# Reduce the array such that each element appears at most K times

Given a sorted array arr of size N, the task is to reduce the array such that each element can appear at most K times.

Examples:

Input: arr[] = {1, 2, 2, 2, 3}, K = 2
Output: {1, 2, 2, 3}
Explanation:
Remove 2 once, as it occurs more than 2 times.

Input: arr[] = {3, 3, 3}, K = 1
Output: {3}
Explanation:
Remove 3 twice, as it occurs more than 1 times.

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

Approach:

1. Traverse the given array arr
2. Maintain the count of each unique element in the array while traversing, using a pointer i
3. If the current frequency of arr[i] till index i is less than or equal to K, add the element arr[i] to the new reduced array and increment the frequency by 1.
4. If the current frequency of arr[i] till index i is more than K, skip till you find the next unique element.
5. After the traversal ends, print the reduced array.

Below is the implementation of the above approach:

 `// C++ program to reduce the array ` `// such that each element appears ` `// at most K times ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to reduce the array ` `void` `reduceArray(``int` `arr[], ``int` `n, ``int` `K) ` `{ ` `    ``// Vector to store the reduced array ` `    ``vector<``int``> vec; ` `    ``int` `size = 0; ` `    ``int` `curr_ele = arr, curr_freq = 1; ` ` `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if` `(curr_ele == arr[i] ` `            ``&& curr_freq <= K) { ` `            ``vec.push_back(arr[i]); ` `            ``size++; ` `        ``} ` `        ``else` `if` `(curr_ele != arr[i]) { ` `            ``curr_ele = arr[i]; ` `            ``vec.push_back(arr[i]); ` `            ``size++; ` `            ``curr_freq = 1; ` `        ``} ` `        ``curr_freq++; ` `    ``} ` ` `  `    ``// Print the reduced array ` `    ``cout << ``"{"``; ` `    ``for` `(``int` `i = 0; i < size; i++) { ` `        ``cout << vec[i] << ``", "``; ` `    ``} ` `    ``cout << ``"}"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] ` `        ``= { 1, 1, 1, 2, ` `            ``2, 2, 3, 3, ` `            ``3, 3, 3, 3, ` `            ``4, 5 }; ` ` `  `    ``int` `n = ``sizeof``(arr) ` `            ``/ ``sizeof``(arr); ` `    ``int` `K = 2; ` ` `  `    ``// Function call ` `    ``reduceArray(arr, n, K); ` ` `  `    ``return` `0; ` `} `

 `// Java program to reduce the array ` `// such that each element appears ` `// at most K times ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to reduce the array ` `static` `void` `reduceArray(``int` `arr[], ``int` `n, ``int` `K) ` `{ ` `    ``// Vector to store the reduced array ` `    ``Vector vec = ``new` `Vector(); ` `    ``int` `size = ``0``; ` `    ``int` `curr_ele = arr[``0``], curr_freq = ``1``; ` `  `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `  `  `        ``if` `(curr_ele == arr[i] ` `            ``&& curr_freq <= K) { ` `            ``vec.add(arr[i]); ` `            ``size++; ` `        ``} ` `        ``else` `if` `(curr_ele != arr[i]) { ` `            ``curr_ele = arr[i]; ` `            ``vec.add(arr[i]); ` `            ``size++; ` `            ``curr_freq = ``1``; ` `        ``} ` `        ``curr_freq++; ` `    ``} ` `  `  `    ``// Print the reduced array ` `    ``System.out.print(``"{"``); ` `    ``for` `(``int` `i = ``0``; i < size; i++) { ` `        ``System.out.print(vec.get(i)+ ``", "``); ` `    ``} ` `    ``System.out.print(``"}"``); ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] ` `        ``= { ``1``, ``1``, ``1``, ``2``, ` `            ``2``, ``2``, ``3``, ``3``, ` `            ``3``, ``3``, ``3``, ``3``, ` `            ``4``, ``5` `}; ` `  `  `    ``int` `n = arr.length; ` `    ``int` `K = ``2``; ` `  `  `    ``// Function call ` `    ``reduceArray(arr, n, K); ` `  `  `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

 `# Python3 program to reduce the array ` `# such that each element appears ` `# at most K times ` ` `  `# Function to reduce the array ` `def` `reduceArray(arr, n, K) : ` ` `  `    ``# List to store the reduced array ` `    ``vec ``=` `[]; ` `    ``size ``=` `0``; ` `    ``curr_ele ``=` `arr[``0``]; curr_freq ``=` `1``; ` ` `  `    ``# Iterate over the array ` `    ``for` `i ``in` `range``(n) : ` ` `  `        ``if` `(curr_ele ``=``=` `arr[i] ` `            ``and` `curr_freq <``=` `K) : ` `            ``vec.append(arr[i]); ` `            ``size ``+``=` `1``; ` ` `  `        ``elif` `(curr_ele !``=` `arr[i]) : ` `            ``curr_ele ``=` `arr[i]; ` `            ``vec.append(arr[i]); ` `            ``size ``+``=` `1``; ` `            ``curr_freq ``=` `1``; ` `         `  `        ``curr_freq ``+``=` `1``; ` ` `  `    ``# Print the reduced array ` `    ``print``(``"{"``,end``=` `""); ` `    ``for` `i ``in` `range``(size) : ` `        ``print``(vec[i] ,end``=` `", "``); ` `     `  `    ``print``(``"}"``,end``=``""); ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``1``, ``1``, ``1``, ``2``, ` `           ``2``, ``2``, ``3``, ``3``, ` `            ``3``, ``3``, ``3``, ``3``, ` `            ``4``, ``5` `]; ` ` `  `    ``n ``=` `len``(arr) ` `    ``K ``=` `2``; ` ` `  `    ``# Function call ` `    ``reduceArray(arr, n, K); ` ` `  `# This code is contributed by AnkitRai01 `

 `// C# program to reduce the array ` `// such that each element appears ` `// at most K times ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG{ ` `   `  `// Function to reduce the array ` `static` `void` `reduceArray(``int` `[]arr, ``int` `n, ``int` `K) ` `{ ` `    ``// List to store the reduced array ` `    ``List<``int``> vec = ``new` `List<``int``>(); ` `    ``int` `size = 0; ` `    ``int` `curr_ele = arr, curr_freq = 1; ` `   `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `   `  `        ``if` `(curr_ele == arr[i] ` `            ``&& curr_freq <= K) { ` `            ``vec.Add(arr[i]); ` `            ``size++; ` `        ``} ` `        ``else` `if` `(curr_ele != arr[i]) { ` `            ``curr_ele = arr[i]; ` `            ``vec.Add(arr[i]); ` `            ``size++; ` `            ``curr_freq = 1; ` `        ``} ` `        ``curr_freq++; ` `    ``} ` `   `  `    ``// Print the reduced array ` `    ``Console.Write(``"{"``); ` `    ``for` `(``int` `i = 0; i < size; i++) { ` `        ``Console.Write(vec[i]+ ``", "``); ` `    ``} ` `    ``Console.Write(``"}"``); ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr ` `        ``= { 1, 1, 1, 2, ` `            ``2, 2, 3, 3, ` `            ``3, 3, 3, 3, ` `            ``4, 5 }; ` `   `  `    ``int` `n = arr.Length; ` `    ``int` `K = 2; ` `   `  `    ``// Function call ` `    ``reduceArray(arr, n, K); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:
```{1, 1, 2, 2, 3, 3, 4, 5, }
```

Time complexity: O(N)
Space complexity: O(1)

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