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

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:

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:

## Python3

 `# Python 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)) `

Output:

```12
```

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

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