.charAt(n - 1) - '0';
return (num & 1);
}
// A function to find the minimum
// number of segments the given string
// can be divided such that every
// segment is a odd number
static int splitIntoOdds(String number)
{
int numLen = number.length();
// Declare a splitdp[] array
// and initialize to -1
int splitDP[] = new int[numLen + 1];
for(int i= 0; i < numLen + 1; i++)
splitDP[i] = -1;
// Build the DP table in
// a bottom-up manner
for (int i = 1; i <= numLen; i++) {
// Initially Check if the entire prefix is odd
if (i <= 9 && (checkOdd(number.substring(0, i)) == 1))
splitDP[i] = 1;
// If the Given Prefix can be split into Odds
// then for the remaining string from i to j
// Check if Odd. If yes calculate
// the minimum split till j
if (splitDP[i] != -1) {
for (int j = 1; j <= 9
&& i + j <= numLen;
j++) {
// To check if the substring from i to j
// is a odd number or not
if (checkOdd(number.substring(i, i + j)) == 1) {
// If it is an odd number,
// then update the dp array
if (splitDP[i + j] == -1)
splitDP[i + j] = 1 + splitDP[i];
else
splitDP[i + j] = Math.min(splitDP[i + j],
1 + splitDP[i]);
}
}
}
}
// Return the minimum number of splits
// for the entire string
return splitDP[numLen];
}
// Driver code
public static void main (String[] args)
{
System.out.println(splitIntoOdds("123456789123456789123"));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to split the given string into odds
# Function to check whether a string
# is an odd number or not
def checkOdd(number):
n = len(number)
num = ord(number[n - 1]) - 48
return (num & 1)
# A function to find the minimum
# number of segments the given string
# can be divided such that every
# segment is a odd number
def splitIntoOdds(number):
numLen = len(number)
# Declare a splitdp[] array
# and initialize to -1
splitDP = [-1 for i in range(numLen + 1)]
# Build the DP table in
# a bottom-up manner
for i in range(1, numLen + 1):
# Initially Check if the entire prefix is odd
if (i <= 9 and checkOdd(number[0:i]) > 0):
splitDP[i] = 1
# If the Given Prefix can be split into Odds
# then for the remaining string from i to j
# Check if Odd. If yes calculate
# the minimum split till j
if (splitDP[i] != -1):
for j in range(1, 10):
if(i + j > numLen):
break;
# To check if the substring from i to j
# is a odd number or not
if (checkOdd(number[i:i + j])):
# If it is an odd number,
# then update the dp array
if (splitDP[i + j] == -1):
splitDP[i + j] = 1 + splitDP[i]
else:
splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i])
# Return the minimum number of splits
# for the entire string
return splitDP[numLen]
# Driver code
print(splitIntoOdds("123456789123456789123"))
# This code is contributed by Sanjit_PrasadC#
// C# program to split the given string into odds
using System;
class GFG {
// Function to check whether a string
// is an odd number or not
static int checkOdd(string number)
{
int n = number.Length;
int num = number[n - 1] - '0';
return (num & 1);
}
// A function to find the minimum
// number of segments the given string
// can be divided such that every
// segment is a odd number
static int splitIntoOdds(string number)
{
int numLen = number.Length;
// Declare a splitdp[] array
// and initialize to -1
int []splitDP = new int[numLen + 1];
for(int i= 0; i < numLen + 1; i++)
splitDP[i] = -1;
// Build the DP table in
// a bottom-up manner
for (int i = 1; i <= numLen; i++) {
// Initially Check if the entire prefix is odd
if (i <= 9 && (checkOdd(number.Substring(0, i)) == 1))
splitDP[i] = 1;
// If the Given Prefix can be split into Odds
// then for the remaining string from i to j
// Check if Odd. If yes calculate
// the minimum split till j
if (splitDP[i] != -1) {
for (int j = 1; j <= 9
&& i + j <= numLen;
j++) {
// To check if the substring from i to j
// is a odd number or not
if (checkOdd(number.Substring(i, j)) == 1) {
// If it is an odd number,
// then update the dp array
if (splitDP[i + j] == -1)
splitDP[i + j] = 1 + splitDP[i];
else
splitDP[i + j] = Math.Min(splitDP[i + j],
1 + splitDP[i]);
}
}
}
}
// Return the minimum number of splits
// for the entire string
return splitDP[numLen];
}
// Driver code
public static void Main (string[] args)
{
Console.WriteLine(splitIntoOdds("123456789123456789123"));
}
}
// This code is contributed by AnkitRai01输出：
3
时间复杂度： O(N)