Minimum steps to make all the elements of the array divisible by 4 Last Updated : 31 Jul, 2022 Comments Improve Suggest changes Like Article Like Report 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 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 <bits/stdc++.h> 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; // increment 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 <stdio.h> #include <string.h> 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; // increment 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; // increment 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 # increment 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; // increment 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 <?php // PHP program for the above approach function getSteps($arr, $n) { // Count to keep track of the number // of steps done. $count = 0; // Modulus array to store all elements mod 4 $modulus = array_fill(0, 4, 0); // sum to check if given task is possible $sum = 0; // Loop to store all elements // mod 4 and calculate sum; for ($i = 0; $i < $n; $i++) { $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. $modulus[2] += (int)($modulus[1] / 2); $modulus[2] += (int)($modulus[3] / 2); // increment 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 ?> JavaScript <script> function getSteps(arr, n) { // Count to keep track of the // number of steps done. let count = 0; // Modulus array to store all elements mod 4 let modulus = new Array(4); modulus.fill(0); // sum to check if given task is possible let sum = 0; // Loop to store all elements mod 4 // and calculate sum; let i; for (i = 0; i < n; i++) { let 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] += parseInt(modulus[1] / 2, 10); modulus[2] += parseInt(modulus[3] / 2, 10); // increment count to half of remaining // modulus[1] or modulus of [3] elements. count += parseInt(modulus[1] / 2, 10); count += parseInt(modulus[3] / 2, 10); // increment count by half of modulus[2] count += parseInt(modulus[2] / 2, 10); return count; } } // size of array let n = 7; // input array let arr = [ 1, 2, 3, 1, 2, 3, 8 ]; let count = getSteps(arr, n); document.write(count); // This code is contributed by divyeshrabadiya07. </script> Output: 3 Time Complexity: O(n) Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Analysis of Algorithms A AbhijeetSridhar Follow Improve Article Tags : DSA Number Divisibility Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. Itâs the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min read Analysis of AlgorithmsAnalysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. Efficiency is measured in terms of time and space.BasicsWhy is Analysis Important?Order of GrowthAsymptotic Analysis Worst, Average and Best Cases Asymptotic NotationsB 1 min read Data StructuresArray Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous 3 min read String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut 2 min read Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The 2 min read Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Hereâs the comparison of Linked List vs Arrays Linked List: 2 min read Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first 2 min read Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems 2 min read Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most 4 min read Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of 3 min read Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this 15+ min read AlgorithmsSearching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input 2 min read Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ 3 min read Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution 14 min read Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get 3 min read Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net 3 min read Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of 3 min read Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read AdvancedSegment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree 3 min read Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i 2 min read GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br 2 min read Interview PreparationInterview Corner: All Resources To Crack Any Tech InterviewThis article serves as your one-stop guide to interview preparation, designed to help you succeed across different experience levels and company expectations. Here is what you should expect in a Tech Interview, please remember the following points:Tech Interview Preparation does not have any fixed s 3 min read GfG160 - 160 Days of Problem SolvingAre you preparing for technical interviews and would like to be well-structured to improve your problem-solving skills? Well, we have good news for you! GeeksforGeeks proudly presents GfG160, a 160-day coding challenge starting on 15th November 2024. In this event, we will provide daily coding probl 3 min read Practice ProblemGeeksforGeeks Practice - Leading Online Coding PlatformGeeksforGeeks Practice is an online coding platform designed to help developers and students practice coding online and sharpen their programming skills with the following features. GfG 160: This consists of most popular interview problems organized topic wise and difficulty with with well written e 6 min read Problem of The Day - Develop the Habit of CodingDo you find it difficult to develop a habit of Coding? If yes, then we have a most effective solution for you - all you geeks need to do is solve one programming problem each day without any break, and BOOM, the results will surprise you! Let us tell you how:Suppose you commit to improve yourself an 5 min read Like