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Minimize length of prefix of string S containing all characters of another string T

  • Last Updated : 08 Jun, 2021

Given two string S and T, the task is to find the minimum length prefix from S which consists of all characters of string T. If S does not contain all characters of string T, print -1.

Examples: 

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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”. 

Naive Approach: 
The simplest approach to solve the problem is to iterate the string S and compare the frequency of each letter in both the prefix and T and return the length traversed if the required prefix is found. Otherwise, return -1. 

Time Complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: 
To optimize the above approach, follow the steps below: 

  1. Store the frequencies of T in a dictionary dictCount.
  2. Store the number of unique letters with a count greater than 0 as nUnique.
  3. Iterate over [0, N], and obtain the ith index character of S as ch.
  4. Decrease the count of ch from dictCount if it exists. If this count goes to 0, decrease nUnique by 1.
  5. If nUnique reaches 0, return the length traversed till then.
  6. After complete traversal of S, if nUnique still exceeds 0, print -1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
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<char, int> 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<Character, Integer> 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<char,
               int> dictCount = new Dictionary<char,
                                               int>();
                                                
    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




<script>
      // 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));
    </script>

 
 

Output: 
12

 

 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

 




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