# Minimum digits to be removed to make either all digits or alternating digits same

Given a numeric string str, the task is to find the minimum number of digits to be removed from the string such that it satisfies either of the below conditions:

• All the elements of the string are the same.
• All the elements at even position are same and all the elements at the odd position are same, which means the string is alternating with the equal occurrence of each digit.

Examples:

Input: s = “95831”
Output:
Explanation:
In this examples, remove any three elements form the string to make it alternating, i.e. “95” has 9 at even index and 5 at odd index and hence it satisfies second condition.
Input: s = “100120013”
Output:
Explanation:
In this case, either make the string 0000 or make the string 1010. In both the cases the minimum element must be removed from the string will be 5

Approach: The idea is to use the Greedy Approach. Below are the steps:

1. Since all the characters in the resultant string are alternating and same then the smallest substring of distinct digits will be of length 2.
2. As, there are only 10 different types of digits that are from 0 to 9. The idea is to iterate every possible string of length 2 and find the occurence of subsequence formed by them.
3. Hence find all possible combinations of the first and the second character of the string of the above two digit string and greedily construct the longest possible sub-sequence of s beginning with those characters.
4. The difference between string length and the maximum length of the subsequence with alternating digit in the above step is the required result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find longest possible` `// subsequence of s beginning with x and y` `int` `solve(string s, ``int` `x, ``int` `y)` `{` `    ``int` `res = 0;`   `    ``// Iterate over the string` `    ``for` `(``auto` `c : s) {` `        ``if` `(c - ``'0'` `== x) {`   `            ``// Increment count` `            ``res++;`   `            ``// Swap the positions` `            ``swap(x, y);` `        ``}` `    ``}`   `    ``if` `(x != y && res % 2 == 1)` `        ``--res;`   `    ``// Return the result` `    ``return` `res;` `}`   `// Function that finds all the` `// possible pairs` `int` `find_min(string s)` `{` `    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < 10; i++) {`   `        ``for` `(``int` `j = 0; j < 10; j++) {`   `            ``// Update count` `            ``count = max(count,` `                        ``solve(s, i, j));` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given string s` `    ``string s = ``"100120013"``;`   `    ``// Find the size of the string` `    ``int` `n = s.size();`   `    ``// Function Call` `    ``int` `answer = find_min(s);`   `    ``// This value is the count of` `    ``// minimum element to be removed` `    ``cout << (n - answer);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG{`   `// Function to find longest possible` `// subsequence of s beginning with x and y` `static` `int` `solve(String s, ``int` `x, ``int` `y)` `{` `    ``int` `res = ``0``;`   `    ``// Iterate over the String` `    ``for` `(``char` `c : s.toCharArray()) ` `    ``{` `        ``if` `(c - ``'0'` `== x) ` `        ``{`   `            ``// Increment count` `            ``res++;`   `            ``// Swap the positions` `            ``x = x+y;` `            ``y = x-y;` `            ``x = x-y;` `        ``}` `    ``}`   `    ``if` `(x != y && res % ``2` `== ``1``)` `        ``--res;`   `    ``// Return the result` `    ``return` `res;` `}`   `// Function that finds all the` `// possible pairs` `static` `int` `find_min(String s)` `{` `    ``int` `count = ``0``;` `    ``for` `(``int` `i = ``0``; i < ``10``; i++) ` `    ``{` `        ``for` `(``int` `j = ``0``; j < ``10``; j++)` `        ``{`   `            ``// Update count` `            ``count = Math.max(count,` `                             ``solve(s, i, j));` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given String s` `    ``String s = ``"100120013"``;`   `    ``// Find the size of the String` `    ``int` `n = s.length();`   `    ``// Function Call` `    ``int` `answer = find_min(s);`   `    ``// This value is the count of` `    ``// minimum element to be removed` `    ``System.out.print((n - answer));` `}` `}`   `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program for the above approach`   `# Function to find longest possible` `# subsequence of s beginning with x and y ` `def` `solve(s, x, y):`   `    ``res ``=` `0`   `    ``# Iterate over the string` `    ``for` `c ``in` `s:` `        ``if``(``ord``(c) ``-` `ord``(``'0'``) ``=``=` `x):`   `            ``# Increment count` `            ``res ``+``=` `1`   `            ``# Swap the positions` `            ``x, y ``=` `y, x`   `    ``if``(x !``=` `y ``and` `res ``%` `2` `=``=` `1``):` `        ``res ``-``=` `1`   `    ``# Return the result` `    ``return` `res`   `# Function that finds all the` `# possible pairs` `def` `find_min(s):`   `    ``count ``=` `0` `    ``for` `i ``in` `range``(``10``):` `        ``for` `j ``in` `range``(``10``):`   `            ``# Update count` `            ``count ``=` `max``(count, solve(s, i, j))`   `    ``# Return the answer ` `    ``return` `count`   `# Driver Code`   `# Given string s` `s ``=` `"100120013"`   `# Find the size of the string` `n ``=` `len``(s)`   `# Function call` `answer ``=` `find_min(s)`   `# This value is the count of` `# minimum element to be removed` `print``(n ``-` `answer)`   `# This code is contributed by Shivam Singh`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG{`   `// Function to find longest possible` `// subsequence of s beginning with x and y` `static` `int` `solve(String s, ``int` `x, ``int` `y)` `{` `    ``int` `res = 0;`   `    ``// Iterate over the String` `    ``foreach` `(``char` `c ``in` `s.ToCharArray()) ` `    ``{` `        ``if` `(c - ``'0'` `== x) ` `        ``{`   `            ``// Increment count` `            ``res++;`   `            ``// Swap the positions` `            ``x = x + y;` `            ``y = x - y;` `            ``x = x - y;` `        ``}` `    ``}`   `    ``if` `(x != y && res % 2 == 1)` `        ``--res;`   `    ``// Return the result` `    ``return` `res;` `}`   `// Function that finds all the` `// possible pairs` `static` `int` `find_min(String s)` `{` `    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < 10; i++) ` `    ``{` `        ``for` `(``int` `j = 0; j < 10; j++)` `        ``{`   `            ``// Update count` `            ``count = Math.Max(count,` `                             ``solve(s, i, j));` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``// Given String s` `    ``String s = ``"100120013"``;`   `    ``// Find the size of the String` `    ``int` `n = s.Length;`   `    ``// Function Call` `    ``int` `answer = find_min(s);`   `    ``// This value is the count of` `    ``// minimum element to be removed` `    ``Console.Write((n - answer));` `}` `}`   `// This code is contributed by gauravrajput1`

Output:

```5

```

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

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