最小化包含另一个字符串T 的所有字符的字符串S 的前缀长度

给定两个字符串S和T ，任务是从S中找到由字符串T的所有字符组成的最小长度前缀。如果S不包含字符串T 的所有字符，则打印-1 。

例子：

Input: S = “MarvoloGaunt”, T = “Tom” 
Output: 12 
Explanation: 
The 12 length prefix “MarvoloGaunt” contains all the characters of “Tom”

Input: S = “TheWorld”, T = “Dio” 
Output: -1 
Explanation: 
The string “TheWorld” does not contain the character ‘i’ from the string “Dio”. 

天真的方法：
解决问题的最简单方法是迭代字符串S并比较前缀和T中每个字母的频率，如果找到所需的前缀，则返回遍历的长度。否则，返回-1。

时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：
要优化上述方法，请按照以下步骤操作：

1. 将T的频率存储在字典dictCount 中。
2. 将计数大于 0 的唯一字母的数量存储为nUnique 。
3. 迭代[0, N] ，得到S 的^第i^个索引字符作为ch 。
4. 如果存在，则从dictCount 中减少ch的计数。如果此计数变为 0，则将 nUnique 减1。
5. 如果nUnique达到 0，则返回到那时遍历的长度。
6. 完全遍历S 后，如果nUnique仍然超过 0，则打印-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int getPrefixLength(string srcStr,
                    string targetStr)
{
     
    // Base Case - if T is empty,
    // it matches 0 length prefix
    if (targetStr.length() == 0)
        return 0;
   
    // Convert strings to lower
    // case for uniformity
    transform(srcStr.begin(),
              srcStr.end(),
              srcStr.begin(), ::tolower);
    transform(targetStr.begin(),
              targetStr.end(),
              targetStr.begin(), ::tolower);
   
    map dictCount;
    int nUnique = 0;
   
    // Update dictCount to the
    // letter count of T
    for(char ch: targetStr)
    {
         
        // If new character is found,
        // initialize its entry,
        // and increase nUnique
        if (dictCount.find(ch) ==
            dictCount.end())
        {
            nUnique += 1;
            dictCount[ch] = 0;
        }
   
        // Increase count of ch
        dictCount[ch] += 1;
    }
   
    // Iterate from 0 to N
    for(int i = 0; i < srcStr.length(); i++)
    {
         
        // i-th character
        char ch = srcStr[i];
   
        // Skip if ch not in targetStr
        if (dictCount.find(ch) ==
            dictCount.end())
            continue;
             
        // Decrease Count
        dictCount[ch] -= 1;
   
        // If the count of ch reaches 0,
        // we do not need more ch,
        // and can decrease nUnique
        if (dictCount[ch] == 0)
            nUnique -= 1;
   
        // If nUnique reaches 0,
        // we have found required prefix
        if (nUnique == 0)
            return (i + 1);
    }
   
    // Otherwise
    return -1;
}
 
// Driver code  
int main()
{
    string S = "MarvoloGaunt";
    string T = "Tom";
   
    cout << getPrefixLength(S, T);
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    public static int getPrefixLength(String srcStr, String targetStr)
    {
       
        // Base Case - if T is empty,
        // it matches 0 length prefix
        if (targetStr.length() == 0)
            return 0;
              
        // Convert strings to lower
        // case for uniformity
        srcStr = srcStr.toLowerCase();
        targetStr = targetStr.toLowerCase();
          
        HashMap dictCount = new HashMap<>();
                                                     
        int nUnique = 0;
        
        // Update dictCount to the
        // letter count of T
        for(char ch : targetStr.toCharArray())
        {
              
            // If new character is found,
            // initialize its entry,
            // and increase nUnique
            if (dictCount.containsKey(ch) != true)
            {
                nUnique += 1;
                dictCount.put(ch, 0);
            }
              
            // Increase count of ch
            dictCount.replace(ch, dictCount.get(ch) + 1);
        }
        
        // Iterate from 0 to N
        for(int i = 0; i < srcStr.length(); i++)
        {
              
            // i-th character
            char ch = srcStr.charAt(i);
        
            // Skip if ch not in targetStr
            if (dictCount.containsKey(ch) != true)
                continue;
                  
            // Decrease Count
            dictCount.replace(ch, dictCount.get(ch) - 1);
        
            // If the count of ch reaches 0,
            // we do not need more ch,
            // and can decrease nUnique
            if (dictCount.get(ch) == 0)
                nUnique -= 1;
        
            // If nUnique reaches 0,
            // we have found required prefix
            if (nUnique == 0)
                return (i + 1);
        }
        
        // Otherwise
        return -1;
    }
   
    // Driver code
    public static void main(String[] args) {
        String S = "MarvoloGaunt";
        String T = "Tom";
        
        System.out.println(getPrefixLength(S, T));
    }
}
 
// This code is contributed by divyesh072019

Python3

# Python3 program for the above approach
 
def getPrefixLength(srcStr, targetStr):
 
    # Base Case - if T is empty,
    # it matches 0 length prefix
    if(len(targetStr) == 0):
        return 0
 
    # Convert strings to lower
    # case for uniformity
    srcStr = srcStr.lower()
    targetStr = targetStr.lower()
 
    dictCount = dict([])
    nUnique = 0
 
    # Update dictCount to the
    # letter count of T
    for ch in targetStr:
 
        # If new character is found,
        # initialize its entry,
        # and increase nUnique
        if(ch not in dictCount):
            nUnique += 1
            dictCount[ch] = 0
 
        # Increase count of ch
        dictCount[ch] += 1
 
    # Iterate from 0 to N
    for i in range(len(srcStr)):
 
        # i-th character
        ch = srcStr[i]
 
        # Skip if ch not in targetStr
        if(ch not in dictCount):
            continue
        # Decrease Count
        dictCount[ch] -= 1
 
        # If the count of ch reaches 0,
        # we do not need more ch,
        # and can decrease nUnique
        if(dictCount[ch] == 0):
            nUnique -= 1
 
        # If nUnique reaches 0,
        # we have found required prefix
        if(nUnique == 0):
            return (i + 1)
 
    # Otherwise
    return -1
 
 
# Driver Code
if __name__ == "__main__":
 
    S = "MarvoloGaunt"
    T = "Tom"
 
    print(getPrefixLength(S, T))

C#

// C# program for the above approach
using System.Collections.Generic;
using System;
 
class GFG{
 
static int getPrefixLength(string srcStr,
                           string targetStr)
{
     
    // Base Case - if T is empty,
    // it matches 0 length prefix
    if (targetStr.Length == 0)
        return 0;
         
    // Convert strings to lower
    // case for uniformity
    srcStr = srcStr.ToLower();
    targetStr = targetStr.ToLower();
     
    Dictionary dictCount = new Dictionary();
                                                
    int nUnique = 0;
   
    // Update dictCount to the
    // letter count of T
    foreach(var ch in targetStr)
    {
         
        // If new character is found,
        // initialize its entry,
        // and increase nUnique
        if (dictCount.ContainsKey(ch) != true)
        {
            nUnique += 1;
            dictCount[ch] = 0;
        }
         
        // Increase count of ch
        dictCount[ch] += 1;
    }
   
    // Iterate from 0 to N
    for(int i = 0; i < srcStr.Length; i++)
    {
         
        // i-th character
        char ch = srcStr[i];
   
        // Skip if ch not in targetStr
        if (dictCount.ContainsKey(ch) != true)
            continue;
             
        // Decrease Count
        dictCount[ch] -= 1;
   
        // If the count of ch reaches 0,
        // we do not need more ch,
        // and can decrease nUnique
        if (dictCount[ch] == 0)
            nUnique -= 1;
   
        // If nUnique reaches 0,
        // we have found required prefix
        if (nUnique == 0)
            return (i + 1);
    }
   
    // Otherwise
    return -1;
}
 
// Driver code  
public static void Main()
{
    string S = "MarvoloGaunt";
    string T = "Tom";
   
    Console.Write(getPrefixLength(S, T));
}
}
 
// This code is contributed by Stream_Cipher

Javascript

      // JavaScript program for the above approach
      function getPrefixLength(srcStr, targetStr) {
        // Base Case - if T is empty,
        // it matches 0 length prefix
        if (targetStr.length === 0) return 0;
 
        // Convert strings to lower
        // case for uniformity
        srcStr = srcStr.toLowerCase();
        targetStr = targetStr.toLowerCase();
 
        var dictCount = {};
 
        var nUnique = 0;
 
        // Update dictCount to the
        // letter count of T
        for (const ch of targetStr) {
          // If new character is found,
          // initialize its entry,
          // and increase nUnique
          if (dictCount.hasOwnProperty(ch) !== true) {
            nUnique += 1;
            dictCount[ch] = 0;
          }
 
          // Increase count of ch
          dictCount[ch] += 1;
        }
 
        // Iterate from 0 to N
        for (var i = 0; i < srcStr.length; i++) {
          // i-th character
          var ch = srcStr[i];
 
          // Skip if ch not in targetStr
          if (dictCount.hasOwnProperty(ch) !== true) continue;
 
          // Decrease Count
          dictCount[ch] -= 1;
 
          // If the count of ch reaches 0,
          // we do not need more ch,
          // and can decrease nUnique
          if (dictCount[ch] === 0) nUnique -= 1;
 
          // If nUnique reaches 0,
          // we have found required prefix
          if (nUnique === 0) return i + 1;
        }
 
        // Otherwise
        return -1;
      }
 
      // Driver code
      var S = "MarvoloGaunt";
      var T = "Tom";
 
      document.write(getPrefixLength(S, T));
    

输出：

12

时间复杂度： O(N)
辅助空间： O(1)

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