用相同的位数从给定的大数中找出最小的可能数
原文:https://www . geesforgeks . org/find-从给定的大数字中找出最小可能的数字，位数相同/
给定一个长度为 N 的数字 K ，任务是通过任意次数的数字交换，找到由 N 个数字中的 K 个数字组成的最小可能数。
示例:
输入: N = 15，K = 325343273113434 输出:1122333334457 说明: 交换给定数字的数字后可能的最小数字是 11223333334457
输入: N = 7，K = 3416781 T3】输出: 1134678
方法:想法是使用哈希。为了实现散列，创建了大小为 10 的数组 arr[] 。给定的数字被迭代，并且每个数字的出现计数被存储在散列中相应的索引处。然后迭代散列数组，并根据其频率打印带有数字的。输出将是所需的最小 N 位数。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// Javascript implementation of the above approach
// Function for finding the smallest
// possible number after swapping
// the digits any number of times
function smallestPoss(s, n)
{
// Variable to store the final answer
var ans = "";
// Array to store the count of
// occurrence of each digit
var arr = Array(10).fill(0);
// Loop to calculate the number
// of occurrences of every digit
for (var i = 0; i < n; i++) {
arr[s[i].charCodeAt(0) - 48]++;
}
// Loop to get smallest number
for (var i = 0; i < 10; i++) {
for (var j = 0; j < arr[i]; j++)
ans = ans + i.toString();
}
// Returning the answer
return ans;
}
// Driver code
var N = 15;
var K = "325343273113434";
document.write( smallestPoss(K, N));
</script>
Output:
112233333344457
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