# Minimum steps to make all the elements of the array divisible by 4

Given an array of size n, the task is to find the minimum number of steps required to make all the elements of the array divisible by 4. A step is defined as removal of any two elements from the array and adding the sum of these elements to the array.

Examples:

Input: array = {1, 2, 3, 1, 2, 3, 8}
Output: 3
Explanation:

As we can see in the image,
combining array[0] and array[2] makes it 4. Similarly for array[1] and array[4] as well as array[3] and array[5]. array[6] is already divisible by 4. So by doing 3 steps, all the elements in the array become divisible by 4.

Input: array = {12, 31, 47, 32, 93, 24, 61, 29, 21, 34}
Output: 4

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

Approach: The idea here is to convert all the elements in the array to modulus 4. First, sum of all the elements of the array should be divisible by 4. If not, this task is not possible.

• Initialize an array modulus with size 4 to 0.
• Initialize a counter count to 0 to keep track of number of steps done.
• Traverse through the input array and take modulus 4 of each element.
• Increment the value of the mod 4 value in the modulus array by 1.
• modulus[0] is the count of elements that are already divisible by 4. So no need to pair them with any other element.
• modulus[1] and modulus[3] elements can be combined to get a number divisible by 4. So, increment count to the minimum value of the both.
• Every 2 elements of modulus[2] can be combined to get an element divisible to 4.
• For the remaining elements, increment value modulus[2] by half of modulus[1] and modulus[3].
• Now, increment count by half modulus[2]. We take half because every two elements are combined as one.
• The final value of count is the number of steps required to convert the all the elements of the input array divisible by 4.

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `int` `getSteps(``int` `arr[], ``int` `n) ` `{ ` `    ``// Count to keep track of the  ` `    ``// number of steps done. ` `    ``int` `count = 0; ` ` `  `    ``// Modulus array to store all elements mod 4 ` `    ``int` `modulus[4] = { 0 }; ` ` `  `    ``// sum to check if given task is possible ` `    ``int` `sum = 0; ` ` `  `    ``// Loop to store all elements mod 4  ` `    ``// and calculate sum; ` `    ``int` `i; ` `    ``for` `(i = 0; i < n; i++) ` `    ``{ ` `        ``int` `mod = arr[i] % 4; ` `        ``sum += mod; ` `        ``modulus[mod]++; ` `    ``} ` ` `  `    ``// If sum is not divisible by 4, ` `    ``// not possible ` `    ``if` `(sum % 4 != 0)  ` `    ``{ ` `        ``return` `-1; ` `    ``} ` `    ``else`  `    ``{ ` ` `  `        ``// Find minimum of modulus[1] and modulus[3] ` `        ``// and increment the count by the minimum ` `        ``if` `(modulus[1] > modulus[3])  ` `        ``{ ` `            ``count += modulus[3]; ` `        ``} ` `        ``else`  `        ``{ ` `            ``count += modulus[1]; ` `        ``} ` `         `  `        ``// Update the values in modulus array. ` `        ``modulus[1] -= count; ` `        ``modulus[3] -= count; ` ` `  `        ``// Use modulus[2] to pair remaining elements. ` `        ``modulus[2] += modulus[1] / 2; ` `        ``modulus[2] += modulus[3] / 2; ` ` `  `        ``// incrememnt count to half of remaining ` `        ``// modulus[1] or modulus of [3] elements. ` `        ``count += modulus[1] / 2; ` `        ``count += modulus[3] / 2; ` ` `  `        ``// increment count by half of modulus[2] ` `        ``count += modulus[2] / 2; ` ` `  `        ``return` `count; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// size of array ` `    ``int` `n = 7; ` `     `  `    ``// input array ` `    ``int` `arr[] = { 1, 2, 3, 1, 2, 3, 8 }; ` ` `  `    ``int` `count = getSteps(arr, n); ` ` `  `    ``cout << count; ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## C

 `#include ` `#include ` ` `  `int` `getSteps(``int` `arr[], ``int` `n) ` `{ ` `    ``// Count to keep track of the number of steps done. ` `    ``int` `count = 0; ` ` `  `    ``// Modulus array to store all elements mod 4 ` `    ``int` `modulus[4] = { 0 }; ` ` `  `    ``// sum to check if given task is possible ` `    ``int` `sum = 0; ` ` `  `    ``// Loop to store all elements mod 4 and calculate sum; ` `    ``int` `i; ` `    ``for` `(i = 0; i < n; i++) { ` `        ``int` `mod = arr[i] % 4; ` `        ``sum += mod; ` `        ``modulus[mod]++; ` `    ``} ` ` `  `    ``// If sum is not divisible by 4, not possible ` `    ``if` `(sum % 4 != 0) { ` `        ``return` `-1; ` `    ``} ` `    ``else` `{ ` ` `  `        ``// Find minimum of modulus[1] and modulus[3] ` `        ``// and increment the count by the minimum ` `        ``if` `(modulus[1] > modulus[3]) { ` `            ``count += modulus[3]; ` `        ``} ` `        ``else` `{ ` `            ``count += modulus[1]; ` `        ``} ` `        ``// Update the values in modulus array. ` `        ``modulus[1] -= count; ` `        ``modulus[3] -= count; ` ` `  `        ``// Use modulus[2] to pair remaining elements. ` `        ``modulus[2] += modulus[1] / 2; ` `        ``modulus[2] += modulus[3] / 2; ` ` `  `        ``// incrememnt count to half of remaining ` `        ``// modulus[1] or modulus of [3] elements. ` `        ``count += modulus[1] / 2; ` `        ``count += modulus[3] / 2; ` ` `  `        ``// increment count by half of modulus[2] ` `        ``count += modulus[2] / 2; ` ` `  `        ``return` `count; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// size of array ` `    ``int` `n = 7; ` `    ``// input array ` `    ``int` `arr[] = { 1, 2, 3, 1, 2, 3, 8 }; ` ` `  `    ``int` `count = getSteps(arr, n); ` ` `  `    ``printf``(``"%d"``, count); ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG  ` `{ ` ` `  `static` `int` `getSteps(``int` `arr[], ``int` `n) ` `{ ` `    ``// Count to keep track of the number of steps done. ` `    ``int` `count = ``0``; ` ` `  `    ``// Modulus array to store all elements mod 4 ` `    ``int` `modulus[] = ``new` `int``[``4``]; ` ` `  `    ``// sum to check if given task is possible ` `    ``int` `sum = ``0``; ` ` `  `    ``// Loop to store all elements ` `    ``// mod 4 and calculate sum; ` `    ``int` `i; ` `    ``for` `(i = ``0``; i < n; i++)  ` `    ``{ ` `        ``int` `mod = arr[i] % ``4``; ` `        ``sum += mod; ` `        ``modulus[mod]++; ` `    ``} ` ` `  `    ``// If sum is not divisible by 4, not possible ` `    ``if` `(sum % ``4` `!= ``0``)  ` `    ``{ ` `        ``return` `-``1``; ` `    ``} ` `    ``else` `{ ` ` `  `        ``// Find minimum of modulus[1] and modulus[3] ` `        ``// and increment the count by the minimum ` `        ``if` `(modulus[``1``] > modulus[``3``])  ` `        ``{ ` `            ``count += modulus[``3``]; ` `        ``} ` `        ``else`  `        ``{ ` `            ``count += modulus[``1``]; ` `        ``} ` `        ``// Update the values in modulus array. ` `        ``modulus[``1``] -= count; ` `        ``modulus[``3``] -= count; ` ` `  `        ``// Use modulus[2] to pair remaining elements. ` `        ``modulus[``2``] += modulus[``1``] / ``2``; ` `        ``modulus[``2``] += modulus[``3``] / ``2``; ` ` `  `        ``// incrememnt count to half of remaining ` `        ``// modulus[1] or modulus of [3] elements. ` `        ``count += modulus[``1``] / ``2``; ` `        ``count += modulus[``3``] / ``2``; ` ` `  `        ``// increment count by half of modulus[2] ` `        ``count += modulus[``2``] / ``2``; ` ` `  `        ``return` `count; ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``// size of array ` `    ``int` `n = ``7``; ` `     `  `    ``// input array ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``1``, ``2``, ``3``, ``8` `}; ` ` `  `    ``int` `count = getSteps(arr, n); ` `    ``System.out.printf(``"%d"``, count);  ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python 3 program for the above approach ` `def` `getSteps(arr, n): ` ` `  `    ``# Count to keep track of the  ` `    ``# number of steps done. ` `    ``count ``=` `0` ` `  `    ``# Modulus array to store all elements mod 4 ` `    ``modulus ``=` `[``0` `for` `i ``in` `range``(``4``)] ` ` `  `    ``# Sum to check if given task is possible ` `    ``Sum` `=` `0` ` `  `    ``# Loop to store all elements mod 4  ` `    ``# and calculate Sum ` `    ``i ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``mod ``=` `arr[i] ``%` `4` `        ``Sum` `+``=` `mod ` `        ``modulus[mod] ``+``=` `1` ` `  `    ``# If Sum is not divisible by 4,  ` `    ``# not possible ` `    ``if` `(``Sum` `%` `4` `!``=` `0``): ` `        ``return` `-``1` `    ``else``: ` ` `  `        ``# Find minimum of modulus[1] and modulus[3] ` `        ``# and increment the count by the minimum ` `        ``if` `(modulus[``1``] > modulus[``3``]): ` `            ``count ``+``=` `modulus[``3``] ` `        ``else``: ` `            ``count ``+``=` `modulus[``1``] ` `             `  `        ``# Update the values in modulus array. ` `        ``modulus[``1``] ``-``=` `count ` `        ``modulus[``3``] ``-``=` `count ` ` `  `        ``# Use modulus[2] to pair remaining elements. ` `        ``modulus[``2``] ``+``=` `modulus[``1``] ``/``/` `2` `        ``modulus[``2``] ``+``=` `modulus[``3``] ``/``/` `2` ` `  `        ``# incrememnt count to half of remaining ` `        ``# modulus[1] or modulus of [3] elements. ` `        ``count ``+``=` `modulus[``1``] ``/``/` `2` `        ``count ``+``=` `modulus[``3``] ``/``/` `2` ` `  `        ``# increment count by half of modulus[2] ` `        ``count ``+``=` `modulus[``2``] ``/``/` `2` ` `  `        ``return` `count ` ` `  `# Driver Code ` ` `  `# size of array ` `n ``=` `7` ` `  `# input array ` `arr ``=` `[``1``, ``2``, ``3``, ``1``, ``2``, ``3``, ``8``] ` ` `  `count ``=` `getSteps(arr, n) ` `print``(count) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `int` `getSteps(``int` `[]arr, ``int` `n) ` `{ ` `    ``// Count to keep track of the number of steps done. ` `    ``int` `count = 0; ` ` `  `    ``// Modulus array to store all elements mod 4 ` `    ``int` `[]modulus = ``new` `int``[4]; ` ` `  `    ``// sum to check if given task is possible ` `    ``int` `sum = 0; ` ` `  `    ``// Loop to store all elements ` `    ``// mod 4 and calculate sum; ` `    ``int` `i; ` `    ``for` `(i = 0; i < n; i++)  ` `    ``{ ` `        ``int` `mod = arr[i] % 4; ` `        ``sum += mod; ` `        ``modulus[mod]++; ` `    ``} ` ` `  `    ``// If sum is not divisible by 4, not possible ` `    ``if` `(sum % 4 != 0)  ` `    ``{ ` `        ``return` `-1; ` `    ``} ` `    ``else` `    ``{ ` ` `  `        ``// Find minimum of modulus[1] and modulus[3] ` `        ``// and increment the count by the minimum ` `        ``if` `(modulus[1] > modulus[3])  ` `        ``{ ` `            ``count += modulus[3]; ` `        ``} ` `        ``else` `        ``{ ` `            ``count += modulus[1]; ` `        ``} ` `         `  `        ``// Update the values in modulus array. ` `        ``modulus[1] -= count; ` `        ``modulus[3] -= count; ` ` `  `        ``// Use modulus[2] to pair remaining elements. ` `        ``modulus[2] += modulus[1] / 2; ` `        ``modulus[2] += modulus[3] / 2; ` ` `  `        ``// incrememnt count to half of remaining ` `        ``// modulus[1] or modulus of [3] elements. ` `        ``count += modulus[1] / 2; ` `        ``count += modulus[3] / 2; ` ` `  `        ``// increment count by half of modulus[2] ` `        ``count += modulus[2] / 2; ` ` `  `        ``return` `count; ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``// size of array ` `    ``int` `n = 7; ` `     `  `    ``// input array ` `    ``int` `[]arr = { 1, 2, 3, 1, 2, 3, 8 }; ` ` `  `    ``int` `count = getSteps(arr, n); ` `    ``Console.Write(``"{0}"``, count);  ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` ``\$modulus``[3])  ` `        ``{ ` `            ``\$count` `+= ``\$modulus``[3]; ` `        ``} ` `        ``else` `        ``{ ` `            ``\$count` `+= ``\$modulus``[1]; ` `        ``} ` `         `  `        ``// Update the values in modulus array. ` `        ``\$modulus``[1] -= ``\$count``; ` `        ``\$modulus``[3] -= ``\$count``; ` ` `  `        ``// Use modulus[2] to pair remaining elements. ` `        ``\$modulus``[2] += (int)(``\$modulus``[1] / 2); ` `        ``\$modulus``[2] += (int)(``\$modulus``[3] / 2); ` ` `  `        ``// incrememnt count to half of remaining ` `        ``// modulus[1] or modulus of [3] elements. ` `        ``\$count` `+= (int)(``\$modulus``[1] / 2); ` `        ``\$count` `+= (int)(``\$modulus``[3] / 2); ` ` `  `        ``// increment count by half of modulus[2] ` `        ``\$count` `+= (int)(``\$modulus``[2] / 2); ` ` `  `        ``return` `\$count``; ` `    ``} ` `} ` ` `  `// Driver Code ` ` `  `// size of array ` `\$n` `= 7; ` ` `  `// input array ` `\$arr` `= ``array``( 1, 2, 3, 1, 2, 3, 8 ); ` ` `  `\$count` `= getSteps(``\$arr``, ``\$n``); ` `print``(``\$count``);  ` ` `  `// This code contributed by mits ` `?> `

Output:

```3
```

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