Given an array nums[ ] of size N, the task is to split the array into groups of size K using the following procedure:
- The first group consists of the first K elements of the array, the second group consists of the next K element of the Array, and so on. Each element can be a part of exactly one group.
- For the last group, if the array does not have K elements remaining, use 0 to complete the group.
Examples:
Input: nums[ ] = {1,2,3,4,5,6,7,8}, K = 4
Output: [[1, 2, 3, 4] [ 5, 6, 7, 8]]
Explanation:
The first 4 element [1, 2, 3, 4] form the first group.
The next 4 elements [ 5, 6, 7, 8] form the second group.
Since all groups can be completely filled by element from the array, no need to use 0.Input: nums[ ] = {3,2,5,7,9,1,3,5,7}, K = 2
Output: [[3, 2] ,[5, 7], [9,1], [3, 5], [7, 0]]
Explanation: The last group was one short of being of size 2. So, one 0 is used.
Approach: This is an easy implementation related problem. Follow the steps mentioned below to solve the problem:
- Maintain a temp vector which represents each group in the string.
- If the index i+1 is divisible by K then it can be concluded that group has ended with this ith index.
- Push the temp into the ans vector if group ends.
- Check whether last group has a size K or not.
- If it is not equal then add K – (len+1) sized fill with 0.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to split the arrayvector<vector<int> > divideArray(int nums[], int K, int N){ vector<vector<int> > ans; vector<int> temp; for (int i = 0; i < N; i++) { temp.push_back(nums[i]); if(((i+1)%K)==0) { ans.push_back(temp); temp.clear(); } } // If last group doesn't have enough // elements then add 0 to it if (!temp.empty()) { int a = temp.size(); while (a != K) { temp.push_back(0); a++; } ans.push_back(temp); } return ans;}// Function to print answervoid printArray(vector<vector<int> >& a){ int n = a.size(); cout << n; for (int i = 0; i < n; i++) { cout << "[ "; for (int j = 0; j < a[i].size(); j++) cout << a[i][j] << " "; cout << "]"; }}// Driver Codeint main(){ int nums[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; int N = sizeof(nums) / sizeof(nums[0]); int K = 4; vector<vector<int> > ans = divideArray(nums, K, N); printArray(ans); return 0;} |
Java
// Java program for the above approachimport java.util.ArrayList;class GFG { // Function to split the array static ArrayList<ArrayList<Integer>> divideArray(int nums[], int K, int N) { ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>(); ArrayList<Integer> temp = new ArrayList<Integer>(); for (int i = 0; i < N; i++) { temp.add(nums[i]); if (((i + 1) % K) == 0) { ans.add(temp); temp = new ArrayList<Integer>(); } } // If last group doesn't have enough // elements then add 0 to it if (temp.size() != 0) { int a = temp.size(); while (a != K) { temp.add(0); a++; } ans.add(temp); } return ans; } // Function to print answer static void printArray(ArrayList<ArrayList<Integer>> a) { int n = a.size(); for (int i = 0; i < n; i++) { System.out.print("[ "); for (int j = 0; j < a.get(i).size(); j++) System.out.print(a.get(i).get(j) + " "); System.out.print("]"); } } // Driver Code public static void main(String args[]) { int nums[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; int N = nums.length; int K = 4; ArrayList<ArrayList<Integer>> ans = divideArray(nums, K, N); printArray(ans); }}// This code is contributed by saurabh_jaiswal. |
Python3
# python3 program for the above approach# Function to split the arraydef divideArray(nums, K, N): ans = [] temp = [] for i in range(0, N): temp.append(nums[i]) if(((i+1) % K) == 0): ans.append(temp.copy()) temp.clear() # If last group doesn't have enough # elements then add 0 to it if (len(temp) != 0): a = len(temp) while (a != K): temp.append(0) a += 1 ans.append(temp) return ans# Function to print answerdef printArray(a): n = len(a) for i in range(0, n): print("[ ", end="") for j in range(0, len(a[i])): print(a[i][j], end=" ") print("]", end="")# Driver Codeif __name__ == "__main__": nums = [1, 2, 3, 4, 5, 6, 7, 8] N = len(nums) K = 4 ans = divideArray(nums, K, N) printArray(ans)# This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to split the array static List<List<int> > divideArray(int[] nums, int K, int N) { List<List<int> > ans = new List<List<int> >(); ; List<int> temp = new List<int>(); for (int i = 0; i < N; i++) { temp.Add(nums[i]); if (((i + 1) % K) == 0) { ans.Add(temp); temp = new List<int>(); } } // If last group doesn't have enough // elements then add 0 to it if (temp.Count != 0) { int a = temp.Count; while (a != K) { temp.Add(0); a++; } ans.Add(temp); } return ans; } // Function to print answer static void printArray(List<List<int> > a) { int n = a.Count; for (int i = 0; i < n; i++) { Console.Write("[ "); for (int j = 0; j < a[i].Count; j++) Console.Write(a[i][j] + " "); Console.Write("]"); } } // Driver Code public static int Main() { int[] nums = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 }; int N = nums.Length; int K = 4; List<List<int> > ans = divideArray(nums, K, N); printArray(ans); return 0; }}// This code is contributed by Taranpreet |
Javascript
<script> // JavaScript code for the above approach // Function to split the array function divideArray(nums, K, N) { let ans = []; let temp = []; for (let i = 0; i < N; i++) { temp.push(nums[i]); if (((i + 1) % K) == 0) { ans.push(temp); temp = []; } } // If last group doesn't have enough // elements then add 0 to it if (temp.length != 0) { let a = temp.length; while (a != K) { temp.push(0); a++; } ans.push(temp); } return ans; } // Function to print answer function printArray(a) { let n = a.length; for (let i = 0; i < n; i++) { document.write("[ "); for (let j = 0; j < a[i].length; j++) document.write(a[i][j] + " "); document.write("]"); } } // Driver Code let nums = [1, 2, 3, 4, 5, 6, 7, 8]; let N = nums.length; let K = 4; let ans = divideArray(nums, K, N); printArray(ans); // This code is contributed by Potta Lokesh </script> |
Output
[ 1 2 3 4 ][ 5 6 7 8 ]
Time Complexity: O(N)
Auxiliary Space: O(N)
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