具有奇数计数的字典最小数字字符串

给定一个正整数N ，任务是生成一个大小为N的字典最小数字字符串，每个数字都有奇数。

例子：

Input: N = 4
Output: 1112
Explanation:
Digit 1 and 2 both have an even count and is the lexicographically smallest string possible.

Input: N = 5
Output: 11111
Explanation:
Digit 1 has an odd count and is the lexicographically smallest string possible.

方法：可以基于以下观察解决给定的问题：如果N的值为even ，则结果字符串包含1s ， (N – 1)次后跟单个2是可能的最小字典字符串。否则，结果字符串包含1s ， N次是可能的最小字典字符串。

下面是上述方法的实现：

C++14

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to construct lexicographically
// smallest numeric string having an odd
// count of each characters
string genString(int N)
{
    // Stores the resultant string
    string ans = "";
 
    // If N is even
    if (N % 2 == 0) {
        ans = string(N - 1, '1') + '2';
    }
 
    // Otherwise
    else {
        ans = string(N, '1');
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int N = 5;
    cout << genString(N);
 
    return 0;
}

Python3

# python program for the above approach
# Function to construct lexicographically
# smallest numeric string having an odd
# count of each characters
def genString(N):
   
    # Stores the resultant string
    ans = ""
 
    # If N is even
    if (N % 2 == 0) :
        ans = "".join("1" for i in range(N-1))
        ans = ans + "2"
     
    # Otherwise
    else :
        ans = "".join("1" for i in range(N))
     
    return ans
   
# Driver code
if __name__ == "__main__":
    N = 5
    print(genString(N))
 
 
# This code is contributed by anudeep23042002

C#

// C# program for the above approach
using System;
class GFG {
 
    // Function to construct lexicographically
    // smallest numeric string having an odd
    // count of each characters
    static string genString(int N)
    {
       
        // Stores the resultant string
        string ans = "";
 
        // If N is even
        if (N % 2 == 0) {
            for (int i = 0; i < N - 1; i++)
                ans += '1';
            ans += '2';
        }
 
        // Otherwise
        else {
            for (int i = 0; i < N; i++)
                ans += '1';
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 5;
        Console.WriteLine(genString(N));
    }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to construct lexicographically
        // smallest numeric string having an odd
        // count of each characters
        function genString(N)
        {
         
            // Stores the resultant string
            let ans = "";
 
            // If N is even
            if (N % 2 == 0) {
                ans = '1'.repeat(N - 1) + '2';
            }
 
            // Otherwise
            else {
                ans = '1'.repeat(N);
            }
 
            return ans;
        }
 
        // Driver code
        let N = 5;
        document.write(genString(N));
 
     // This code is contributed by Potta Lokesh
    

输出：

11111

时间复杂度： O(N)
辅助空间： O(N)