Given an array arr[], of size N and a positive integer K, the task is to find a subarray of size K whose elements can be used to generate a number which is divisible by 3. If no such subarray exists, then print -1.
Examples:
Input: arr[] = {84, 23, 45, 12 56, 82}, K = 3
Output: 12, 56, 82
Explanation:
Number formed by the subarray {12, 56, 82} is 125682, which is divisible by 3.Input: arr[] = {84, 23, 45, 14 56, 82}, K = 3
Output : -1
Naive Approach: The simplest approach is to generate all possible subarrays of size K from the given array and for each subarray, check if the number formed by that subarray is divisible by 3 or not.
Time Complexity: O(N * K)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is based on the following observation:
A number is divisible by 3 if and only if the summation of the digits of the number is divisible by 3.
Follow the steps below to solve the problem:
- Store the sum of first K elements of the array in a variable, say sum.
- Traverse the remaining elements of the array
- Using the Sliding window technique, subtract the first element of the subarray from the sum and add the next array element into the subarray.
- At each step, check if the sum is divisible by 3 or not.
- If found to be true, then print the current K size subarray.
- If no such subarray is found, then print -1.
Below is the implementation of the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the// K size subarrayvoid findSubArray(vector<int> arr, int k){ pair<int, int> ans; int i, sum = 0; // Check if the first K elements // forms a number which is // divisible by 3 for (i = 0; i < k; i++) { sum += arr[i]; } int found = 0; if (sum % 3 == 0) { ans = make_pair(0, i - 1); found = 1; } // Using Sliding window technique for (int j = i; j < arr.size(); j++) { if (found == 1) break; // Calculate sum of next K // size subarray sum = sum + arr[j] - arr[j - k]; // Check if sum is divisible by 3 if (sum % 3 == 0) { // Update the indices of // the subarray ans = make_pair(j - k + 1, j); found = 1; } } // If no such subarray is found if (found == 0) ans = make_pair(-1, 0); if (ans.first == -1) { cout << -1; } else { // Print the subarray for (i = ans.first; i <= ans.second; i++) { cout << arr[i] << " "; } }}// Driver's codeint main(){ // Given array and K vector<int> arr = { 84, 23, 45, 12, 56, 82 }; int K = 3; // Function Call findSubArray(arr, K); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;import java.awt.Point;class GFG{ // Function to find the // K size subarray public static void findSubArray(Vector<Integer> arr, int k) { Point ans = new Point(0, 0); int i, sum = 0; // Check if the first K elements // forms a number which is // divisible by 3 for(i = 0; i < k; i++) { sum += arr.get(i); } int found = 0; if (sum % 3 == 0) { ans = new Point(0, i - 1); found = 1; } // Using Sliding window technique for(int j = i; j < arr.size(); j++) { if (found == 1) break; // Calculate sum of next K // size subarray sum = sum + arr.get(j) - arr.get(j - k); // Check if sum is divisible by 3 if (sum % 3 == 0) { // Update the indices of // the subarray ans = new Point(j - k + 1, j); found = 1; } } // If no such subarray is found if (found == 0) ans = new Point(-1, 0); if (ans.x == -1) { System.out.print(-1); } else { // Print the subarray for(i = ans.x; i <= ans.y; i++) { System.out.print(arr.get(i) + " "); } } } // Driver code public static void main(String[] args){ // Given array and K Vector<Integer> arr = new Vector<Integer>(); arr.add(84); arr.add(23); arr.add(45); arr.add(12); arr.add(56); arr.add(82); int K = 3; // Function call findSubArray(arr, K); }}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 implementation of the # above approach# Function to find the# K size subarraydef findSubArray(arr, k): ans = [(0, 0)] sm = 0 i = 0 found = 0 # Check if the first K elements # forms a number which is # divisible by 3 while (i < k): sm += arr[i] i += 1 if (sm % 3 == 0): ans = [(0, i - 1)] found = 1 # Using Sliding window technique for j in range(i, len(arr), 1): if (found == 1): break # Calculate sum of next K # size subarray sm = sm + arr[j] - arr[j - k] # Check if sum is divisible by 3 if (sm % 3 == 0): # Update the indices of # the subarray ans = [(j - k + 1, j)] found = 1 # If no such subarray is found if (found == 0): ans = [(-1, 0)] if (ans[0][0] == -1): print(-1) else: # Print the subarray for i in range(ans[0][0], ans[0][1] + 1, 1): print(arr[i], end = " ")# Driver codeif __name__ == '__main__': # Given array and K arr = [ 84, 23, 45, 12, 56, 82 ] K = 3 # Function call findSubArray(arr, K)# This code is contributed by SURENDRA_GANGWAR |
C#
// C# implementation of // the above approachusing System;using System.Collections.Generic;class GFG{class Point{ public int x, y; public Point(int first, int second) { this.x = first; this.y = second; } } // Function to find the // K size subarray public static void findSubArray(List<int> arr, int k) { Point ans = new Point(0, 0); int i, sum = 0; // Check if the first K elements // forms a number which is // divisible by 3 for(i = 0; i < k; i++) { sum += arr[i]; } int found = 0; if (sum % 3 == 0) { ans = new Point(0, i - 1); found = 1; } // Using Sliding window technique for(int j = i; j < arr.Count; j++) { if (found == 1) break; // Calculate sum of next K // size subarray sum = sum + arr[j] - arr[j - k]; // Check if sum is // divisible by 3 if (sum % 3 == 0) { // Update the indices of // the subarray ans = new Point(j - k + 1, j); found = 1; } } // If no such subarray is found if (found == 0) ans = new Point(-1, 0); if (ans.x == -1) { Console.Write(-1); } else { // Print the subarray for(i = ans.x; i <= ans.y; i++) { Console.Write(arr[i] + " "); } } } // Driver code public static void Main(String[] args){ // Given array and K List<int> arr = new List<int>(); arr.Add(84); arr.Add(23); arr.Add(45); arr.Add(12); arr.Add(56); arr.Add(82); int K = 3; // Function call findSubArray(arr, K); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the above approach// Function to find the// K size subarrayfunction findSubArray(arr, k){ var ans = []; var i, sum = 0; // Check if the first K elements // forms a number which is // divisible by 3 for(i = 0; i < k; i++) { sum += arr[i]; } var found = 0; if (sum % 3 == 0) { ans = [0, i - 1]; found = 1; } // Using Sliding window technique for(var j = i; j < arr.length; j++) { if (found == 1) break; // Calculate sum of next K // size subarray sum = sum + arr[j] - arr[j - k]; // Check if sum is divisible by 3 if (sum % 3 == 0) { // Update the indices of // the subarray ans = [j - k + 1, j]; found = 1; } } // If no such subarray is found if (found == 0) ans = [-1, 0]; if (ans.first == -1) { cout << -1; } else { // Print the subarray for(i = ans[0]; i <= ans[1]; i++) { document.write( arr[i] + " "); } }}// Driver code// Given array and Kvar arr = [ 84, 23, 45, 12, 56, 82 ];var K = 3;// Function CallfindSubArray(arr, K);// This code is contributed by importantly</script> |
Output:
12 56 82
Time Complexity: O(N)
Auxiliary Space: O(1)
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