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# Find the maximum length of the prefix

Given an array arr[] of N integers where all elements of the array are from the range [0, 9] i.e. a single digit, the task is to find the maximum length of the prefix of this array such that removing exactly one element from the prefix will make the occurrence of the remaining elements in the prefix same.
Examples:

Input: arr[] = {1, 1, 1, 2, 2, 2}
Output:
Required prefix is {1, 1, 1, 2, 2}
After removing 1, every element will have equal frequency i.e. {1, 1, 2, 2}

Input: arr[] = {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5}
Output: 13

Input: arr[] = {10, 2, 5, 4, 1}
Output:

Approach: Iterate over all the prefixes and check for each prefix if we can remove an element so that each element has same occurrence. In order to satisfy this condition, one of the following conditions must hold true:

• There is only one element in the prefix.
• All the elements in the prefix have the occurrence of 1.
• Every element has the same occurrence, except for exactly one element which has occurrence of 1.
• Every element has the same occurrence, except for exactly one element which has the occurrence exactly 1 more than any other elements.

Below is the implementation of the above approach:

## C++14

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the maximum` `// length of the required prefix` `int` `Maximum_Length(vector<``int``> a)` `{` `    `  `    ``// Array to store the frequency` `    ``// of each element of the array` `    ``int` `counts[11] = {0};`   `    ``// Iterating for all the elements` `    ``int` `ans = 0;` `    ``for``(``int` `index = 0; ` `            ``index < a.size(); ` `            ``index++) ` `    ``{` `        `  `        ``// Update the frequency of the` `        ``// current element i.e. v` `        ``counts[a[index]] += 1;`   `        ``// Sorted positive values ` `        ``// from counts array` `        ``vector<``int``> k;` `        ``for``(``auto` `i : counts)` `            ``if` `(i != 0)` `                ``k.push_back(i);`   `        ``sort(k.begin(), k.end());`   `        ``// If current prefix satisfies` `        ``// the given conditions` `        ``if` `(k.size() == 1 || ` `           ``(k[0] == k[k.size() - 2] &&` `            ``k.back() - k[k.size() - 2] == 1) || ` `           ``(k[0] == 1 and k[1] == k.back()))` `            ``ans = index;` `    ``}` `    `  `    ``// Return the maximum length` `    ``return` `ans + 1;` `}`   `// Driver code` `int` `main()` `{` `    ``vector<``int``> a = { 1, 1, 1, 2, 2, 2 };`   `    ``cout << (Maximum_Length(a));` `}`   `// This code is contributed by grand_master`

## Java

 `// Java implementation of the approach` `import` `java.util.*;` `public` `class` `Main` `{` `    ``// Function to return the maximum` `    ``// length of the required prefix` `    ``public` `static` `int` `Maximum_Length(Vector a)` `    ``{` `          `  `        ``// Array to store the frequency` `        ``// of each element of the array` `        ``int``[] counts = ``new` `int``[``11``];` `      `  `        ``// Iterating for all the elements` `        ``int` `ans = ``0``;` `        ``for``(``int` `index = ``0``; ` `                ``index < a.size(); ` `                ``index++) ` `        ``{` `              `  `            ``// Update the frequency of the` `            ``// current element i.e. v` `            ``counts[a.get(index)] += ``1``;` `      `  `            ``// Sorted positive values ` `            ``// from counts array` `            ``Vector k = ``new` `Vector();` `            ``for``(``int` `i : counts)` `                ``if` `(i != ``0``)` `                    ``k.add(i);` `      `  `            ``Collections.sort(k);  ` `      `  `            ``// If current prefix satisfies` `            ``// the given conditions` `            ``if` `(k.size() == ``1` `|| ` `               ``(k.get(``0``) == k.get(k.size() - ``2``) &&` `                ``k.get(k.size() - ``1``) - k.get(k.size() - ``2``) == ``1``) || ` `               ``(k.get(``0``) == ``1` `&& k.get(``1``) == k.get(k.size() - ``1``)))` `                ``ans = index;` `        ``}` `          `  `        ``// Return the maximum length` `        ``return` `ans + ``1``;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) {` `        ``Vector a = ``new` `Vector(); ` `        ``a.add(``1``);` `        ``a.add(``1``);` `        ``a.add(``1``);` `        ``a.add(``2``);` `        ``a.add(``2``);` `        ``a.add(``2``);` `       `  `        ``System.out.println(Maximum_Length(a));` `    ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the maximum ` `# length of the required prefix` `def` `Maximum_Length(a):`   `    ``# Array to store the frequency ` `    ``# of each element of the array` `    ``counts ``=``[``0``]``*``11`   `    ``# Iterating for all the elements` `    ``for` `index, v ``in` `enumerate``(a):`   `        ``# Update the frequency of the ` `        ``# current element i.e. v` `        ``counts[v] ``+``=` `1`   `        ``# Sorted positive values from counts array` `        ``k ``=` `sorted``([i ``for` `i ``in` `counts ``if` `i])`   `        ``# If current prefix satisfies ` `        ``# the given conditions` `        ``if` `len``(k)``=``=` `1` `or` `(k[``0``]``=``=` `k[``-``2``] ``and` `k[``-``1``]``-``k[``-``2``]``=``=` `1``) ``or` `(k[``0``]``=``=` `1` `and` `k[``1``]``=``=` `k[``-``1``]):` `            ``ans ``=` `index`   `    ``# Return the maximum length` `    ``return` `ans ``+` `1`   `# Driver code` `if` `__name__``=``=``"__main__"``:` `    ``a ``=` `[``1``, ``1``, ``1``, ``2``, ``2``, ``2``]` `    ``n ``=` `len``(a)` `    ``print``(Maximum_Length(a))`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic; ` `class` `GFG {` `    `  `    ``// Function to return the maximum` `    ``// length of the required prefix` `    ``static` `int` `Maximum_Length(List<``int``> a)` `    ``{` `         `  `        ``// Array to store the frequency` `        ``// of each element of the array` `        ``int``[] counts = ``new` `int``[11];` `     `  `        ``// Iterating for all the elements` `        ``int` `ans = 0;` `        ``for``(``int` `index = 0; ` `                ``index < a.Count; ` `                ``index++) ` `        ``{` `             `  `            ``// Update the frequency of the` `            ``// current element i.e. v` `            ``counts[a[index]] += 1;` `     `  `            ``// Sorted positive values ` `            ``// from counts array` `            ``List<``int``> k = ``new` `List<``int``>();` `            ``foreach``(``int` `i ``in` `counts)` `                ``if` `(i != 0)` `                    ``k.Add(i);` `     `  `            ``k.Sort();` `     `  `            ``// If current prefix satisfies` `            ``// the given conditions` `            ``if` `(k.Count == 1 || ` `               ``(k[0] == k[k.Count - 2] &&` `                ``k[k.Count - 1] - k[k.Count - 2] == 1) || ` `               ``(k[0] == 1 && k[1] == k[k.Count - 1]))` `                ``ans = index;` `        ``}` `         `  `        ``// Return the maximum length` `        ``return` `ans + 1;` `    ``}`   `  ``static` `void` `Main() {` `    ``List<``int``> a = ``new` `List<``int``>(``new` `int``[]{ 1, 1, 1, 2, 2, 2 });` `    ``Console.Write(Maximum_Length(a));` `  ``}` `}`   `// This code is contributed by divyesh072019`

## Javascript

 ``

Output:

`5`

Time Complexity: O(aloga * a) where a is the length of the array

Auxiliary Space: O(a) where a is the length of the array

#### Approach:

1. Initialize the maximum prefix length to be the length of the first string in the set.
2. Compare the first character of each subsequent string with the corresponding character of the first string.
3. If the characters match, increment a prefix counter.
4. If the characters do not match, update the maximum prefix length to be the minimum of the current maximum prefix length and the prefix counter.
5. Repeat steps 2-4 for all strings in the set.
6. The final maximum prefix length is the result.

## C

 `#include ` `#include `   `int` `max_prefix_length(``char``** strings, ``int` `num_strings) {` `    ``int` `max_prefix = ``strlen``(strings[0]);` `    ``for` `(``int` `i = 1; i < num_strings; i++) {` `        ``int` `prefix_len = 0;` `        ``while` `(strings[i][prefix_len] == strings[0][prefix_len]) {` `            ``prefix_len++;` `        ``}` `        ``max_prefix = (prefix_len < max_prefix) ? prefix_len : max_prefix;` `    ``}` `    ``return` `max_prefix;` `}`   `int` `main() {` `    ``char``* strings[] = {``"hello"``, ``"hell"``, ``"help"``, ``"helm"``, ``"he"``};` `    ``int` `num_strings = 5;` `    ``int` `max_prefix = max_prefix_length(strings, num_strings);` `    ``printf``(``"The maximum prefix length is %d\n"``, max_prefix);` `    ``return` `0;` `}`

## C++

 `#include ` `using` `namespace` `std;`   `int` `max_prefix_length(``char``** strings, ``int` `num_strings) {` `    ``int` `max_prefix = std::``strlen``(strings[0]);` `    ``for` `(``int` `i = 1; i < num_strings; i++) {` `        ``int` `prefix_len = 0;` `        ``while` `(strings[i][prefix_len] == strings[0][prefix_len]) {` `            ``prefix_len++;` `        ``}` `        ``max_prefix = (prefix_len < max_prefix) ? prefix_len : max_prefix;` `    ``}` `    ``return` `max_prefix;` `}`   `int` `main() {` `    ``char``* strings[] = {``"hello"``, ``"hell"``, ``"help"``, ``"helm"``, ``"he"``};` `    ``int` `num_strings = 5;` `    ``int` `max_prefix = max_prefix_length(strings, num_strings);` `    ``cout << ``"The maximum prefix length is "` `<< max_prefix << std::endl;` `    ``return` `0;` `}`

## Python3

 `def` `max_prefix_length(strings):` `    ``max_prefix ``=` `len``(strings[``0``])` `    ``for` `i ``in` `range``(``1``, ``len``(strings)):` `        ``prefix_len ``=` `0` `        ``while` `(prefix_len < ``len``(strings[i])` `               ``and` `prefix_len < ``len``(strings[``0``])` `               ``and` `strings[i][prefix_len]` `                      ``=``=` `strings[``0``][prefix_len]):` `            ``prefix_len ``+``=` `1` `        ``max_prefix ``=` `min``(max_prefix, prefix_len)` `    ``return` `max_prefix`   `strings ``=` `[``"hello"``, ``"hell"``, ``"help"``, ``"helm"``, ``"he"``]` `max_prefix ``=` `max_prefix_length(strings)` `print``(``"The maximum prefix length is"``, max_prefix)`

## C#

 `using` `System;`   `class` `Program {` `    ``static` `int` `MaxPrefixLength(``string``[] strings)` `    ``{` `        ``int` `maxPrefix = strings[0].Length;` `        ``for` `(``int` `i = 1; i < strings.Length; i++) {` `            ``int` `prefixLen = 0;` `            ``while` `(prefixLen < strings[i].Length` `                   ``&& prefixLen < strings[0].Length` `                   ``&& strings[i][prefixLen]` `                          ``== strings[0][prefixLen]) {` `                ``prefixLen++;` `            ``}` `            ``maxPrefix = Math.Min(maxPrefix, prefixLen);` `        ``}` `        ``return` `maxPrefix;` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``string``[] strings` `            ``= { ``"hello"``, ``"hell"``, ``"help"``, ``"helm"``, ``"he"` `};` `        ``int` `maxPrefix = MaxPrefixLength(strings);` `        ``Console.WriteLine(` `            ``"The maximum prefix length is {0}"``, maxPrefix);` `    ``}` `}`

## Javascript

 `// Javascript code addition `   `// A function which calculates maximum prefix length. ` `function` `max_prefix_length(strings, num_strings) {` `    `  `    ``// make first string length as max_prefix` `    ``let max_prefix = strings[0].length;` `    `  `    ``for` `(let i = 1; i < num_strings; i++) {` `        `  `        ``// calculate for curr prefix length. ` `        ``let prefix_len = 0;` `        ``while` `(strings[i][prefix_len] == strings[0][prefix_len]) {` `            ``prefix_len++;` `        ``}` `        `  `        ``// update max prefix if prefix length is less than max_prefix. ` `        ``max_prefix = (prefix_len < max_prefix) ? prefix_len : max_prefix;` `    ``}` `    ``return` `max_prefix;` `}`   `// Declare a strings array ` `let strings = [``"hello"``, ``"hell"``, ``"help"``, ``"helm"``, ``"he"``];`   `// size of strings array` `let num_strings = 5;`   `// calculating max prefix length` `let max_prefix = max_prefix_length(strings, num_strings);`   `// printing the maximum prefix length` `console.log(``"The maximum prefix length is "` `+ max_prefix);`   `// The code is contributed by Nidhi goel.`

Output

`The maximum prefix length is 2`

Time complexity: O(N*M), where N is the number of strings and M is the length of the longest string in the set.
Space complexity: O(1), as we only use a fixed number of variables regardless of the size of the input.

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