# Minimum characters to be replaced to remove the given substring

Given two strings str1 and str2. The task is to find the minimum number of characters to be replaced by \$ in string str1 such that str1 does not contain string str2 as any substring.

Examples:

```Input: str1 = "intellect", str2 = "tell"
Output: 1
4th character of string "str1" can be replaced by \$
such that "int\$llect" it does not contain "tell"
as a substring.

Input: str1 = "google", str2 = "apple"
Output: 0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach is similar to Searching for Patterns | Set 1 (Naive Pattern Searching).
The idea is to find the leftmost occurrence of the string ‘str2’ in the string ‘str1’. If all the characters of ‘str1’ match with ‘str2’, we will replace (or increment our answer with one) the last symbol of occurrence and increment the index of string ‘str1’, such that it checks again for the substring after the replaced character(that is index i will be equal to i+length(b)-1).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// function to calculate minimum ` `// characters to replace ` `int` `replace(string A, string B) ` `{ ` ` `  `    ``int` `n = A.length(), m = B.length(); ` `    ``int` `count = 0, i, j; ` ` `  `    ``for` `(i = 0; i < n; i++) { ` `        ``for` `(j = 0; j < m; j++) { ` ` `  `            ``// mismatch occurs ` `            ``if` `(A[i + j] != B[j])  ` `                ``break``; ` `        ``} ` ` `  `        ``// if all characters matched, i.e, ` `        ``// there is a substring of 'a' which ` `        ``// is same as string 'b' ` `        ``if` `(j == m) { ` `            ``count++; ` ` `  `             ``// increment i to index m-1 such that ` `            ``// minimum characters are replaced ` `            ``// in 'a' ` `            ``i += m - 1; ` `            `  `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string str1 = ``"aaaaaaaa"``; ` `    ``string str2 = ``"aaa"``; ` ` `  `    ``cout << replace(str1 , str2); ` ` `  `  ``return` `0; ` `} `

## Java

 `// Java implementation of ` `// above approach ` `import` `java.io.*; ` ` `  `// function to calculate minimum ` `// characters to replace ` `class` `GFG ` `{ ` `static` `int` `replace(String A, String B) ` `{ ` ` `  `    ``int` `n = A.length(), m = B.length(); ` `    ``int` `count = ``0``, i, j; ` ` `  `    ``for` `(i = ``0``; i < n; i++) ` `    ``{ ` `        ``for` `(j = ``0``; j < m; j++)  ` `        ``{ ` ` `  `            ``// mismatch occurs ` `            ``if``(i + j >= n) ` `            ``break``; ` `            ``else` `if` `(A.charAt(i + j) != B.charAt(j))  ` `                ``break``; ` `        ``} ` ` `  `        ``// if all characters matched, i.e, ` `        ``// there is a substring of 'a' which ` `        ``// is same as string 'b' ` `        ``if` `(j == m)  ` `        ``{ ` `            ``count++; ` ` `  `            ``// increment i to index m-1 such that ` `            ``// minimum characters are replaced ` `            ``// in 'a' ` `            ``i += m - ``1``; ` `             `  `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``String str1 = ``"aaaaaaaa"``; ` `    ``String str2 = ``"aaa"``; ` ` `  `    ``System.out.println(replace(str1 , str2)); ` `} ` `} ` ` `  `// This code is contributed by Subhadeep `

## Python3

 `# Python3 implementation of the  ` `# above approach  ` ` `  `# Function to calculate minimum  ` `# characters to replace  ` `def` `replace(A, B):  ` ` `  `    ``n, m ``=` `len``(A), ``len``(B)  ` `    ``count, i ``=` `0``, ``0` `    ``while` `i < n:  ` `         `  `        ``j ``=` `0` `        ``while` `j < m:  ` `             `  `            ``# mismatch occurs  ` `            ``if` `i ``+` `j >``=` `n ``or` `A[i ``+` `j] !``=` `B[j]: ` `                ``break` `             `  `            ``j ``+``=` `1` `             `  `        ``# If all characters matched, i.e,  ` `        ``# there is a substring of 'a' which  ` `        ``# is same as string 'b'  ` `        ``if` `j ``=``=` `m: ` `            ``count ``+``=` `1` `                 `  `            ``# increment i to index m-1 such that  ` `            ``# minimum characters are replaced  ` `            ``# in 'a'  ` `            ``i ``+``=` `m ``-` `1` `             `  `        ``i ``+``=` `1` `             `  `    ``return` `count  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``str1 ``=` `"aaaaaaaa"` `    ``str2 ``=` `"aaa"` ` `  `    ``print``(replace(str1 , str2))  ` ` `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# implementation of above approach  ` `using` `System; ` ` `  `// function to calculate minimum  ` `// characters to replace  ` `class` `GFG ` `{ ` `public` `static` `int` `replace(``string` `A,  ` `                          ``string` `B) ` `{ ` ` `  `    ``int` `n = A.Length, m = B.Length; ` `    ``int` `count = 0, i, j; ` ` `  `    ``for` `(i = 0; i < n; i++) ` `    ``{ ` `        ``for` `(j = 0; j < m; j++) ` `        ``{ ` ` `  `            ``// mismatch occurs  ` `            ``if` `(i + j >= n) ` `            ``{ ` `            ``break``; ` `            ``} ` `            ``else` `if` `(A[i + j] != B[j]) ` `            ``{ ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// if all characters matched, i.e,  ` `        ``// there is a substring of 'a'  ` `        ``// which is same as string 'b'  ` `        ``if` `(j == m) ` `        ``{ ` `            ``count++; ` ` `  `            ``// increment i to index m-1  ` `            ``// such that minimum characters  ` `            ``// are replaced in 'a'  ` `            ``i += m - 1; ` ` `  `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code  ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``string` `str1 = ``"aaaaaaaa"``; ` `    ``string` `str2 = ``"aaa"``; ` ` `  `    ``Console.WriteLine(replace(str1, str2)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Shrikant13 `

## PHP

 `= ``\$n``) ` `            ``{ ` `                ``break``; ` `            ``} ` `            ``else` `if` `(``\$A``[``\$i` `+ ``\$j``] != ``\$B``[``\$j``]) ` `            ``{ ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// if all characters matched, i.e,  ` `        ``// there is a substring of 'a'  ` `        ``// which is same as string 'b'  ` `        ``if` `(``\$j` `== ``\$m``) ` `        ``{ ` `            ``\$count``++; ` ` `  `            ``// increment i to index m-1  ` `            ``// such that minimum characters  ` `            ``// are replaced in 'a'  ` `            ``\$i` `= ``\$i` `+ ``\$m` `- 1; ` ` `  `        ``} ` `    ``} ` ` `  `    ``return` `\$count``; ` `} ` ` `  `// Driver Code  ` `\$str1` `= ``"aaaaaaaa"``; ` `\$str2` `= ``"aaa"``; ` ` `  `echo` `(replace(``\$str1``, ``\$str2``)); ` ` `  `// This code is contributed ` `// by Kirti_Mangal ` `?> `

Output:

```2
```

Time Complexity: O(len1 * len2), where len1 is the length of first string and len2 is the length of second string.

Also, this problem can be solved directly by using Python’s in-built function-string1.count(string2)

 `/``/` `Python program to find minimum numbers ` `/``/` `of characters to be replaced to ` `/``/` `remove the given substring ` `str1 ``=` `"aaaaaaaa"` `str2 ``=` `"aaa"` ` `  `# inbuilt function ` `answer ``=` `str1.count(str2) ` `print``(answer) `

Output:

```2
```

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