检查给定字符串的任何排列在字典上是否大于另一个给定字符串

给定两个长度为N的字符串str1和str2 ，任务是检查任何给定字符串中是否存在任何可能的排列，使得一个字符串的每个字符都大于或等于另一个字符串的每个字符，对应的指数。如果排列存在则返回真，否则返回假。

例子：

Input: str1 = “adb”, str2 = “cda”
Output: true
Explanation: After permutation str1 = “abd” and str2 = “acd”, so every character from str2 is greater or equals to every character from s1.

Input: str1 = “gfg”, str2 = “agd”
Output: true

方法：上述问题可以通过对两个字符串进行排序然后按字典顺序比较它们来解决。
请按照以下步骤了解如何：

• 将给定的字符串转换为 char 数组
• 对两个字符数组进行排序
• 现在，遍历这些数组并检查一个数组的每个字符是否大于或等于另一个数组中的对应字符
• 如果发现字典序较大，则打印 true，否则打印 false。

以下是上述方法的实现：

C++

// C++ implementation for the above approach
#include 
#include 
#include 
using namespace std;
bool checkGreaterOrNot(string str1,
                       string str2)
{
    // Sorting both strings
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
   
    // Checking if any string
      //is greater or not
    bool flag = true;
   
    for (int i = 0; i < str1.length(); i++) {
        if (str1[i] < str2[i]) {
            flag = false;
            break;
        }
    }
   
    // If str1 is greater returning true
    if (flag)
        return true;
   
    flag = true;
    for(int i = 0; i < str2.length(); i++){
        if (str1[i] > str2[i]) {
            return false;
        }
    }
   
    // If str2 is greater returning true
    return true;
}
int main()
{
    string str1 = "adb";
    string str2 = "cda";
    bool ans =
      checkGreaterOrNot(str1, str2);
    if (ans) {
        cout << "true";
    }
    else {
        cout << "false";
    }
    return 0;
}
 
// This code is contributed by Kdheeraj.

Java

// Java implementation for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
    public static boolean
    checkGreaterOrNot(String str1,
                      String str2)
    {
        // Sorting strings
        char[] arr1 = str1.toCharArray();
        Arrays.sort(arr1);
        char[] arr2 = str2.toCharArray();
        Arrays.sort(arr2);
        boolean flag = true;
 
        // str1 is greater
        // if it does not break the loop
        for (int i = 0; i < arr1.length; i++) {
            if (arr1[i] < arr2[i]) {
                flag = false;
                break;
            }
        }
 
        // If str1 is greater returning true
        if (flag)
            return true;
        flag = true;
 
        // If characters of str1 is greater
        // then none of the strings have all
        // corresponding characters greater
        // so return false
        for (int i = 0; i < arr2.length; i++) {
            if (arr1[i] > arr2[i]) {
                return false;
            }
        }
 
        // If str2 is greater returning true
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str1 = "adb";
        String str2 = "cda";
        boolean ans = checkGreaterOrNot(str1, str2);
        System.out.println(ans);
    }
}

Python3

# Python 3 implementation for the above approach
def checkGreaterOrNot(str1, str2):
   
    # Sorting both strings
    str1  = sorted(str1)
    str1 = "".join(str1)
    str2  = sorted(str2)
    str2 = "".join(str2)
   
    # Checking if any string
      #is greater or not
    flag = True
   
    for i in range(len(str1)):
        if(str1[i] < str2[i]):
            flag = False
            break
   
    # If str1 is greater returning true
    if (flag):
        return True
    flag = True
    for i in range(len(str2)):
        if (str1[i] > str2[i]):
            return False
   
    # If str2 is greater returning true
    return True
 
  # Driver code
if __name__ == '__main__':
    str1 = "adb"
    str2 = "cda"
    ans = checkGreaterOrNot(str1, str2)
    if (ans):
        print("true")
    else:
        print("false")
         
        # This code is contributed by ipg2016107.

C#

// C# implementation for the above approach
using System;
 
class GFG {
    public static bool checkGreaterOrNot(string str1,
                                         string str2)
    {
        // Sorting strings
        char[] arr1 = str1.ToCharArray();
        Array.Sort(arr1);
        char[] arr2 = str2.ToCharArray();
        Array.Sort(arr2);
        bool flag = true;
 
        // str1 is greater
        // if it does not break the loop
        for (int i = 0; i < arr1.Length; i++) {
            if (arr1[i] < arr2[i]) {
                flag = false;
                break;
            }
        }
 
        // If str1 is greater returning true
        if (flag)
            return true;
        flag = true;
 
        // If characters of str1 is greater
        // then none of the strings have all
        // corresponding characters greater
        // so return false
        for (int i = 0; i < arr2.Length; i++) {
            if (arr1[i] > arr2[i]) {
                return false;
            }
        }
 
        // If str2 is greater returning true
        return true;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        string str1 = "adb";
        string str2 = "cda";
        bool ans = checkGreaterOrNot(str1, str2);
        Console.WriteLine(ans);
    }
}
 
// This code is contributed by ukasp.

Javascript

       // JavaScript Program to implement
       // the above approach
 
       function checkGreaterOrNot(str1,
           str2)
       {
        
           // Sorting both strings
           str1.sort(function (a, b) { return a.charCodeAt(0) - b.charCodeAt(0); });
           str2.sort(function (a, b) { return a.charCodeAt(0) - b.charCodeAt(0); });
            
           // Checking if any string
           //is greater or not
           let flag = true;
 
           for (let i = 0; i < str1.length; i++) {
               if (str1[i].charCodeAt(0) > str2[i].charCodeAt(0)) {
                   flag = false;
                   break;
               }
           }
 
           // If str1 is greater returning true
           if (flag)
               return true;
 
           flag = true;
           for (let i = 0; i < str2.length; i++) {
               if (str1[i].charCodeAt(0) > str2[i].charCodeAt(0)) {
                   return false;
               }
           }
 
           // If str2 is greater returning true
           return true;
       }
 
       let str1 = ['a', 'd', 'b'];
       let str2 = ['c', 'd', 'a']
       let ans =
           checkGreaterOrNot(str1, str2);
       if (ans) {
           document.write("true");
       }
       else {
           document.write("false");
       }
 
    // This code is contributed by Potta Lokesh
   

输出：

true

时间复杂度： O(n*log n)
辅助空间： O(n)

方法 2：上述方法可以使用给定字符串的频率图进行优化。

• 为给定的字符串制作频率图
• 创建变量count1和count2以指示各个字符串的累积频率
• 遍历频率图并检查任何字符串的值是否大于另一个。
• 如果是，则打印 true。否则打印错误。

以下是上述方法的实现：

C++

// C++ implementation for the above approach
#include 
#include 
using namespace std;
bool checkGreaterOrNot(string str1,
                       string str2)
{
    int arr1[26] = { 0 };
    int arr2[26] = { 0 };
 
    // Making frequency map for both strings
    for (int i = 0;
         i < str1.length(); i++) {
        arr1[str1[i] - 'a']++;
    }
    for (int i = 0;
         i < str2.length(); i++) {
        arr1[str2[i] - 'a']++;
    }
 
    // To check if any array
    // is greater to the other or not
    bool str1IsSmaller = false,
          str2IsSmaller = false;
 
    int count1 = 0, count2 = 0;
    for (int i = 0; i < 26; i++) {
        count1 += arr1[i];
        count2 += arr2[i];
 
        if (count1 > count2) {
 
         // None of the strings have
         // all corresponding characters
         // greater than other string
            if (str2IsSmaller)
                return false;
 
            str1IsSmaller = true;
        }
 
        if (count1 < count2) {
 
         // None of the strings have
         // all corresponding characters
         // greater than other string
            if (str1IsSmaller)
                return false;
 
            str2IsSmaller = true;
        }
    }
    return true;
}
 
// Driver code
int main()
{
    string str1 = "geeks";
    string str2 = "peeks";
    bool ans =
      checkGreaterOrNot(str1, str2);
    if (ans) {
        cout << "true";
    }
    else {
        cout << "false";
    }
}
 
// This code is contributed by Kdheeraj.

Java

// Java implementation for the above approach
import java.util.*;
 
class GFG {
 
    public static boolean checkGreaterOrNot(
        String str1, String str2)
    {
        int[] freq1 = new int[26];
        int[] freq2 = new int[26];
 
        // Making frequency map
        // for both strings
        for (int i = 0;
             i < str1.length(); i++) {
 
            freq1[str1.charAt(i) - 'a']++;
        }
 
        for (int i = 0;
             i < str2.length(); i++) {
            freq2[str2.charAt(i) - 'a']++;
        }
 
        boolean str1IsSmaller = false;
        boolean str2IsSmaller = false;
        int count1 = 0, count2 = 0;
 
        // Checking if any array
        // is strictly increasing or not
        for (int i = 0; i < 26; i++) {
 
            count1 += freq1[i];
            count2 += freq2[i];
            if (count1 > count2) {
 
                // None of the strings have
                // all corresponding characters
                // greater than other string
                if (str2IsSmaller)
                    return false;
 
                str1IsSmaller = true;
            }
            else if (count2 > count1) {
 
                // None of the strings have
                // all corresponding characters
                // greater than other string
                if (str1IsSmaller)
                    return false;
 
                str2IsSmaller = true;
            }
        }
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str1 = "geeks";
        String str2 = "peeks";
        boolean ans = checkGreaterOrNot(str1, str2);
        System.out.println(ans);
    }
}

Python3

# python implementation for the above approach
def checkGreaterOrNot(str1, str2):
    arr1 = [0 for x in range(26)]
    arr2 = [0 for x in range(26)]
 
    # Making frequency map for both strings
    for val in str1:
        arr1[ord(val)-97] += 1
    for val in str2:
        arr1[ord(val)-97] += 1
         
    # To check if any array
    # is greater to the other or not
    str1IsSmaller = False
    str2IsSmaller = False
 
    count1 = 0
    count2 = 0
    for i in range(0, 26):
        count1 += arr1[i]
        count2 += arr2[i]
        if (count1 > count2):
 
            #  None of the strings have
            #  all corresponding characters
            #  greater than other string
            if str2IsSmaller == True:
                return False
            str1IsSmaller = True
 
        if (count1 < count2):
 
            #  None of the strings have
            # all corresponding characters
            # greater than other string
            if str1IsSmaller == True:
                return False
            str2IsSmaller = True
    return True
 
# Driver code
str1 = "geeks"
str2 = "peeks"
ans = checkGreaterOrNot(str1, str2)
if ans == True:
    print("true")
else:
    print("false")
 
    # This code is contributed by amreshkumar3.

C＃

// C# program for the above approach
using System;
 
class GFG {
 
    public static bool checkGreaterOrNot(
        string str1, string str2)
    {
        int[] freq1 = new int[26];
        int[] freq2 = new int[26];
 
        // Making frequency map
        // for both strings
        for (int i = 0;
             i < str1.Length; i++) {
 
            freq1[str1[(i)] - 'a']++;
        }
 
        for (int i = 0;
             i < str2.Length; i++) {
            freq2[str2[(i)] - 'a']++;
        }
 
        bool str1IsSmaller = false;
        bool str2IsSmaller = false;
        int count1 = 0, count2 = 0;
 
        // Checking if any array
        // is strictly increasing or not
        for (int i = 0; i < 26; i++) {
 
            count1 += freq1[i];
            count2 += freq2[i];
            if (count1 > count2) {
 
                // None of the strings have
                // all corresponding characters
                // greater than other string
                if (str2IsSmaller)
                    return false;
 
                str1IsSmaller = true;
            }
            else if (count2 > count1) {
 
                // None of the strings have
                // all corresponding characters
                // greater than other string
                if (str1IsSmaller)
                    return false;
 
                str2IsSmaller = true;
            }
        }
        return true;
    }
 
    // Driver Code
    public static void Main()
    {
        string str1 = "geeks";
        string str2 = "peeks";
        bool ans = checkGreaterOrNot(str1, str2);
        Console.WriteLine(ans);
    }
}
 
// This code is contributed by avijitmondal1998.

Javascript

 
// Javascript implementation for the above approach
 
function checkGreaterOrNot(str1, str2)
{
    var arr1 = Array(26).fill(0);
    var arr2 = Array(26).fill(0);
 
    // Making frequency map for both strings
    for (var i = 0;
         i < str1.length; i++) {
        arr1[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }
    for (var i = 0;
         i < str2.length; i++) {
        arr1[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }
 
    // To check if any array
    // is greater to the other or not
    var str1IsSmaller = false,
          str2IsSmaller = false;
 
    var count1 = 0, count2 = 0;
    for (var i = 0; i < 26; i++) {
        count1 += arr1[i];
        count2 += arr2[i];
 
        if (count1 > count2) {
 
         // None of the strings have
         // all corresponding characters
         // greater than other string
            if (str2IsSmaller)
                return false;
 
            str1IsSmaller = true;
        }
 
        if (count1 < count2) {
 
         // None of the strings have
         // all corresponding characters
         // greater than other string
            if (str1IsSmaller)
                return false;
 
            str2IsSmaller = true;
        }
    }
    return true;
}
 
// Driver code
var str1 = "geeks";
var str2 = "peeks";
var ans =
  checkGreaterOrNot(str1, str2);
if (ans) {
    document.write("true");
}
else {
    document.write("false");
}
 
// This code is contributed by rutvik_56.

输出：

true

时间复杂度： O(n)
辅助空间： O(1)