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:
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:
- Store the frequencies of T in a dictionary dictCount.
- Store the number of unique letters with a count greater than 0 as nUnique.
- Iterate over [0, N], and obtain the ith index character of S as ch.
- Decrease the count of ch from dictCount if it exists. If this count goes to 0, decrease nUnique by 1.
- If nUnique reaches 0, return the length traversed till then.
- 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 approachdef 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 Codeif __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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