给定一个长度为N的数字K ,任务是通过任意次数交换数字,找到可以由 N 个数字中的 K 个组成的最小可能数字。
例子:
Input: N = 15, K = 325343273113434
Output: 112233333344457
Explanation:
The smallest number possible after swapping the digits of the given number is 112233333344457
Input: N = 7, K = 3416781
Output: 1134678
方法:想法是使用哈希。为了实现散列,创建了一个大小为 10 的数组arr[] 。迭代给定的数字,并将每个数字的出现次数存储在相应索引处的散列中。然后迭代哈希数组并根据其频率打印第i 个数字。输出将是所需的最小 N 位数。
下面是上述方法的实现:
CPP
// C++ implementation of the above approach
#include
using namespace std;
// Function for finding the smallest
// possible number after swapping
// the digits any number of times
string smallestPoss(string s, int n)
{
// Variable to store the final answer
string ans = "";
// Array to store the count of
// occurrence of each digit
int arr[10] = { 0 };
// Loop to calculate the number
// of occurrences of every digit
for (int i = 0; i < n; i++) {
arr[s[i] - 48]++;
}
// Loop to get smallest number
for (int i = 0; i < 10; i++) {
for (int j = 0; j < arr[i]; j++)
ans = ans + to_string(i);
}
// Returning the answer
return ans;
}
// Driver code
int main()
{
int N = 15;
string K = "325343273113434";
cout << smallestPoss(K, N);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function for finding the smallest
// possible number after swapping
// the digits any number of times
static String smallestPoss(String s, int n)
{
// Variable to store the final answer
String ans = "";
// Array to store the count of
// occurrence of each digit
int arr[] = new int[10];
// Loop to calculate the number
// of occurrences of every digit
for (int i = 0; i < n; i++)
{
arr[s.charAt(i) - 48]++;
}
// Loop to get smallest number
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < arr[i]; j++)
ans = ans + String.valueOf(i);
}
// Returning the answer
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 15;
String K = "325343273113434";
System.out.print(smallestPoss(K, N));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 implementation of the above approach
# Function for finding the smallest
# possible number after swapping
# the digits any number of times
def smallestPoss(s, n):
# Variable to store the final answer
ans = "";
# Array to store the count of
# occurrence of each digit
arr = [0]*10;
# Loop to calculate the number
# of occurrences of every digit
for i in range(n):
arr[ord(s[i]) - 48] += 1;
# Loop to get smallest number
for i in range(10):
for j in range(arr[i]):
ans = ans + str(i);
# Returning the answer
return ans;
# Driver code
if __name__ == '__main__':
N = 15;
K = "325343273113434";
print(smallestPoss(K, N));
# This code is contributed by 29AjayKumar
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function for finding the smallest
// possible number after swapping
// the digits any number of times
static String smallestPoss(String s, int n)
{
// Variable to store the readonly answer
String ans = "";
// Array to store the count of
// occurrence of each digit
int []arr = new int[10];
// Loop to calculate the number
// of occurrences of every digit
for (int i = 0; i < n; i++)
{
arr[s[i] - 48]++;
}
// Loop to get smallest number
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < arr[i]; j++)
ans = ans + String.Join("",i);
}
// Returning the answer
return ans;
}
// Driver code
public static void Main(String[] args)
{
int N = 15;
String K = "325343273113434";
Console.Write(smallestPoss(K, N));
}
}
// This code is contributed by PrinciRaj1992
Javascript
输出:
112233333344457
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